Dedekind, Prüfer and Skolem Rings

Transcription

1 Mathematisches Institut Friedrich-Alexander-Universität Erlangen-Nürnberg Sommersemester 2006 Supervisor: Dr. W. Ruppert Dedekind, Prüfer and Skolem Rings Diploma Thesis by Floris Ernst July 6, 2006 revised on September 21, 2006 There are many magic rings in this world, Bilbo Baggins, and none of them should be used lightly. J.R.R. Tolkien: The Lord of the Rings

2 Acknowledgements I wish to express my gratefulness towards my supervisor, Dr. W. Ruppert, and towards Dr. J. Clark, the supervisor of my Honours Project, which was the basis for this thesis. Furthermore, I am in great debt towards my parents and my friends for patiently reading my work to eliminate all the rough edges that escaped my eyes. ii

3 Contents Introduction 1 1 Preliminaries Rings and Fields Ideals Quotient Rings Inverting elements Minimal Primes Localisation Fractional Ideals Extension Rings and Factorisation of Ideals Integral Extensions Algebraic Extensions Dedekind Domains Basics of Algebraic Number Theory Prüfer Domains An Introduction to Valuation Theory Absolute Values Places Valuations Skolem Rings A-Valued Polynomials Skolem s Theorem Generalisations and Consequences A Zorn s Lemma 103 B The Chinese Remainder Theorem 104 C Notation 105 D Noetherian Domains 107 Index 109 Bibliography 112 iii

4 Introduction While working on proofs of Fermat s Last Theorem, several fascinating mathematical discoveries were made. One famous mistake, which arguably is the reason for the work shown in this thesis, was made by the French mathematician Gabriel Lamé in 1847: when trying to factor X n + Y n in Z[e 2πi/n ], Lamé assumed uniqueness of factorisation to prove Fermat s Last Theorem. This work was shown to be false, though, in a paper by Ernst Kummer which had appeared three years earlier: this factorisation is not unique for n > 23. While trying to save Lamé s proof, Kummer invented ideal primes to restore uniqueness in the factorisation. With this he was able to prove the theorem for more n than Lamé, in fact, for all prime exponents p such that the class number 1 h(p) of the cyclotomic ring Z[e 2πi/p ] is not divided by p. In doing so, he together with Richard Dedekind, who introduced ideals as we know them today laid the foundation for a whole new part of Algebra: ideal theory. In this thesis, chapter 1 serves as a stepping stone: key concepts of Algebra are introduced and ideals defined. Then, in chapter 2, things become more interesting. Quotient rings and fields are introduced together with fractional ideals which, in turn, enable us to speak of factorisation of ideals. Furthermore, an important example for a ring without unique factorisation is given. In chapter 3 we then delve deeper into unique factorisation: Dedekind domains are introduced, rings in which all ideals can be uniquely factored into products of prime ideals and are multiplicatively invertible. This is then generalised to Prüfer domains where this invertibility needs to be possible for finitely generated ideals only. In between, section 3.4 introduces concepts from Algebraic Number Theory used later in this thesis. Chapter 4 goes away from the general line by explaining the concept of absolute values, places and valuations. These results are then used in chapter 5 to prove some theorems on special polynomial rings, so-called Skolem rings, and to show the final result of this thesis, Theorem 5.3.8, to find more examples for Prüfer rings. 1 The class number of an integral domain R is the number of equivalence classes of on the set of ideals of R by defining a b : a, b R, a, b 0, such that a a = b b for non-zero ideals a, b. 1

5 Chapter 1 Preliminaries This chapter is intended to provide the reader with the necessary background for understanding this project. The following will give a brief overview of the concepts of fields, rings and ideals. Additionally, a good introduction can be found in [11]. Throughout the whole project, certain notational conventions are used. A short summary can be found in Appendix C. 1.1 Rings and Fields This section comprises a brief reminder of the basic structure of rings and fields. Definition Ring A ring (R, +, ) is a set of elements together with two operations +, called addition, and, called multiplication. (R, +) is an additive Abelian group 1, (R, ) is a multiplicative semigroup 2. Furthermore, the distributive laws 3 hold. If there is an element 1 R such that x 1 = 1 x = x for all x R, R is said to be a ring with identity or unity. If (R, ) is a commutative semigroup, the ring is called commutative. For abbreviation purposes, a ring will be referred to as R if the operations are clear and a b will be denoted by ab. A subset S of R is called a subring of R if (S, + S S, S S ) is a ring. If R is a ring with unity, the group (R, ) := ({r R s R such that rs = 1}, ) is called the group of units of R. Elements from R are called units. If for r, s R there is a unit u such that r = us, we say that r and s are associates. 1 A triple (G,, e) is called a group if e G, is a binary operation on G and for all a, b, c G, (a b) c = a (b c), a e = e a = a and there is an element a 1 G such that a a 1 = a 1 a = e. If a b = b a for all a, b G, G is called Abelian. 2 A tuple (G, ) is called a semigroup if is a binary operation on G and for all a, b, c G, (a b) c = a (b c). 3 This means that a (b + c) = a b + a c and (b + c) a = b a + c a. 2

6 CHAPTER 1. PRELIMINARIES Example 1.1: (a) R is the ring of real numbers under the familiar operations. (b) (Z p, + p, p), p N, is the ring of integers modulo p where a + p b := (a + b) mod p and a p b := (a b) mod p. (c) M(n, R) (or M n (R)) is the ring of n n matrices with real entries, n 2. (d) C (R, R) is the ring of continuous functions f : R R. (e) F[X] is the ring of polynomials over a field 4 F. (f) 2Z is the ring of even integers. (g) If R is a commutative ring with identity, R[X], the set of all polynomials over R, also is a ring. Examples (a), (b), (d), (e), (g) are commutative rings with identity, (c) is non-commutative with identity, (f) is commutative without identity. Definition Zero-divisors A non-zero element r in a ring R is said to be a left zero-divisor if there exists a nonzero element s R such that rs = 0. Right zero-divisors are defined analogously. If an element is both a left and a right zero-divisor, it is simply called a zero-divisor. An element is called regular if it is neither a left nor right zero-divisor. A ring with unity and no zero-divisors is called a domain; if it is commutative, it is called an integral domain. The set of all left zero-divisors of a ring R is denoted by z l (R). Similarly, z r (R) denotes the set of all right zero-divisors of R and z(r) := z l (R) z r (R). Example 1.2: Let R = M 2 (Z). Then [ 0 1 ] 0 0 is a zero-divisor since [ ] [ ] [ = 0 = ] [ ]. 4 See Definition

7 1.1. RINGS AND FIELDS Let R = C (R, R). Define two elements f, g R as follows: R R { f : 0 if x 0 x x if x > 0 R R { and g : x if x 0 x 0 if x > 0 Then, clearly, f g = 0 = g f, showing that both f and g are zero divisors. Note 1.3: Although z l (R) is closed under left multiplication from R, it need not be closed under right multiplication: consider the ring R := M (Z), the ring of all matrices over the integers where every column and every row contains only a finite number of non-zero entries (this assures that standard matrix multiplication works). Let A := Then A is a left zero-divisor, since AB = 0 for 1 0 B := Now, if z l (R) were closed under right multiplication, AA T would have to be a left zero-divisor. But, quite obviously, AA T = 1, the identity matrix, and therefore certainly not a zero-divisor. Since 0 = 0 T = (AB) T = B T A T = BA T, A T is a right zero-divisor and therefore this argument also shows that z r (R) is not closed under left multiplication. Definition Annihilator Let R be a commutative ring, r R. Then the set ann(r) := {s R rs = 0} is called the annihilator of r. In a similar way, the annihilator of a subset S R is defined as ann(s) := ann(s). s S This gives reason to look at rings, where all non-zero elements are invertible. 4

8 CHAPTER 1. PRELIMINARIES Definition Fields Let R be a ring with unity and no zero-divisors. Then it is called a field if (R\ {0}, ) is an Abelian group. Example 1.4: For any prime p, Z p is a field (see [11]). Q, R and C are fields. Definition Characteristic of a Ring Let R be a ring. If there is a minimal integer n such that n 1 = = 0, }{{} n times we say that R has characteristic n, otherwise we say that it has characteristic 0. We write χ(r) for the characteristic of R. It is clear that χ(q) = χ(r) = χ(c) = 0. Furthermore, for all n Z, χ(z n ) = n. Note that there are infinite rings with finite characteristic: χ(z p [X]) = p. 1.2 Ideals In the 19th century, while trying to extend the concept of unique factorisation to rings beyond the integers, Kummer and Dedekind encountered several problems (see Example 3.2), leading them to examine a certain class of subsets of rings, called ideals. Definition Ideals A subset a of a ring R, a R, is called a left ideal of R if, for all x, y a, x y lies in a and, for all r R and x a, rx lies in a. Similarly, it is called a right ideal of R if, for all x, y a and r R, x y and xr lie in a. If a is both a left and right ideal, a is called a two-sided ideal of R. If a R, we say that a is a proper ideal. Note 1.5: If R is commutative, all its ideals are two-sided. Example 1.6: (a) R and 0 := {0} are always ideals of any ring R. (b) The set of even integers, 2Z, is a two-sided ideal of Z. (c) The set a := { f C (R, R) } f(π) = 0 is a two-sided ideal of C (R, R). 5

9 1.2. IDEALS Definition Factor Rings Let R be a ring and let a be a two-sided ideal. Define two operations and on the set of all cosets 5 of a by setting (x + a) (y + a) := (x + y) + a and (x + a) (y + a) := xy + a. Then ( / ) R a,, := ({x + a x R},, ) is a ring. It is called the factor ring of R with respect to a. Furthermore, let S := {a i i I} for some index set I and some a i R, i I. Then, if a i + a = a j + a if and only if i = j and for all r R there is an i I such that r + a = a i + a, S is called a complete system of residues of R with respect to a or, shorter, a CSR of a in R. Example 1.7: For any n N, Z n = { 0, 1,..., n 1 } = Z / nz, x denotes the coset x + Z. The set {0, 1,..., n 1} is a CSR of Z n as is {n, n + 1,..., 2n 1}. Definition Special ideals Let R be a ring. A proper ideal m of R is said to be a left maximal ideal if, whenever n is a left ideal of R such that m n R, then n = m or n = R. If a ring R has only one maximal ideal, R is said to be local; it is said to be quasi-local if it has a finite number of maximal ideals. A proper two-sided ideal p of R is said to be prime if, for any two elements a, b R, arb p implies that a p or b p. It is called completely prime if ab p implies that a p or b p. Example 1.8: 2Z is a maximal ideal of Z. It turns out that the set of ideals of a ring R is closed under certain set-theoretic operations. The following proposition gives the details of how to define sums, products, intersections and quotients of ideals. 5 The cosets of a subset A of a group (G,, e) are the sets g A := {g a a A} for all g G. 6

10 CHAPTER 1. PRELIMINARIES Proposition Binary operations on ideals Given two left ideals a and b in a ring R. Then the following subsets of R are left ideals: (a) a b (b) a + b := { } a + b a a, b b (a + b is called the sum of a and b.) i=1 { n } (c) ab := a i b j n N, a i a, b j b, i, j = 1,..., n (ab is called the product of a and b.) (d) a : b := {r R rb a} = {r R r b a b b} (a : b is called the divisional ideal of a and b.) (a) For x, y a b, x y a and x y b. Therefore, x y a b. Furthermore, for r R and x a b, r x a and r x b. Hence, r x a b. (b) a + b is an ideal since: a 1, a 2 a, b 1, b 2 b : a 1 + b 1 (a 2 + b 2 ) = a 1 a }{{} 2 + b 1 b 2 a + b }{{} a b and r R, x = a + b a + b, a a, b b : r x = r a + r b a + b (c) Let x, y ab, x = n m a i b i, y = c j d j i=1 j=1 where a i, c j a, b i, d j b for i = 1,..., n and j = 1,..., m. Then x y = n+m i=1 a i b i where a n+j := c j, b n+j := d j, j = 1,..., m and so x y ab. Now consider r x for some r R: r x = r n a i b i = i=1 n (r a i ) b i ab }{{} a i=1 (d) Let x, y a : b, i.e. x z, y z a for all z b. Then xz yz = (x y)z a for all z b. Now, if r R, rxz a for all z b and thus rx a:b. 7

11 1.2. IDEALS Note 1.9: Part (a) of Proposition can be extended to the intersection j J a j of an arbitrary family {a j j J} of left ideals. By induction, parts (b) and (c) can be shown to hold for any finite family of left ideals {a 1,..., a n }, giving their sum a a n and product a 1 a 2... a n. The product of two ideals introduced in part (c) can be used to define powers of ideals by setting, for any ideal a and any n N, a n := } a a {{... a }, a 0 := R. n terms Now the question arises how the ideals of a given ring can be characterised. First, we look at a certain family of ideals, the so-called principal ideals. Then this concept is extended to finitely generated ideals: Definition Principal and finitely generated ideals Let R be a ring. A left ideal a in R is called principal if there is an element x R such that a = Rx := x := {rx r R}. A commutative ring with unity is called a principal ideal domain, or PID, if it contains no zero-divisors and all its ideals are principal (i.e., a PID is an integral domain with only principal ideals). Let = S = {s i i I} R. We say a left ideal a is generated by S if every element a a can be expressed as a = r i s i, r i R, i I such that almost all r i = 0. We write a =: S. If I is finite, say I = {1,..., n} for some n N, we say that a is finitely generated and we write a =: s 1,..., s n. It is straightforward to show that in this case a is the sum n i=1 s i of the principal left ideals s i, i = 1,..., n. Example 1.10: Let R = Z. Then 2 = 2Z and 3 = 3Z are both principal ideals as is their sum 2Z + 3Z = 1Z = Z (since 1 = , 3 ). Their intersection (and product) is the principal ideal 6 (since x 6 is equivalent to 2 x and 3 x, e.g. x 2 3 ). Although the intersection of two ideals is again an ideal, we now show that the intersection of two finitely generated ideals is not necessarily finitely generated. Consider the ring R of polynomials over Q, R := Q[X], and the subring D of R, D := Z + X 2 R = { a 0 + a 2 X a n X n a 0 Z, a i Q for i 2, n 2 }. The two ideals i := X 2 and j := X 3 of D are obviously finitely generated. Now 8

12 CHAPTER 1. PRELIMINARIES consider an element F a := i j. Then, for some a 0 Z, a 2,..., a n Q and b 0 Z, b 2,..., b m Q, we see: F = X 2 ( a 0 + a 2 X a n X n) = X 3 ( b 0 + b 2 X b n X m) a 0 X 2 + a 2 X 4 + a 3 X 5 + a n X n+2 = b 0 X 3 + b 2 X b n X m+3 a 0 = a 2 = b 0 = 0 F X 5 R a X 5 R Since obviously X 5 R a, we see that a = X 5 R. We will now show that a is not finitely generated as an ideal of D. Assume it were finitely generated, say by the n elements F 1,..., F n, where F i = a 5i X 5 + a 6i X 6 +, a ki Q, k N, i = 1,..., n. Say a 5i = c5i d 5i where c 5i, d 5i Z, d 5i 0 and the fraction is in lowest form. Now consider the element ( G := 1 n ) d X5 a where d := d 5i + 1. i=1 Since for every element H F 1,..., F n, H = h 5 X 5 +, we know that h 5 n i=1 d 5i Z, whence we see that G F 1,..., F n. This shows that a cannot be finitely generated. Note 1.11: Similarly, the divisional ideal of two finitely generated ideals need not be finitely generated. See Example Now we can easily prove the following results. Proposition Basic results Let R be a ring with identity. Then: (a) A proper two-sided ideal p of R is prime whenever a and b are left ideals of R such that ab p, either a p or b p. (b) If R is commutative, an ideal p is prime whenever ab p, either a p or b p. (E.g. if R is commutative, prime completely prime.) (c) R is an integral domain the zero ideal is prime. (d) If a, b R, then Ra Rb a = rb for some r R. (e) Every two-sided maximal ideal is prime. (f) A commutative ring R is a field the zero ideal is maximal. (g) If R is commutative, an ideal p is prime R/p is an integral domain. 9

13 1.2. IDEALS (h) If R is commutative, an ideal m is maximal R/m is a field. (i) Let a be a left ideal and let {b i } i=1,...,n be a family of left ideals of R. Then n n a: b i = (a:b i ). i=1 i=1 (j) If a 1,..., a s R are ideals and p R is a prime ideal such that p a 1 a s, then there exists a k {1,..., s} such that a k p. (a) Suppose p is prime and a, b are ideals such that ab p but a p and b p. Then a a, b b such that a p, b p but RaR RbR = RaRbR RpR = p. Contradiction! Suppose p is not prime. Then there are a, b R such that arb p but a p and b p. Let a = RaR and b = RbR. Then a, b are left ideals of R with ab = RaRbR RpR = p but a p and b p. Contradiction! (b) Let p be prime, let a, b R such that ab p. Then abr = arb r R and hence arb p. Therefore, either a p or b p. Conversely, let a, b R such that arb p. Then ab = a 1 b p and therefore either a p or b p. (c) R is an integral domain if and only if ab = 0 a = 0 or b = 0. This is obviously equivalent to the zero ideal being prime. (d) Since Ra Rb, a Rb. Hence, r R such that a = r b. If a Rb then for all r R, r a Rb. Hence, Ra Rb. (e) Let m be maximal among the proper two-sided ideals. Assume m is not prime. Then there are two-sided ideals a, b such that ab m and a m, b m. Thus a + m = R = b + m (a + m)(b + m) = R but also (a + m)(b + m) = ab + am + mb + m m R. Contradiction! Now assume m is maximal among either the left or right ideals. Then m is also maximal among the two-sided ideals and by above therefore is prime. (f) Let 0 a R. Since 0 is maximal, Ra = R. Then b R such that b a = a b = 1. Hence, a is invertible and therefore R is a field. Let 0 m be an ideal in R. Then 0 a m. Since R is a field, b R such that 1 = b a. Hence, 1 m and thus m = R. (g) Let a + p, b + p R/p such that ab + p = 0 + p. Therefore, ab p. Since p is prime, at least one of a, b p, say a p. Then a + p = 0 + p. Therefore 10

14 CHAPTER 1. PRELIMINARIES R/p is an integral domain. Let a, b R such that ab p. Then (a + p)(b + p) = ab + p = 0 + p. Therefore, at least one of a + p, b + p are zero, say a + p = 0. This shows that a p. Therefore, p is prime. (h) Suppose that m is maximal and let a R \ m. Now consider the ideal a := ar + m. Since m is maximal, it is clear that a = R, whence 1 = ar + m for some r R, m m. This shows that 1 + m = ar + m + m = ar + m = (a + m)(r + m), establishing that a + m is invertible in R/m, whence R/m is a field. Conversely, suppose that R/m is a field and a m is an ideal of R. Let a a \ m. Then a + m 0 + m whence there is a b R such that (a + m)(b + m) = 1 + m. Then, since a a, we see that ab a. And since 1 ab m a, this shows that 1 a, whence a = R and thus m must be maximal. (i) Since a : n i=1 b i = {r R r ( n i=1 b i) a}, we see that for all i = 1,..., n, a: n i=1 b i {r R rb i a} = a:b i. Therefore, a: n i=1 b i n i=1 (a:b i). Conversely, let r n i=1 (a:b i). Then for all i = 1,..., n, rb i a. This shows that n i=1 rb i a and so s n i=1 b i a. This shows that r a: n i=1 b i. (j) Assume this is not true. Then there exist elements a k a k \p, k = 1,..., s. Then their product a 1 a s a 1 a s p. Since p is prime, this shows that at least one a k p. Contradiction! Note 1.12: Two-sidedness in part (e) of Proposition is necessary, not only because of the definition of prime ideals: consider the ring R := M 2 (Z) and the left ideal [ ] Z 2Z m =. Z 2Z m is maximal: Let n m be a left ideal in R. Let M n\m. Then [ ] a c M = b d where at least one of c, d has to be odd. Assume c odd, d even. Then [ ] [ ] [ a c + 1 a c 0 1 = b d b d 0 0 ] n. 11

15 1.2. IDEALS Since [ 0 0 ] [ 0 0 ] [ 0 1 ] [ 0 0 ] 1 0 R, we see that = 0 1 n. Assume c even, d odd. Then [ ] [ ] [ a c a c 0 0 = b d + 1 b d 0 1 Assume both c and d are odd. Then [ ] [ ] [ a c + 1 a c 0 1 = b d + 1 b d 0 1 ] n. ] n. Since [ 0 0 ] [ 0 0 ] [ 0 1 ] [ 0 0 ] 1 0 R, we see that = 0 1 n. Therefore, in each case, [ ] [ = ] n, and therefore n = R. This shows that m is maximal. m is not two-sided: Consider the following: [ ] [ ] [ = ] m. Hence, m is clearly not prime and not even contained in a prime ideal. Note 1.13: The converse of (e) in Proposition is not true: consider the ring R := Z[X] and the ideal X of R. X is prime since Z[X] / X = {z + X z Z[X]} = {z + X z Z} = Z is an integral domain. X is clearly not maximal since X 2, X Z[X]. Definition Noetherian Rings A ring R is called left (right) Noetherian if every left (right) ideal a in R is finitely generated. 12

16 CHAPTER 1. PRELIMINARIES Example 1.14: (a) Let R be the ring of functions from R to R and a be the set of those functions f : R R such that there is n f N (depending on f!) for which f(x) = 0 if x [ n f, n f ]. a is an ideal but it is not finitely generated: Suppose there were a set of generators f 1,..., f k with associated interval bounds n 1,..., n k. Then if r i R, r i f i is associated with n i. This shows that k i=1 r if i corresponds to the interval bound n 0 := max {n 1,..., n k }. Let g R be non-zero on [ n 0 1, n 0 + 1]. Then g f 1,..., f k. (b) Example 1.10 shows that D = Z + X 2 Q[X] is not Noetherian. We now introduce a concept which will provide another way of characterising Noetherian Domains. Definition Ascending Chain Condition A collection C of ideals, C = {a s s S}, is called a chain (with respect to inclusion) if, given any s 1, s 2 S, we have either a s1 a s2 or a s2 a s1. A ring R is said to satisfy the Ascending Chain Condition (ACC), if, given any chain C of ideals, indexed by N, C = {a n n N}, where a n a n+1 n, there exists a k N such that a k = a k+i for all i 1. In this case we say that the chain becomes stationary. Note 1.15: If C = {a s s S} is a chain, s S a s is an ideal. We now show that there is a connection between the ACC and a ring being Noetherian: Theorem Characterisation of Noetherian Rings A ring R is left Noetherian if and only if R satisfies the ACC on left ideals. Suppose that R is Noetherian and let {a n } n N be an ascending chain of left ideals in R. Let a := n N a n. Then a is a left ideal of R. Furthermore, since R is Noetherian, a is finitely generated, say by x 1,..., x k where x i a ni. Let n 0 := max i {n i }. Then x 1,..., x k a n0. Thus Rx 1 + Rx Rx k a n0 and so a a n0. Hence a = a n0 (since a n0 a ) and so R satisfies the ACC on left ideals (the chain becomes stationary at a n0 13

17 1.2. IDEALS or earlier). Suppose that R is not Noetherian. Then there is a left ideal a of R which is not finitely generated. Pick x 1 a. Then a 1 := Rx 1 a. So there exists x 2 a\a 1. Then a 2 := a 1 + Rx 2 a. So there exists x 3 a\a 2. Continue in this way to find a strictly ascending chain a 1 a 2. This is a contradiction since R was required to satisfy the ACC on left ideals. Example 1.16: Consider the ring of integers, Z. We will show that Z is a principal ideal domain, e.g. that all its ideals are principal ideals. Firstly, we note that Z obviously is an integral domain. Now, let a be an ideal of Z, say a = S for some subset S of Z. Define z := gcd(s). Note that z must exist: there is only a finite number of possible values for z since 1 z s for any s S. Now assume that s a, i.e. there are indices n 1,..., n m such that m s = s ni a i i=1 for some a i Z. Then, obviously, z s since z s ni for all i. Thus, S z. On the other hand, let s = za z. Then, since z is the gcd of S, by the Extended Euclidean Algorithm there must be indices n 1,..., n m such that z = s n1 a s nm a m for some a i Z. Hence, s = za = s n1 a 1 a +... s nm a m a S, establishing that z S. Thus we have equality, showing that all ideals of Z are principal ideals. This also shows that Z is Noetherian. The notion of Definition can be generalised: Definition ACCP Let R be a ring. We say that R satisfies the ascending chain condition on principal ideals, or the ACCP, if every ascending chain of principal ideals of R eventually becomes stationary. Note 1.17: There are one-sided Noetherian rings as the following example shows. 14

18 CHAPTER 1. PRELIMINARIES Example 1.18: Consider the ring [ ] {[ Z Q z p R := = Q q ] } z Z, p, q Q where the blank entry denotes 0. Let l be a left ideal of R. Then there are subsets s Z, u, t Q such that [ ] {[ ] } s u s u l = s s, u u, t t. t t It is important to note that the sets s, u, t are not independent from each other, i.e. the element [ ] s u, s s, u u, t t, t does not necessarily lie in l. It is obvious that s (Z, +), u, t (Q, +). Now let [ ] n q1 q 2 R. Then [ n q1 ] [ s u ] [ ns nu + q1 t ] [ s u ] q 2 t = q 2 t t. This shows that s must be an ideal of Z and that t must be an ideal of Q. This leaves two cases, t = 0 and t = Q. We only look at the first case, which yields [ ] s u l =. 0 Consider the left ideal [ ] 0 Q l :=. 0 If l were finitely generated as a left ideal of R, there would have to be elements q 1,..., q n Q for some n N such that for all q Q there exist r 1,..., r n R such that [ 0 q 0 ] = r 1 [ 0 q1 0 ] + + r n [ 0 qn 0 ]. Say [ ni k i r i = l i ] so that r i [ 0 qi 0 ] [ 0 ni q i = 0 ]. 15

19 1.2. IDEALS Since the Abelian group Q is not finitely generated (see [10]), l cannot be finitely generated. This shows that R is not left Noetherian. Now let d be a right ideal of R, say [ ] s u d =. t Again, arbitrarily choosing s s, u u and t t does not necessarily yield an element from d. Now, for all n Z, q 1, q 2 Q, we see [ ] [ ] [ ] [ ] s u n q1 ns q1 s + q 2 u s u =, t q 2 t t q 2 yielding that s is an ideal of Z and that t is an ideal of Q. Consider these three cases: (a) s = 0, t = 0: Then, obviously, [ ] [ 0 Q 0 0 d = or d = 0 0 ]. (b) s = 0, t 0: Then [ 0 u d u/t = t ] {[ 0 uq = tq ] q Q } where u/t Q. Quite obviously, for two different fractions u 1 /t 1 and u 2 /t 2 (both in lowest form), neither [ ] [ ] [ ] [ ] 0 u1 0 u2 0 u2 0 u1 nor holds. t 1 t 2 t 2 t 1 (c) s 0: Then [ s Q ] [ s Q ] d = 0 or d = Q. Together, this yields the lattice of ideals shown in Figure 1.1. Since Z is a principal ideal domain, it is Noetherian and therefore its lattice satisfies the ACC (by Theorem 1.2.9). This shows that the lattice in Figure 1.1 also satisfies the ACC. Therefore, R is right Noetherian. 16

20 CHAPTER 1. PRELIMINARIES 0 0 Q 0 Q Q 0 0 Q 0 Figure 1.1: The lattice of ideals in R. denotes a lattice of ideals isomorphic to the lattice of ideals of Z. The non-labelled s represent the d u/t s for u 0. Definition Norm of an Ideal Let R be a ring and let a be an ideal of R. Then the number ( / ) N (a) := # R a N { } is called the absolute norm of a in R. Example 1.19: Let a be a non-zero ideal of Z. Since Z is a PID, a = n for some n N. Hence, N (a) = n. Note that if a = 0, then N (a) = #Z =, showing that N is not necessarily finite. 17

21 Chapter 2 Quotient Rings From now on, we will require a ring R to contain an identity. 2.1 Inverting elements In a ring R, for any given element r R, there need not be an inverse s R such that rs = 1. The question arises as to how the ring can be extended in a way such that there is an inverse for every regular element. Note that for a zero-divisor r there cannot be an inverse: assume that rs = 0, s 0, and r is an inverse for r. Then: 0 = r 0 = r rs = 1 s = s, a contradiction. To be able to further investigate this, we require some definitions. Definition Multiplicatively closed set Let R be a commutative ring. A subset S of R is called multiplicatively closed (a multiplicatively closed set, or MCS) if (a) 1 S, (b) 0 S, (c) for all s, t S, we get st S. Now, a multiplicatively closed set gives reason for the definition of the following equivalence relation : 18

22 CHAPTER 2. QUOTIENT RINGS Proposition Let R be a commutative ring and S R be an MCS. Then is an equivalence relation on Q := R S by defining (r 1, s 1 ) (r 2, s 2 ) : t S : t(r 1 s 2 r 2 s 1 ) = 0 for (r 1, s 1 ), (r 2, s 2 ) Q. We need to show that is reflexive, transitive and symmetric. Let (r 1, s 1 ), (r 2, s 2 ) Q. Then: (r 1, s 1 ) (r 1, s 1 ) by selecting t = 1. If (r 1, s 1 ) (r 2, s 2 ) using t 0, then (r 2, s 2 ) (r 1, s 1 ), again using t 0. Suppose (r 1, s 1 ) (r 2, s 2 ) and (r 2, s 2 ) (r 3, s 3 ) with (a) t(r 1 s 2 ) = t(r 2 s 1 ) (b) u(r 2 s 3 ) = u(r 3 s 2 ) where u, t S. Then: uts 2 (r 1 s 3 ) = us 3 (tr 1 s 2 ) (a) = us 3 tr 2 s 1 = (ur 2 s 3 )ts 1 (b) = ur 3 s 2 ts 1 = = uts 2 (r 3 s 1 ) and so (r 1, s 1 ) (r 3, s 3 ) with t := uts 2. This completes the proof. Note 2.1: If R is an integral domain, t can always be cancelled and may hence be omitted in the definition of. Note that S may contain zero-divisors, but if s S is a zero-divisor, e.g. rs = 0 for some r R, r 0, r will not be in S. Definition Quotient Rings For any commutative ring R and an MCS S R we define (r, s) to be the equivalence class of (r, s) in Q, i.e. (r, s) := {(x, y) Q (r, s) (x, y)}. 19

23 2.1. INVERTING ELEMENTS Then we define R S := Q/ := {(r, s) } (r, s) Q, the set of all equivalence classes of in Q. The class (r, s) can also be denoted by r s. Then it can easily be checked that (R S, +, ) is a commutative ring if we define r 1 + r 2 := r 1s 2 + r 2 s 1 s 1 s 2 s 1 s 2 r 1 r2 := r 1r 2. s 1 s 2 s 1 s 2 and The ring R S is called the localisation of R at S. If S is the set of all regular elements of R, R S is called the total quotient ring or classical ring of quotients of R and is denoted by T. We also write quot (R) := T. Example 2.2: Q is the total quotient ring of Z. Note 2.3: Obviously, if p is a prime ideal of R, R\p is an MCS. For abbreviation purposes, it is common to denote R R\p by R p. Although technically R is not contained in R S, R can be embedded into R S by means of a homomorphism: Definition Embedding Let R be a commutative ring and S be an MCS. Define the following map: ψ : { R RS r (r, 1). Then ψ is called the embedding homomorphism from R into R S and ψ(r) is called the embedding of R into R S. Note 2.4: Quite clearly, ψ is a ring homomorphism: let r, s R. Then ψ(r + s) = (r + s, 1) = (r, 1) + (s, 1) = ψ(r) + ψ(s) and ψ(rs) = (rs, 1) = (r, 1) (s, 1) = ψ(r) ψ(s). Note that this mapping is only injective if S contains no zero-divisors: let t S be a zero-divisor, e.g. 0 r R such that tr = 0. Then ψ(r + 1) = ψ(1) since t(r + 1 1) = tr = 0, although r (since otherwise r = 0). If S contains no zero-divisors, though, we see that ψ is injective: let r, s R 20

24 CHAPTER 2. QUOTIENT RINGS such that ψ(r) = ψ(s). Then t S such that t(r s) = 0. Assume r s 0. Then, since 0 S and therefore t 0, t and r s must be zero-divisors. Contradiction! Quite often we will not mention the embedding homomorphism ψ but just identify the elements r and (r, 1) to be able to define multiplication of elements from R and R S as well as to simplify notation when dealing with fractional ideals in section 2.4. Theorem Quotient Rings Let R be an integral domain. Then the set S := R \ {0} is an MCS and the localisation R S is a field. R S is also called the total field of quotients of R. S obviously is an MCS. Let 0 r R. Then r S and therefore 1 r R S and r 1 1 r = r r = 1 1, hence 1 r is the inverse of r in R S. Therefore, R S is a field. 2.2 Minimal Primes The following two results are required for a generalisation of the concept of quotient rings. Definition Minimal Primes Let a be a proper left ideal of a ring R. A prime ideal p of R is said to be minimal over a (or a minimal prime of a) if (a) a p and (b) if q is a prime ideal of R such that a q p, then q = p. We denote the set of all minimal primes of a by min(a). Example 2.5: In Z, p is a minimal prime over n for all n Z such that p n. Proposition Existence of maximal super-ideals Let l be a proper left ideal of some ring R. Then there exists a maximal left ideal m of R such that l m. 21

25 2.3. LOCALISATION We use Zorn s Lemma (see Appendix A). Let P be the set of all proper left ideals of R which contain l. P since l l and hence l P. Let be a partial ordering on P given by. Let C be a chain in (P, ), say C = {C λ λ Λ}. Then C := λ Λ C λ. Because C is a chain, C is a left ideal of R, clearly containing l. Since 1 C λ for all λ, 1 cannot be in C. Hence, C is a proper ideal and therefore an upper bound for C in P. This shows that we can apply Zorn s Lemma, yielding the required left maximal ideal m. Proposition Existence of min(a) For each proper ideal a R of some commutative ring R, min(a). Let P be the set of all prime ideals of R which contain a. Since R is commutative, every maximal ideal of R is prime (see Proposition 1.2.6) and hence P (by Proposition 2.2.2). Define a partial ordering on P by (reverse inclusion). Let C be a chain in (P, ), say C = {C λ λ Λ}. Let C := λ Λ C λ a. Then C is an ideal of R. Moreover, C λ C for all λ Λ. Suppose a, b R with ab C. Then λ Λ, ab C λ. So, since C λ is prime, we have either a C λ or b C λ. Suppose that a C λ. Given δ Λ, we have either C δ C λ or C λ C δ. For C δ C λ, b C δ. Then, for C δ C λ we have a C δ. Hence, b C δ for all C δ C λ. Therefore, b C. Hence, C is prime and so is an upper bound for C. Thus we can apply Zorn s Lemma. This gives a minimal prime over a. Example 2.6: In Z, min( 6 ) = { 2, 3 }. In general, if a = p k1 1 pk pkn n for some pairwise distinct primes p i and integers k i, i = 1,..., n, then min(a) = { p 1,..., p n }. Note 2.7: If R is not commutative, min(a) may be empty! See Note Localisation Now we will generalise the concept of a quotient ring. In this section we shall assume every ring to be commutative. Definition Localisation Let a be an ideal in a ring R and let S R be an MCS. Let be defined as in 22

26 CHAPTER 2. QUOTIENT RINGS Definition Then the set { } a S := (a, s) a a, s S R/ is called the localisation of a with respect to S. Definition Extension and Contraction Let a be an ideal in a ring R and let S R be an MCS. Then the localisation a S of a is also called the extension of a with respect to S (or to R S ), denoted by a e. Vice versa, if b is an ideal of R S, the set of elements { } b c := b R b 1 b is called the contraction of b. Quite obviously, the contraction b c of an ideal b is an ideal itself. Again, if p is a prime ideal we will denote a R\p by a p. We will now state some straightforward properties of localisation: Proposition Properties of Localisation Let a, b be ideals of a ring R, S R be an MCS and c be an ideal of R S. Then: (a) a S b S = (ab) S (b) a S b S = (a b) S (c) a S + b S = (a + b) S (d) If b is finitely generated, (a:b) S = a S :b S (e) c = c ce (a) Each element in a S b S is a sum of products of the form a s, b t where a a, b b, s, t S, hence it is a sum of elements of the form ab st, which are in (ab) S. Suppose x s (ab) S, x ab, s S. Hence, x is a sum of products of the form ab, a a, b b. So x ab s is a sum of products of the form s where a a, b b, s S, each of which is in a S b S, since ab s = a b s 1. 23

27 2.3. LOCALISATION (b) Let d s (a b) S, d a b, s S. Then d s a S and d s b S. Therefore, d s a S b S. Let d s a S b S. Then d s = a u = b t exist s 1, s 2 S such that where a a, b b, u, t S. Then there s 1 du = s 1 as and s 2 dt = s 2 bs. Then d s = ds 1s 2 ut ss 1 s 2 ut = s 1 ass 2 t ss 1 s 2 ut s 2 bss 1 u ss 1 s 2 ut a S b S and so d s a S b S. (c) This is obvious. (d) The containment (a : b) S a S : b S always holds, even if b is not finitely generated: Let r s 1 r s 1 (a:b) S and let b s 2 b S. Then rb a and we note: b = rb a S since rb a. s 2 s 1 s 2 This shows that r s 1 a S :b S, establishing (a:b) S a S :b S. We now prove the reverse inclusion. Assume that b = b is a principal ideal. Then if x a S :b S, x = r s for some r R, s S. Furthermore, r 1 a S :b S as well. Therefore, rb 1 a S and so rb (a S ) c. Assume that rb a. Then there exist a 0 a, s 0 S such that (rb, 1) (a 0, s 0 ). This yields rbs 0 t = a 0 t a for some t S. Therefore, there always exists an element s 1 S (possibly s 1 = 1) such that rbs 1 a. Consequently, rs 1 a : b and x = rs1 ss 1 (a : b) S. Now, if b = b 1,..., b k is a finitely generated ideal of R, we see: ( k ) a S :b S = a S : b i = k i=1 i=1 (a: b i ) S (b) = S ( k (c) = a S : i=1 ( k ) a: b i i=1 b i S ) S ( ) = = (a:b) S, k (a S : b i S ) = completing the proof (note that ( ) holds because of Proposition 1.2.6, part (i)). (e) It is clear that c ce c. Now consider x c, x = r s for some r R, s S. Then r 1 = xs c and so r cc and consequently x = r s cce. Therefore, c ce c. i=1 24

28 CHAPTER 2. QUOTIENT RINGS Note 2.8: Let R be a ring and let p be a non-zero prime ideal of R. Then R p is local: consider the set { } a p p = b a p, b R \ p = p R \ p, the localisation of p at R \ p. It is clear that p p is an ideal of R p : let m 1, m 2 p p and r R p. Then and m 1 m 2 = a 1 b 1 a 2 b 2 = a 1b 2 a 2 b 1 b 1 b 2 p R \ p = p p rm 1 = aa 1 bb 1 p R \ p = p p since p is a prime ideal and R \ p is an MCS. p p is maximal: assume there is an ideal R n p p. Then there must be an x n, x = a b, such that x p p, i.e. a R \ p. But then b a R p and hence 1 1 = b a a b n, establishing that n = R p, a contradiction. p p is the only maximal ideal: assume there is another maximal ideal, say n. Then, as above, there must be an x n such that x p p, establishing that n = R S. We will now proceed to the most important result of localisation: Theorem The Localisation Principle Let a, b be ideals of a ring R. If a m = b m for all maximal ideals m of R, then a = b. Let a a. Then a 1 b m for all maximal ideals m of R. Therefore, for each such ideal m there exists an element c m R\m such that c m a b. Now consider the ideal c generated by {c m m maximal ideal of R}. Assume that c R. Then, by Proposition 2.2.2, there exists a maximal ideal m 0 such that c m 0. By definition of c this would require an element c m0 R\m 0 to be in c m 0, leading to a contradiction. Hence, c = R. Therefore, 1 c and thus there are maximal ideals m 1,..., m k for some k N such that 1 = x 1 c m1 + + x k c mk for some x 1,..., x k R. This yields a = ax 1 c m1 + + ax k c mk b, 25

29 2.4. FRACTIONAL IDEALS since each ac mi b (by definition of the c m s). Hence a b. b a is shown similarly. 2.4 Fractional Ideals Again, we assume every ring to be commutative. To be able to define fractional ideals, we require some more algebraic background: Definition R-Module, R-Submodule Let R be a ring. Then an abelian group (M, +) is called an R-module if there is a binary operation : R M M such that, for all x, y M and r, s R, (a) r (x + y) = r x + r y, (b) (r + s) x = r x + s x and (c) (rs) x = r (s x). (d) 1 x = x. If M is an R-module and (N, +) (M, +) is a subgroup of M, then N is an R- module of its own and we say that it is an R-submodule of M. Usually, we omit the in r x and just write rx. Note that in this definition we can just as well define : M R: but this would only make a difference if R wasn t commutative. Definition Finitely Generated R-Module Let M be an R-module over some ring R. Then a finite set of elements m 1,..., m t M for some t N is said to generate M if m M r 1,..., r t R such that m = t i=1 r im i. If such a set exists, M is said to be finitely generated. Now, if R is a ring and T is its total ring of quotients, T obviously is an R-module. Furthermore, we can think of R being contained in T by identifying r R with r 1 (the embedding homomorphism ψ from Definition 2.1.4). Then R is a subring of T. Note that each ideal of R is an R-submodule of the R-module R. Hence, if an R-submodule A of T is a subset of R, A is an ideal of R. Definition Fractional Ideals An R-submodule A of T is called a fractional ideal of R if there exists a regular 26

30 CHAPTER 2. QUOTIENT RINGS element d R such that da R. If A is an ideal of R, it is also called an integral ideal. Example 2.9: For n N, non-zero a i Z, A := n i=1 1 a n Z is a fractional ideal of Z, since da Z taking d := a 1... a n. The total quotient ring of Z is Q. A := { a 2 n a Z, n N0 } is not a fractional ideal of Z. Any integral ideal A of R is a fractional ideal of R (since da R with d = 1). The sum, product and intersection of two and hence any finite number of fractional ideals are defined equivalently to the respective operations on integral ideals (see Proposition 1.2.4). Only the divisional ideal of two fractional ideals is defined (somewhat) differently: Definition Division of fractional ideals Let R be a ring and let A and B be two fractional ideals of R with respect to its total ring of quotients T. Then [A:B] T := {t T tb A} = {t T t x A x B} is called the divisional ideal of A and B. Whenever the context is clear we will write A:B instead of [A:B] T. Proposition Division of fractional ideals Let R be a ring and let A and B be two fractional ideals of R. Then [A:B] T is a fractional ideal if there exists a regular element b B. It is obvious that [A:B] T is an R-module. Since A is a fractional ideal, there exists a regular element d such that da R. Since b B, it is clear that b [A:B] T A. Since b is regular, db is regular and therefore db [A:B] T R. This shows that [A:B] T is a fractional ideal of R. Example 2.10: Let R be a ring with unity and let A = ar and B = br two principal fractional ideals of R where 0 a, b T = quot (R) and b is regular. Then: [A:B] T = a b R 27

31 2.4. FRACTIONAL IDEALS To see this, let t [A : B] T. Then tx B A for all x B B. Since b = b 1 B we see that tb = ar for some r R, establishing that t = a b r. Hence, [A : B] T a b R. On the other hand, suppose that t a b R. Then t = a b r 1 for some r 1 R. It is obvious that for x B B, tx B = a b r 1br 2 = ar 1 r 2 A for some r 2 R. This shows that a b R [A:B] T. As an example of this, [ 1 2 Z: 3 5 Z] = 5 Q 6Z, showing that the term divisional ideal is justified. Note 2.11: The set F (R) of all fractional ideals of a ring R is a commutative monoid 1 under multiplication, with the ideal R as its identity. We will now inquire what elements are the units in F (R). Definition Invertible Ideals If A is a fractional ideal of a ring R and there is a fractional ideal B of R such that AB = R, then A is called invertible and B is said to be its inverse. Proposition Uniqueness of inverses If the inverse B of a fractional ideal A exists, it is unique and B = [R : A] T. Furthermore, every principal fractional ideal C = cr, c T, c regular, is invertible. Its inverse is 1 c R. Uniqueness: Assuming there were a second inverse, C, we see: B = BR = BAC = RC = C Form of B: Let B := [R:A] T = {t T ta R}. Since BA = R, we have B B. Also, B = B R = B AB RB = B. Therefore, B = B. Clearly, by definition, 1 c R is a fractional ideal and [R : C] T = 1 c R by Example Hence, C 1 c R = R. Note 2.12: Suppose A is a fractional ideal of R and d is a non-zero element of R such that da R. Then da is an integral ideal of R since 1 A triple (G,, e) is called a monoid if e G, is a binary operation on G and for all a, b, c G, (a b) c = a (b c) and a e = e a = a. If a b = b a for all a, b G, G is called a commutative monoid. 28

32 CHAPTER 2. QUOTIENT RINGS (a) if x, y da then x = dx, y = dy for some x, y A and so x y = dx dy = d (x y) da. (b) if x da, r R, then x = dx for some x A and so rx = d (rx) da. Example 2.13: Let R := Z[X]. (a) 2 and X are invertible where 2 1 = 1 2 and X 1 = 1 X : Since { n 1 X = X i=1 Xf i 1 X g i } f i, g i R for i = 1,..., n, n 0 R and 1 = X 1 X X 1 X, we see that X 1 X = R. A similar argument proves that indeed = R. (b) 2, X is not invertible: Assume it were. By Proposition 2.4.7, its inverse must be [R: 2, X ] T. Then we see: [R: 2, X ] T = {t T t 2, X R} = = {t T t(2f + Xg) R f, g R} This yields (by (a)) that and [R: 2, X ] T [R: X ] T = [R: 2, X ] T [R: 2 ] T = 1 X 1. 2 Let t X. Then there are f, g R such that 1 2 f = t = 1 X g. Multiplying this by 2X yields Xf = 2Xt = 2g. This shows that 2g(0) = 0 f(0) = 0, therefore showing that g = Xg for some g R. Hence t = 1 X Xg = g R. Since obviously R 1 2, R 1 X and R [R: 2, X ]T, we see that [R: 2, X ] T = R. But since 1 2, X, we see that 2, X [R: 2, X ] T = 2, X R = 2, X R. Contradiction! Note 2.14: Together with Note 1.9 this allows the definition of A n, n Z, for (invertible) fractional ideals. It turns out that the invertibility of an ideal can also be characterized by using module theory. To see this, we first require three definitions. 29

33 2.4. FRACTIONAL IDEALS Definition Direct Sum Let R be a ring and let {M i i I} be a family of R-submodules of the R-module T. Then the sum i I M i := { i I m i mi = 0 for almost all i I } also is an R-submodule of R. We say that this sum is direct and write it as i I M i if every element m i I M i has a unique representation in the form m = i I m i where m i M i and m i = 0 for almost all i I. Definition Free Module Let R be a ring and let M be an R-module. If M has a basis (i.e. if there exists a linearly independent subset N of M such that M = N ), M is called free. Definition Projective Module Let R be a ring. An R-submodule P of R is called projective if there exists a free module F such that F = P Q for some module Q. Now we can state the aforementioned characterisation: Theorem Characterisation of invertible ideals Let A be an ideal of some integral domain R. Then A is invertible if and only if A is projective as an R-module. : Let A be a non-zero ideal of R. Suppose that A is projective, i.e. that there exist a module M and a free module F such that F = A M as above. Say that F has the basis {e λ λ Λ}, so that F = λ Λ e λ. For each λ, let π λ be defined as follows: { F R π λ : µ Λ r. µe µ r λ Quite obviously, π λ is a module homomorphism for all λ. For each λ, let e λ = a λ + m λ where a λ A, m λ M. Given x A, x = λ Λ π λ(x)e λ, we see that x = ( ) ( ) π λ (x)(a λ + m λ ) = π λ (x)a λ + π λ (x)m λ = π λ (x)a λ. λ Λ λ Λ If 0 x, y A, then π λ (xy) = xπ λ (y) = yπ λ (x). λ Λ λ Λ 30

34 CHAPTER 2. QUOTIENT RINGS So in the total quotient field T we have b λ := π λ(x) x = π λ(y) y for all 0 x, y A. Note that since π λ (x) = 0 for almost all λ, we have b λ = 0 for almost all λ. Also x A, b λ x = π λ(x)x x so b λ A R. Now we see: x A, x = λ Λ = π λ (x) R, π λ (x)a λ = λ Λ (xb λ )a λ = x λ Λ b λ a λ and so, since x 0, we get 1 = λ Λ b λ a λ. Set A := {b λ λ Λ}, then A A = R and so A is invertible. : Suppose A is invertible with AA = R. Let 1 = n j=1 a jb j where a j A, b j A. Define { n A i=1 R and w : π : x (b 1 x, b 2 x,..., b n x) { n i=1 R A (r 1,..., r n ) n j=1 r ja j. Then both w and π are module homomorphisms. Now π(w(x)) = π(b 1 x,..., b n x) = n n b j xa j = x a j b j = x. j=1 Therefore, A is isomorphic to a direct summand of n i=1 R. This shows that A is projective. j=1 Example 2.15: Since, by Example 2.13, 2, X is not an invertible ideal of R := Z[X], Theorem shows that 2, X is not projective as an R-module. 31

35 Chapter 3 Extension Rings and Factorisation of Ideals 3.1 Integral Extensions From now on, we assume all rings to be commutative with identity. In a similar way as extending Z to Q, R or C, rings in general can be extended. A first example, which we have already encountered, is the ring R[X] of polynomials over R. Obviously, R is contained in R[X] which therefore is an extension of R. In this section, we will explain the idea of a certain type of ring extensions, so-called integral extensions. Definition Integral Extension Let R be a ring. An element t T for some extension ring T R is called integral over R if there exists a monic polynomial p R[X] such that p(t) = 0. If this polynomial is of minimal degree, it is called the minimal polynomial of t over R. If all elements of T are integral over R, T is said to be an integral ring extension of R. In general, if t T R, R[t] denotes the smallest subring of T containing both R and t. Before we can proceed to the next result, we require some definitions and a wellknown result from Linear Algebra: Definition Minors, Cofactors, Adjoint Matrix Let A be an n n matrix over some ring R. Then the matrix obtained by dropping row i and column j from the matrix A is called the ij-th minor of A, usually denoted 32

36 CHAPTER 3. EXTENSION RINGS AND FACTORISATION OF IDEALS by M ij. The determinant 1 of M ij multiplied by ( 1) i+j is called the ij-th cofactor of A, usually denoted by A ij : A ij := ( 1) i+j M ij The matrix ( (Aij ) adj(a) := i=1,...,n j=1,...,n ) T is called the adjoint matrix of A. Theorem Laplace Expansion Let A be an n n matrix with entries a ij over some ring R. Denote the ij-th cofactor by A ij. Then ( n ) A adj(a) = a ik A jk k=1 i=1,...,n j=1,...,n ( n ) = A 1 = A ki a kj k=1 i=1,...,n j=1,...,n = adj(a) A. A proof shall not be given here. It can be found in [36, Th ] and in many other texts on Linear Algebra. Theorem Integral Extension Let T be a ring extension of some ring R. Then the following are equivalent: (a) t T is integral over R. (b) The ring R[t] is finitely generated as an R-module, i.e. there is an n N such that R[t] = { r 0 + r 1 t + + r n t n r 0, r 1,..., r n R }. (c) R[t] is contained in a subring S T that is finitely generated as an R-module. (d) There exists a finitely generated R-module S T such that (i) ts S and (ii) if u R[t] and us = 0, then u = 0. (a) (b): Let t be integral over R with minimal polynomial p, deg(p) = m for some m N. Then consider the finitely generated R-submodule N of T defined as follows: N := R + tr t m 1 R 1 See [36, Kapitel IV] for an explanation. 33

37 3.1. INTEGRAL EXTENSIONS Since p(t) = 0, we know that t m N. Now assume that for some integer k m, t k N. Hence, t k = m 1 i=0 r i ti for some r i R, i = 0,..., m 1. And therefore, in accordance with the above, t k+1 = tt k = r m 1t m + m 1 i=1 r i 1 ti has to lie in N. This shows that for all k N, t k N and therefore N = R[t]. (b) (c): Obvious. (c) (d): We take S to be the subring required to exist by condition (c). Then, ts S since t S. Furthermore, if us = 0, u = u 1 = 0, since 1 S. (d) (a): (By [4, Lemma 1.3.2]) Let S be the finitely generated R-module with properties (i) and (ii), required to exist by (d). By property (i), it is also finitely generated as an R[t]-module. Then, by property (ii), ann(s) = {0}. Assume S is generated by n N elements from R, say S = s 1,..., s n. Hence, for all k {1,..., n}: and so n ts k = a kj s j for some a ij R, i, j = 1,..., n j=1 n (δ kj t a kj )s j = 0 k. j=1 Now consider the matrix A := ( δ ij t a ij ) i=1,...,n j=1,...,n. Let D := det(a) = A and let M ij be the ij-th minor of A and A ij := ( 1) i+j M ij, the ij-th cofactor of A. Then for all i 0 = = n n A ki k=1 j=1 (δ kj t a kj )s j = ( n n ) A ki (δ kj t a kj ) j=1 k=1 n k=1 j= s j = n A ki (δ kj t a kj )s j = n Dδ ij s j = Ds i. j=1 This shows that D ann(s), therefore D = 0. But we also know that for some a 0,..., a n 1 R D = A = t n + a n 1 t n a 1 t + a 0. Hence t is integral over R. Definition Integral Closure Let T be a ring extension of a ring R. Then the set R = { t T t integral over R } is called the integral closure of R in T. 34

38 CHAPTER 3. EXTENSION RINGS AND FACTORISATION OF IDEALS Definition Normal Ring A ring R is called normal if it is integrally closed, i.e. if R = R, in its total ring of fractions. Theorem Integral Closure The integral closure R of some commutative ring R in some commutative superring T of R is a ring. Consider two elements t 1, t 2 R. Then, by Theorem 3.1.4, the ring R[t 1, t 2 ] = R[t 1 ][t 2 ] is a finitely generated R[t 1 ]-module and hence a finitely generated R- module. Since R[t 1 t 2 ], R[t 1 + t 2 ] and R[t 1 t 2 ] are all contained in R[t 1, t 2 ], it follows from Theorem 3.1.4, (c), that t 1 t 2, t 1 + t 2 and t 1 t 2 are integral over R. Hence R is a ring. Example 3.1: Let d Z be a square-free integer. We wish to find the integral closure R of R := Z in Q( d). Firstly, note that obviously Z[ d] R. Now, consider an element 0 z = p1 q 1 + p2 q 2 d R where the fractions p 1 q 1 lowest form. We see that the minimal polynomial of z must be f(x) = q 2 1q 2 2X 2 2p 1 q 1 q 2 2X + p 2 1q 2 2 q 2 1p 2 2d. Since z is required to be integral over Z, there are four possibilities: and p2 q 2 are in p 1 = 0, establishing that f(x) = q 2 2X 2 p 2 2d, requiring that q 2 2 p 2 2d whence, since d is squarefree, q 2 p 2, yielding that z Z[ d]. p 2 = 0, establishing that f(x) = q 2 1X 2 2p 1 q 1 X + p 2 1, requiring that q 1 p 1, yielding that z Z. (q 1 q 2 ) 2 = 1, establishing that q 1 = ±1 and q 2 = ±1, whence z Z[ d]. (q 1 q 2 ) 2 2p 1 q 1 q 2 2 and (q 1 q 2 ) 2 (q 2 1p 2 2d p 2 1q 2 2), requiring that q 1 2p 1 whence, since gcd(p 1, q 1 ) = 1, that either q 1 = ±1 or q 1 = ±2. If q 1 = ±1, then q 2 2 (p 2 2d p 2 1q 2 2), requiring that q 2 2 p 2 2d whence, since d is squarefree, q 2 = ±1. Then z Z[ d]. If q 1 = ±2, then 4q 2 2 (4p 2 2d p 2 1q 2 2), requiring that q 2 2 4p 2 2d, establishing that q 2 = ±2. Then, in turn, we have that 16 (4p 2 2d 4p 2 1), i.e. that 4 (p 2 2d p 2 1). Now, again, there are two possibilities. 4 p 2 2d and 4 p 2 1, whence p 1 = ±2k 1 and p 2 = ±2k 2 for some k 1, k 2 Z. This shows that z Z[ d]. 35

39 3.1. INTEGRAL EXTENSIONS 4 p 2 2d and 4 p 2 1, whence p 1 = 2k + 1 and p 2 = 2l + 1 for some k, l Z, establishing that 4 (4(dk 2 + dk 4l 2 4l) + (d 1)). This, in turn, can ( only be true if 4 (d 1). Assume this to be true. Then z = 2l ) [ 1 d + k l Z 2 (1 + ] d). On the other hand, if ( z = a + b ) [ ( 1 d Z )] d \ Z[ d] (i.e. 2 a), we can [ ( select l = a 1 2 and k = b + a 1 2, showing that Z )] d R. ( Concluding, we see that if 4 (d 1), R = Z 1 + )] d, otherwise R = Z[ d]. [ 1 2 Example 3.2: Furthermore, the ring R is an intriguing example: it shows that the property of unique factorisation is lost in this simple extension ring: ( ) ( 1 5 ) = 6 = 2 3 (3.1) Even worse, in this ring prime numbers (i.e. the prime elements of Z) are not necessarily irreducible: ( ) ( 6 5 ) = 41 We also see that 2, 3, and 1 5 are irreducible. To show this, consider the map { N : R N z = a + 5b z 2 = z z = a 2 + 5b 2. It is obvious that 2, 3 N(R) and that N preserves multiplication, i.e. N(xy) = N(x)N(y) for any x, y R. Since N(2) = 4 = 2 2, N(3) = 9 = 3 3 and N(1 + 5) = N(1 5) = 6 = 2 3, we see that the four aforementioned elements are in fact irreducible and no two of them are associates. See also [13]. This shows together with Equation 3.1 that irreducible elements are not necessarily prime! For a proper definition of prime and irreducible elements, refer to [11]. This phenomenon was examined in detail for the first time by Ernst Eduard Kummer in 1843 while trying to prove Fermat s Last Theorem. He introduced so-called ideal numbers, imaginary primes, to explain Equation 3.1 by setting 2 = π 1 π 2, 3 = π 3 π 4, = π 1 π 3, 1 5 = π 2 π 4, (3.2) where N(π 1 ) = N(π 2 ) = 2 and N(π 3 ) = N(π 4 ) = 3. The achievement of Julius Wilhelm Richard Dedekind then was to interpret these ideal numbers as ideals: π 1 = 2, 1 + 5, π 2 = 2, 1 5, π 3 = 3, 1 + 5, π 4 = 3, 1 5 It can be shown that these ideals in fact are prime ideals in R and that Equation 3.1 hold as equations of ideals: 2 = π 1 π 2, 3 = π 3 π 4, = π 1 π 3, 1 5 = π 2 π 4 36

40 CHAPTER 3. EXTENSION RINGS AND FACTORISATION OF IDEALS In section 3.3 we will determine in what rings it is possible to decompose an ideal into a product of prime ideals. 3.2 Algebraic Extensions When regarding the field R it becomes clear that some elements of R are not integral over Z and others are: for example, the minimal polynomial of 2 is X 2 2, but the minimal polynomial of 3 3 is 3X2 1. Hence, 2 is integral over Z yet 3 3 is not. This gives reason to the following definition. Definition Algebraic Extension Let R be some ring inside some extension ring T. An element t T is said to be algebraic over R if there is a polynomial f R[X] such that f(t) = 0. If all elements of T are algebraic over R, we say that T is an algebraic ring extension of R. Furthermore, if α is an element algebraic over R, R[α] is the smallest ring containing both R and α. It turns out that fields can be extended this way as well: in fact, C = R(i). On the other hand we see that the field R is not algebraic over Q: there is no polynomial f Q[X] such that f(π) = 0. Note that a field extension is indicated by ( ), a ring extension by [ ]. Definition Finite Extension Let R be a ring and let T be another ring algebraic over R. We say that T is a finite algebraic extension of R if there are elements t 1,..., t n T such that R[t 1,..., t n ] = T. Otherwise, we say that T is an infinite algebraic extension of R. If we consider the ring of all elements of C algebraic over Q we find that the resulting ring Q is an infinite algebraic extension of Q. 37

41 3.2. ALGEBRAIC EXTENSIONS Definition Degree of a Field Extension Let E be an extension field of a field F. Then the dimension of E regarded as a vector space over F is called the degree of E over F. We write [E : F ] = n if the dimension is finite and say that E is a finite extension of F and, if the dimension is infinite, that E is an infinite extension of F. If E = F (α) for some α algebraic over F we say that E is a simple extension of F. It turns out that the degree of a simple algebraic field extension F (α) is equal to the degree of the minimal polynomial of α. See [11, Th. 20.3]. Definition Separable Extension Let E be an extension field of some field F. We say that E is a separable extension of F if for all e E the minimal polynomial f e F [X] does not have multiple roots (we say that the polynomial (or the element itself) is separable). A field is called perfect if all its algebraic extensions are separable. It is an important result that all fields of characteristic 0 and all finite fields of characteristic p, p prime, are perfect. See [13, III, Satz and III, Satz 4.2.1]. There is an interesting result about separable field extensions of finite degree: Theorem Primitive Element Theorem Let E be a finite separable extension of F. Then there is an element α E such that E = F (α). This is a standard result, see [11, Th. 21.6]. Definition Number Ring Let K := Q(α) be an algebraic extension of Q (an algebraic number field). Then the integral closure of Z in Q(α) is denoted by O K or O α and is called the number ring of K. In fact, we note that quot (O K ) = K. As an important result, Stewart shows in [33, Ch. 4.3] that O Q( d), d N, has unique factorisation if and only if d = 1, 2, 3, 7, 11, 19, 43, 67, 163. Furthermore, Kummer showed in 1844 that factorisation into irreducibles in Z[e 2πi/n ] = O Q(e 2πi/n ) is not unique for n >

42 CHAPTER 3. EXTENSION RINGS AND FACTORISATION OF IDEALS To avoid confusion, from now on elements from Z may be called rational integers, elements from O K algebraic integers. The following is a nice result for the norm of prime ideals in Number Rings: Theorem Norm and degree of prime ideals Let p be a non-zero prime ideal in the Number Ring O α. Then there exists exactly one prime p Z and an integer n N such that N (p) = p n. Furthermore, there is an ideal a of O α such that pa = po α. n is called the degree of p and is denoted by deg(p). See [33, Th. 5.11]. It is clear that any number field K = Q(α) with f the minimal polynomial of α is isomorphic to Q[X]/ f. Hence, it is possible to embed K in the field of complex numbers. But note: if α 1,..., α n are the roots of f in C, there are n distinct (since f is separable) such embeddings σ i : K C such that σ i (Q) = Q and σ i (α) = α i. This allows us to define an important invariant of algebraic number fields, the discriminant. Definition Discriminant of Algebraic Number Fields Let K = Q(α) be an algebraic number field and let σ i, i = 1,..., n, be the embeddings of K into C. If b 1,..., b n is an integral basis 2 of K over Q, then (K) := det ((t ij ) i,j=1,...,n ) is called the discriminant of K. Here, t ij = T (b i b j ) and T is the trace form of K, n T (xy) := σ i (x)σ i (y). i=1 As an example of these embeddings, consider the field K = Q[X]/ X = { a + bx } a, b Q. L L 2 n If there are elements b 1,..., b n O K such that O K = i=1 Zb n i and K = i=1 Qb i, we say that the b 1,..., b n form an integral basis of K over Q. 39

43 3.3. DEDEKIND DOMAINS Then there are two roots of X in C: i and i. Hence, we see that K = Q(i) by means of σ 1 (a + bx) = a + bi and K = Q( i) = Q(i) by means of σ 2 (a + bx) = a bi. In a similar way, if O K is a ring of algebraic integers, the discriminant of O K is defined as the determinant of the trace matrix of the embeddings of O K into C. The same works for any free module Z[α] where α O K. Another important quantity is the discriminant (f) of a polynomial f, which is defined as (f) = a 2n 2 n n (r i r j ) 2, i,j i<j where a n is the leading coefficient of f and the r i are the roots of the polynomial f in the corresponding splitting field Dedekind Domains Since as mentioned before not every ring supports the property of (unique) irreducible factorisation, the question arises how much of this can be saved. The solution is to look at classes of ring elements, ideals, and try to decompose these into products of prime ideals. We will now determine in which rings this kind of prime factorisation is possible. Definition Dedekind Domain Let R be a commutative ring. Then R is called a Dedekind domain if (a) R is a Noetherian domain, (b) R is normal, and (c) non-zero prime ideals of R are maximal. Example 3.3: The standard examples of Dedekind domains are principal ideal domains. To see this, let R be a PID. Then Noetherianness is clear. Now let p be a non-zero prime ideal of R and let m be a maximal ideal of R containing p (existence is clear from Proposition 2.2.2). Then p = pr and m = mr for some p, m R, implying that p = mr for some r R. Now, since p is prime, either m p or r p. 3 The splitting field of a polynomial f K[X] is the smallest field L K such that in L[X] the polynomial f splits into a product of polynomials of degree 1. 40

44 CHAPTER 3. EXTENSION RINGS AND FACTORISATION OF IDEALS If m p, m = p and we re done. Otherwise, there is an s R such that r = ps, whence p = pms. This shows that m must be a unit in R, establishing that m = R, a contradiction. Now, let a b quot (R) be integral over R, b 0, in lowest form. Then there is a monic polynomial f with coefficients c i R such that f ( ) a b = 0, i.e. (a ) n ( a ) n 1 a + cn c1 b b b + c 0 = 0. Then, by multiplying out the denominators, we see that b a n. Since a b was in lowest form, this shows that b is a unit in R, whence a b R. Hence, R is a Dedekind domain. Definition leads to the following interesting result: Theorem Invertibility of Prime Ideals Let R be a Dedekind domain and p be a non-zero prime ideal of R. invertible in T and its inverse p 1 = [R:p] T. Then p is By Definition 2.4.4, [R:p] T is a fractional ideal of R. Since R is Noetherian, p is finitely generated, say by a 1,..., a n. We now show that [R:p] T R. Pick some 0 p p. Now we show that Rp contains a product p 1 p s, s N, of non-zero prime ideals of R. Suppose not. Then there exists an ideal i maximal among those that do not contain a product of non-zero prime ideals, since R is Noetherian and hence satisfies the ACC (see Theorem 1.2.9). Now, i cannot be prime. Hence (by Proposition 1.2.6), there exist two ideals a, b of R such that ab i but a i and b i. Therefore, (i + a)(i + b) = ii + ia + ib + ab i. But i i + a i and i i + b i and by definition of i there are non-zero prime ideals p 1,..., p k and q 1,..., q l, k, l N, such that i + a p 1 p k and i + a q 1 q l. However, this means that i (i + a)(i + b) p 1 p k q 1 q l. A contradiction! Now, assume the number s to be minimal. Since p Rp p 1 p s, p contains one of the p i s, say p 1. Since R is Dedekind, p 1 is maximal and hence p = p 1. Next, assume s = 1. Then, p = Rp = p 1 and [R:p] T = R 1 p is the unique inverse of p. This fractional ideal is not R since p is not a unit (p is a prime ideal and hence cannot contain units). Now assume s 1. Now, p Rp pp 2 p s. By the minimality of s we have Rp p 2 p s. Therefore, there exists an element r p 2 p s such that r Rp. This yields r p T \R, i.e. r p R. Since r p p = 1 p rp 1 p pp 2 p s 1 Rp = R, p we get r p [R:p] T and therefore [R:p] T R. By definition of [R:p] T we immediately see that p p [R:p] T R. Since p is maxi- 41

45 3.3. DEDEKIND DOMAINS mal, to show that p [R:p] T = R, it suffices to show that p [R:p] T p. Suppose that p [R:p] T = p. Then for every x [R:p] T \R we have xp p. Since R is Noetherian, p is finitely generated as an R-module. And since R is an integral domain, up = 0 requires u = 0 since p 0. Therefore, by Theorem (d), we know that x is integral over R. Since R is normal, we establish x R, a contradiction. Therefore, p [R:p] T = R. Now we can prove that indeed the concept of unique factorisation which is not necessarily possible in a Dedekind domain can be saved by generalising it to the unique factorisation of ideals. Theorem Unique factorisation Every non-zero fractional ideal a of a Dedekind domain R has a unique (up to ordering) representation in the form of a product of prime ideals, i.e. there are rational integers n p (a), almost all zero, (where p runs through all prime ideals of R) such that a = p p np(a). (3.3) Let R be a Dedekind domain. Let A be a fractional ideal of R. Then there is a regular element d R such that da R. Then a := da = d A is an ideal of R. Now, assume that for the ideals a and d there is a factorisation as in Equation 3.3. Then A has the factorisation A = p p np(a), n p (A) = n p (a) n p ( d ). Hence, in this proof it suffices to only consider integral ideals of R. Denote by A the family of those non-zero ideals in R which do not have a representation as a product of prime ideals. Suppose that A. Then since R is Noetherian there exists a maximal element a in A. a cannot be maximal, since all maximal ideals are prime (Proposition 1.2.6). Therefore it is properly contained in some maximal ideal p a. By Theorem 3.3.2, p 1 := [R : p] T is the unique inverse of p. From the condition a p we obtain ap 1 pp 1 = R, while p 1 R yields ap 1 a. Furthermore we see that a ap 1 : Assume they were equal. Then for each x p 1 we get xa a and therefore, by Theorem (d) and R being a Noetherian domain, x is integral over R. Since R is normal, this yields p 1 R. This is not possible by the proof of Theorem Hence, ap 1 a and ap 1 is a proper ideal since the equality ap 1 = R would imply that a = app 1 = Rp = p, contradicting the assumption a A. Since 42

46 CHAPTER 3. EXTENSION RINGS AND FACTORISATION OF IDEALS a ap 1 R, ap 1 must be a product of prime ideals say ap 1 = p 1 p 2 p k. Then a = pp 1 p 2 p k, showing that a A. It remains to show that this representation is unique. Assume that a = p 1 p s = q 1 q r (3.4) for some proper non-zero primes p i, q j, i = 1,..., s, j = 1,..., r. Since p 1 p 1 p s = q 1 q r, Proposition (i) shows that there exists j {1,..., s} such that p 1 q j. In the same way it can be shown that q j p i for some i. Since R is Dedekind, p 1 is maximal. This shows that p 1 = q j = p i. By Theorem 3.3.2, p 1 is invertible and therefore it can be cancelled from Equation 3.4, yielding p 2 p s = q 1 q j 1 q j+1 q r. (3.5) The same process can be continued to further reduce Equation 3.5, thus showing that r = s and that q 1,..., q r is just a permutation of p 1,..., p s. Let R be a Dedekind domain. We will say that an ideal a divides an ideal b, in signs a b, if there is an ideal c such that b = ac. Then, we get this nice result: Proposition Let R be a Dedekind domain and let a, b be ideals of R. Then, a b if and only if a b. If a b, there is an ideal c of R such that a ac = b. Conversely, let a b. Then, since R is Dedekind, a is invertible. Hence, R = a 1 a a 1 b =: c, whence c is an ideal of R. Then, ac = aa 1 b = Rb = b, i.e. a b. Then we have some immediate consequences of Theorem 3.3.3: Corollary Let R be a Dedekind domain and let a and b be two ideals of R. Then, if we define the rational integers n p ( ) as in Theorem 3.3.3, we have: n p (ab) = n p (a) + n p (b), n p (a + b) = min(n p (a), n p (a)) Let a = p pnp(a), b = p pnp(b) be ideals of R. Then, clearly, ab = p p np(a) p p np(b) = p p np(a)+np(b), 43

47 3.3. DEDEKIND DOMAINS showing that indeed n p (ab) = n p (a) + n p (b). Now, since a a + b and b a + b we see that (a + b) a and (a + b) b, establishing that n p (a + b) min(n p (a), n p (b)). On the other hand, consider the ideal c := p p min(np(a),np(b)). We see that c a and c b, whence c (a + b). This shows that min(n p (a), n p (b)) n p (a + b), whence, by the above, we have equality. We will now give an example of the computation of the factorisation of an ideal: Example 3.4: Consider the ring R := Z [ 17 ]. Then R is Dedekind as we will see in Theorem Let a := 18 be an ideal of R. We wish to find the factorisation of a. First, note that = 18 = (1 + 17)(1 17). Now consider the ideal p 1 := 2, Since 1 17 = 2 (1+ 17) p 1, we see that 18 = (1+ 17)(1 17) p 2 1, whence, by the above, p 2 1 a. Similarly, by defining p 2 := 3, and p 3 := 3, 1 17 we see that, since 18 = p 2 2 p 2 3, both p 2 2 and p 2 3 must divide a. Thus we have a p 2 1p 2 2p 2 3. (3.6) It is straightforward to see that p 1, p 2 and p 3 are all proper prime ideals of R. It now remains to show that we have equality in Equation 3.6. To do this, consider the norm of a: ( / ) N (a) = # R a Since every element from R can uniquely be written in the form a + b 17 + x, a, b {0,..., 17}, x 18, we see that N (a) = Furthermore, we show that an element u R lies either in p 1 or is of the form 1 + x, x p 1, establishing that N (p 1 ) = 2. To see this, let u = a + b 17 for some a, b Z. Assume u p 1. Then a b must be odd, since otherwise u = a + b 17 = 2 a b 2 + b(1 + 17) p 1, 44

48 CHAPTER 3. EXTENSION RINGS AND FACTORISATION OF IDEALS a contradiction. But then a b 1 must be even, whence u = a b b(1 + 17) = 1 + x, x p 1. Similarly, we see that N (p 2 ) = N (p 3 ) = 3 (by showing that R = {0, 1, 2} + p 2 = {0, 1, 2}+p 3 ). Note that this also shows that p 1, p 2 and p 3 are maximal ideals: their residue rings are the fields F 2 and F 3, respectively. Now assume that a = p 2 1p 2 2p 2 3q for some ideal q of R. Then, since the norm is multiplicative, we see that 18 2 = N (a) = N (p 1 ) 2 N (p 2 ) 2 N (p 3 ) 2 N (q) = N (q), requiring that N (q) = 1, whence q = R, establishing that a = p 2 1p 2 2p 2 3. From Theorem 3.3.3, the following corollary follows easily. Corollary Invertible Ideals Every non-zero ideal of a Dedekind domain is invertible. Let a be a non-zero ideal of a Dedekind domain R. By Theorem 3.3.3, a = p 1 p 2 p n for some prime ideals p 1,..., p n of R. Therefore, we see that ap 1 n p 1 n 1 p 1 1 = p 1 p 2 p n p 1 n By Theorem 3.3.2, each p i is invertible. p 1 n 1 p 1 1 = R, establishing that a is invertible. Example 3.5: Although Z[X] is Noetherian (see [20, Th. 2.17]), Z[X] is not Dedekind: on the one hand, not all prime ideals are maximal (see Note 1.13), contradicting Definition On the other hand, not all finitely generated ideals are invertible (see Example 2.13), contradicting Corollary Again, there is a different approach to the definition of Dedekind domains by using module theory: Theorem Dedekind Equivalences Let R be a commutative integral domain. Then the following are equivalent: (a) R is Dedekind. (b) Every ideal of R is projective as an R-module. (c) Every submodule of a projective R-module is projective (i.e. R is a hereditary ring). 45

49 3.3. DEDEKIND DOMAINS The equivalency (a) (b) has partially been shown in Theorem That a ring in which all ideals are invertible is Dedekind is shown in [37, Ch. 5, 6, Th. 13]. Due to the complexity of the result, a proof for the equivalence (a) (c) shall not be given here. It can be found in [25, p. 124]. There are some more important results of localisations of Dedekind domains. Theorem Let R be a Dedekind domain, let S be a multiplicatively closed subset of R and let p be a non-zero prime ideal of R. Then: (a) R S is a Dedekind domain. (b) R p is a local Dedekind domain. (c) If R is local, R is a principal ideal domain. Furthermore, there is an m R such that every fractional ideal A of R can be written as A = Rm n for some n Z. (a) We have to show that R S is normal, Noetherian and non-zero prime ideals of R S are maximal. Normality: It is clear that quot (R S ) = quot (R). Then, since R is normal, so is R S. Noetherianness: Let a be an ideal of R S. Then, by Proposition 2.3.3, (d), a c is an ideal of R such that R S a c = S 1 a c = (a c ) e = a. Then, since R is Noetherian, a c = n i=1 Ra i for some a i R, n N. This shows that a = n i=1 R Sa i, establishing that R S is Noetherian. Non-zero prime ideals are maximal: Let p be a non-zero prime ideal of R S. Then, by the above, p = S 1 p c. It is clear that p c must be a prime ideal of R. Now, let m be a maximal ideal of R S containing p. Then m c p c. Since R is Dedekind, we have that either m c = R or m c = p c. The former is a contradiction, since we would then have that m = R S. The latter shows, though, that m = p, completing the proof. (b) This follows from (a) and Note

50 CHAPTER 3. EXTENSION RINGS AND FACTORISATION OF IDEALS (c) Let m be the unique maximal ideal of R. Then, since R is Dedekind, m also is the only non-zero prime ideal of R. Hence, by Theorem 3.3.3, every non-zero fractional ideal A of R has the factorisation A = m n for some n Z. It is clear that m \ m 2 : assume m = m 2. Then Rm = m 2 whence, by Corollary 3.3.6, R = m, a contradiction. Now let m m \ m 2. Then Rm = m k for some k N. Since m n m 2 for all n 2, we see that k = 1, i.e. Rm = m. Then, m n = Rm n for all n Z, establishing that all fractional ideals of R are principal. The follwing is a nice corollary of Theorem 3.3.8, (c). It will prove to be useful when dealing with valuations on Dedekind domains. See Example 4.8. Corollary Let R be a local Dedekind domain. Then there is an element m R such that each x quot (R) \ {0} can be written as x = εm n where ε is a unit in R and n Z. Let m be the unique maximal ideal of R. Then, by Theorem 3.3.8, (c), there is an m R such that m = m and each fractional ideal A of R is principal with A = m n for some n Z. Now, let x quot (R), x 0. Then Rx = m n = Rm n for some n Z. Hence, x = r 1 m n and m n = r 2 x for some r 1, r 2 R, establishing that r 1 = r 1 2, i.e. r 1 is a unit in R, completing the proof. Note 3.6: This can be generalised to the non-local case: Let R be a Dedekind domain. Then, by Theorem 3.3.8, R p is a local Dedekind domain, whence, by Corollary 3.3.9, there is an m R p such that every element x R p \ {0} can be written as x = um n where u is a unit in R p and n N 0. Hence, m = a b a R and b R \ p. This shows that every x 0 can be written as x = εt n, for some where ε := u b is a unit in R p, n N 0 and t := a R. Note that we identify elements r from R with r 1 K. Now, since R pm = p p, we know that m p p \ (p p ) 2. Hence, ( t = mb p p \ (p p ) 2) R = p \ p 2. It is clear that if R is local, there is only one prime ideal p and R p = R, yielding that t = m. It turns out that the Dedekind-property is stable under integral extensions : 47

51 3.4. BASICS OF ALGEBRAIC NUMBER THEORY Theorem Integral Extensions and Dedekind Domains Let R be a Dedekind domain, T its quotient field. If Q is a finite extension of T, the integral closure S of R in Q is Dedekind as well. See [37, Ch. V, 8, Th. 19]. As an immediate consequence of this we see that we have already encountered a very important class of Dedekind domains: Theorem Algebraic Number Rings Let Q be an algebraic number field and let O Q be its number ring. Then O Q is Dedekind. Since Q is an algebraic number field, Q = Q(α) is a finite extension of Q for some α. Then, since Z is Dedekind and since O Q is the integral closure of Z in Q, O K is Dedekind by Theorem The concept of Dedekind domains can be further generalised. This shall be done in section Basics of Algebraic Number Theory Throughout the next chapters we will need some definitions and results from Algebraic Number Theory. Mostly, proofs will not be given here. For further information and more background theory, the reader is referred to [11], [14], [18] and [33]. Before we can show an important consequence of Definition 3.2.8, we need an intermediate result. Lemma Let A be a Dedekind domain, K its quotient field, L a finite separable extension of K and B the integral closure of A in L. Then there is an element α B such that L = K(α). 48

52 CHAPTER 3. EXTENSION RINGS AND FACTORISATION OF IDEALS Since L is a finite separable extension of K, there is an element β L such that L = K(β). Hence, there is a polynomial of minimal degree in K[X], say f, such that f(β) = 0. Without loss of generality, f can be selected from A[X], just multiply with the lcm of the denominators of the coefficients. Hence, let f(x) = a 0 + a 1 X + + a n X n, a n 0, a i A for i = 0,..., n. Now consider the polynomial f(x) := a 0 a n 1 n + a 1 a n 2 n X + + a n 1 X n 1 + X n. It is clear that this polynomial has the root α := a n β B. Furthermore, { n } L = b i β i, b i K. i=0 But since a n 0, every such sum can also be expressed in terms of α, n n b i β i 1 = b i a i α i, n i=0 i=0 establishing that L = K(α). Note 3.7: By the same argument we see that if A is a Dedekind domain, K its quotient field and α is algebraic over K, there is an element c A such that cα is integral over A. Now, to the promised consequences. Proposition Let K be an algebraic number field and let (K) be the discriminant of K. Then there is an element α O K such that Q(α) = K. Let f be the minimal polynomial of α. Then: (a) (K) is a rational integer and independent of the choice of the integral basis B = (b 1,..., b n ). (b) (O K ) = (K) (c) (f) = (Z[α]) = [O K : Z[α]] 2 (K) (d) [O K : Z[α]] O K Z[α] Example 3.8: Let α = 7 and consider the algebraic number field K := Q(α). We know that O K = Z [ 1 2 (1 + α) ]. Furthermore, the minimal polynomial f of α is f(x) = X Obviously, ( B K = 1, 1 ( ) ) T =: (b 1, b 2 ) T 2 49

53 3.4. BASICS OF ALGEBRAIC NUMBER THEORY is an integral basis for K. The embeddings of K into C are σ 1 : xb 1 + yb 2 x y ( ), σ 2 : xb 1 + yb 2 x y ( 1 7 ). Then, clearly, σ 1 (b 1 ) = σ 2 (b 1 ) = 1, σ 1 (b 2 ) = 1 ( ) 1 + 7, 2 σ2 (b 2 ) = 1 ( ) This shows that [ 2 1 T K = 1 3 ], whence (K) = det (T K ) = 7. And the discriminant of f is (f) = 1 ( ) 2 = 28. Considering the Z-module Z[α] we see that B Z[α] = ( 1, 7 ) T =: (b 1, b 2) T is an integral basis for Z[α]. Here, the embeddings of Z[α] into C are σ 1 : xb 1 + yb 2 x + y 7, σ 2 : xb 1 + yb 2 x y 7, yielding that [ 2 0 T Z[α] = 0 14 ], whence (Z[α]) = det ( T Z[α] ) = 24. To complete this example, we need to find out the index of Z[α] in O K. Therefore, / O K Z[α] {0 = + Z[α], } 7 + Z[α], 2 establishing that [O K : Z[α]] = 2. Then, finally, we see that (f) = 28 = (Z[α]) = 4 ( 7) = [O K : Z[α]] 2 (K). The existence of α follows from Lemma The other consequences are shown in [14, Ch. 3], [26, Ch. 2] and, partially, in [18, Ch. 3]. Note that [O K : Z[α]] denotes the order of the (finite, Abelian) factor group O K /Z[α]. With this spadework, we can state a more fundamental theorem which we will require in the course of the proof of Theorem Theorem Prime Ideals of Degree 1 50

54 CHAPTER 3. EXTENSION RINGS AND FACTORISATION OF IDEALS Let K = Q(α), α O K, be an algebraic number field, let f be the minimal polynomial of α and let p be a prime such that p (f). Then there is a prime ideal p of O K of degree one with p p if and only if there is an x p Z such that f(x p ) 0 (mod p). Then p = O K p + O K (α x p ). See [27, Satz 1]. Lemma Let f Z[X] be a non-constant polynomial. Then there are infinitely many primes p such that f(x) 0 (mod p) has a solution. See [27, Satz 2]. Corollary Prime Ideals of Degree 1 Let K be an algebraic number field. Then, in O K, there are infinitely many prime ideals of degree 1. This is a consequence of Theorem and Lemma Theorem Unramified Prime Ideals Let A be a Dedekind Domain, K its field of fractions, L a finite separable extension of K and B the integral closure of A in L. Then all but a finite number of prime ideals P of B are unramified over A, i.e. (P A) B = PP e Pen n for n + 1 distinct prime ideals of B. See [31, 5, Th. 1 and Cor. 2]. In [23, Th. 4.14], this is extended to the fact that all e i = 1. Finally, we state some other important and very well-known results we will use later on. They are all consequences of Lagrange s Theorem: Theorem Lagrange s Theorem 51

55 3.4. BASICS OF ALGEBRAIC NUMBER THEORY If G is a finite group 4 and H is a subgroup of G, then ord(h) ord(g). Moreover, the number of distinct left (right) cosets of H in G is ord(g)/ord(h). See [11, Th. 7.1]. Theorem Group-order Theorem Let G be a (multiplicatively written) group. Then, if a G, a ord(g) = 1. See [11, Ch. 7, Corollary 4]. Theorem Euler s Theorem Let a, b be two relatively prime rational integers. Then, if φ is Euler s function, a φ(b) 1 (mod b). This is a consequence of the Group-order Theorem: let a, b Z such that gcd(a, b) = 1. Then a is a unit in Z b and hence an element of G := Z b. Furthermore, φ(b) = ord (G). Hence, a φ(b) 1 (mod b), completing the proof. 4 Here, ord(g) denotes the order of the group G, i.e. the number of elements in G. 52

56 CHAPTER 3. EXTENSION RINGS AND FACTORISATION OF IDEALS Theorem Fermat s Little Theorem Let p be a rational prime. Then, for a 0 (mod p), we have that a p 1 1 (mod p). This is an immediate consequence of Euler s Theorem: φ(p) = p 1, hence a p 1 = a φ(p) 1 (mod p). A fundamental result of algebraic number theory is the following theorem, Hilbert s Nullstellensatz: Theorem Hilbert s Nullstellensatz Let K be an algebraically closed field 5 and a be an ideal of K[X 1,..., X n ], the polynomial ring in n indeterminates. Let V (a) := {a K n such that f(a) = 0 for all f a}. Suppose that f(a) = 0 for all a V (a). Then there is an integer r > 0 such that f r a. A proof can be found in [19, Ch. X, 2], in [35, Ch. XI, 75] and in [38, Ch. VII, 3, Th. 14]. Corollary Weak Nullstellensatz Let K be an algebraically closed field and let a be a proper ideal of K[X 1,..., X n ]. Then V (a). This is equivalent to Theorem See [38, loc. cit.]. An important special case, which we will use twice in chapter 5, is the following: Corollary Let K be a field. If the polynomials f 1,..., f m K[X 1,..., X n ] have no common zero in the algebraic closure of K, there are polynomials A 1,..., A m K[X 1,..., X n ] 5 A field K is algebraically closed if it is a splitting field for every polynomial f K[X]. 53

57 3.5. PRÜFER DOMAINS such that A 1 f A m f m = 1. See [35, p. 10]. 3.5 Prüfer Domains The well-known Chinese Remainder Theorem from Number Theory (Appendix B) can be transferred to a situation where one does not deal with individual elements of a ring but with ideals. Definition Chinese Remainder Theorem Let R be a commutative ring, let a 1,..., a k be a finite number of ideals of R and let x 1,..., x k be elements in R. Consider the system of congruences x x 1 (mod a 1 ) x x 2 (mod a 2 ) ( ). x x k (mod a k ) where x x i (mod a i ) means that x x i a i. If this system has a solution x R, then for any i, j we have x x i a i, x x j a j, so x i x j a i + a j. If this condition is also sufficient for all a 1,..., a k ideals of R and for all x 1,..., x k R, then R is said to satisfy the Chinese Remainder Theorem, or CRT. Definition Prüfer Domains An integral domain R is called a Prüfer domain if it satisfies the CRT. This definition immediately leads to some other characterisations of Prüfer domains: Theorem Properties of the CRT Let R be a ring. Then the following are equivalent: (a) The CRT holds in R. (b) If a, b, c are ideals in R then a (b + c) = a b + a c. (c) If a, b, c are ideals in R then a + (b c) = (a + b) (a + c). 54

58 CHAPTER 3. EXTENSION RINGS AND FACTORISATION OF IDEALS (b) (c) Assume (b) holds. Then for all a, b, c ideals of R we have (a + b) (a + c) = ((a + b) a) + ((a + b) c) = = a + (c a + c b) = a + (b c). (c) (b) Assume (c) holds. Then for all a, b, c ideals of R we have (a b) + (a c) = (a b + a) (a b + c) = a (c + a b) ( ) and (a b) + (a c) = (a + a c) (b + a c) = a (b + a c). This shows that c + a b = b + a c. Since for any ideal I, I = I + I, we notice that c + a b = c + a b + b + a c = b + c. Together with ( ), this yields a b + a c = a (b + c). (a) (c) In any ring, we always have a + (b c) (a + b) (a + c) since if r a + (b c), then r = a + b = a + c where a a, b b, c c and b = c. Hence, r (a + b) (a + c). The reverse inclusion is shown as follows: Let t (a + b) (a + c). We wish to write t = x + y where x a, y b c. So we wish to solve the system of congruences x 0 (mod a) x t (mod b) x t (mod c). Since t 0 = t a + b, t 0 = t a + c and t t = 0 b + c, x exists since the CRT holds in R. (c) (a) Consider the system ( ) of Definition for which x i x j a i +a j for all i, j k. 55

59 3.5. PRÜFER DOMAINS We proceed by induction on k. k=2 Then x 1 x 2 = a 1 + a 2 for some a 1 a 1, a 2 a 2. Let x = x 1 a 1 = x 2 + a 2. Then x x 1 (mod a 1 ) and x x 2 (mod a 2 ). Hence, k = 2 is true without using (3). k k + 1 Assume any system of k congruences to be solvable. Consider the system of k + 1 congruences x x i (mod a i ), i = 1,..., k, k + 1 where x i x j a i + a j for all i, j k + 1. By induction, there exists y R such that y x i (mod a i ) for all i = 1,..., k. Consider the system of two congruences x y ( mod ) k a i i=1 x x k+1 (mod a k+1 ). Then, from the k = 2 case, we know there exists a solution t R of this system, provided y x k+1 a k+1 + k a i. i=1 Then t y a i for each i = 1,..., k and t x k+1 a k+1. But y x i a i for i = 1,..., k, so we also get t x i = (t y) + (y x i ) a i. Hence, t will be a required solution for the system of k + 1 congruences. Since (3) holds in R, we have ( n ) n a + b i = (a + b i ) (by induction) and hence a k+1 + i=1 k a i = i=1 We also have: i=1 k (a k+1 + a i ). ( 2 ) i=1 y x k+1 = (y x i ) + (x i x k+1) a i + a k+1 i = 1,..., k }{{}}{{} a i a i+a k+1 k and so, by ( 2 ), y x k+1 a k+1 + i=1 a i This completes the proof. 56

60 CHAPTER 3. EXTENSION RINGS AND FACTORISATION OF IDEALS Example 3.9: Let R = Z and let a, b, c be ideals of R such that a = R p a1 1 pa k k, b = R p b1 1 pb k k, c = R p c1 1 pc k k where p i prime and a i, b i, c i 0 for i = 1,..., k. Then a b = R p d1 1 pd k k e i = max{a i, c i }. Hence, a b + a c = R p f1 1 pf k k where where d i = max{a i, b i } and a c = R p e1 1 pe k k f i = min{d i, e i } = min{max{a i, b i }, max{a i, c i }} = max{a i, min{b i, c i }}. Therefore, a b + a c = a (b + c). Hence, the CRT holds in R. where Proposition Let R be a commutative ring in which the ideals form a chain with respect to inclusion. Then the CRT holds in R. Let a, b, c be ideals of R. Without loss of generality, we suppose that b c. Then (a + b) (a + c) = a + b = a + (b c). Hence, by Theorem 3.5.3, the CRT holds in R. Definition Chain Ring A commutative ring R is called a chain ring if the ideals of R form a chain with respect to inclusion. An integral domain R is called a chain domain if it is a chain ring. In fact, we realise that this chain property needs only be true for principal ideals: Proposition Chain Ring A ring R is a chain ring if and only if for all a, b R either ar br or br ar. : obvious. : Let a, b be ideals of R. Suppose a b and b a. Then there exist a a\b, b b\a, but either ar br or br ar, leading to a contradiction. Note 3.10: It is obvious that in a chain ring all finitely generated ideals must be principal: Assume not. Let a = a, b be an ideal in the chain ring R which 57

61 3.5. PRÜFER DOMAINS cannot be generated by one element. Then, by the above, without loss of generality a b. Hence, a, b = a, a contradiction. This argument can be extended by induction to show that all finitely generated ideals in R must be principal. In fact, we have even more: it is immediate from Proposition that R has exactly one maximal ideal. Hence, a chain ring is also a local ring. Note, though, that chain rings need not be principal ideal rings: the ring R from Note 3.12 is an example. An integral domain where all finitely generated ideals are principal is called a Bézout Domain. Proposition Units in a Chain Ring Let R be a chain ring with unique maximal ideal m. Then R\m is the set of units in R. By the above, R has exactly one maximal ideal, say m. Obviously, since m R, m cannot contain units. Now, let s R \ m. Assume s is not a unit. Then Rs R. Therefore, by Proposition 2.2.2, Rs m, yielding s m. Contradiction! Example 3.11: We will now give an example where the divisional ideal of two principal ideals is not finitely generated, let alone principal (this was promised in section 1.2). See also [16]. Let K be any field. Now we will consider the ring R of power series over K in the indeterminate X. Define R as follows: { } R := K[[X]] := a i X i a i K i N 0 The units in R are exactly the elements where a 0 0: Assume f = i N 0 a i X i R is a unit, i.e. g = i N 0 b i X i R such that fg = 1. This can only be the case if a 0 b 0 = 1. Therefore b 0 = a 1 0, i.e. a 0 0. The remaining coefficients of g can now be calculated easily. Now if f, g R are non-zero non-units, there exist m, n N and units k, h R such that f = X m k and g = X n h. Without loss of generality, we assume that m n. Then g = X n h = (X n m hk 1 )X m k = (X n m hk 1 )f. Therefore, g f. Proposition then shows that R is a chain ring and this yields (by Proposition 3.5.7) that the unique maximal ideal n = X. 58

62 CHAPTER 3. EXTENSION RINGS AND FACTORISATION OF IDEALS This result can in a similar way be generalised to the ring R of generalised power series over K, { } R := a i X si a i K, s i R + 0, i I where I N 0, i < j s i < s j and either I is finite or lim i I s i =, yielding that R as well is a chain domain with unique maximal ideal m. Then m is idempotent in R, i.e. m 2 = m: Consider X s k m, k a unit. Then X s k = X s 2 X s 2 k m 2. Now we show that m cannot be finitely generated. Assume it were. Then since R is a chain ring m = mr for some m R. Since m is idempotent, we have m m 2 = m 2 R, so m = m 2 x, (x 0), yielding mx = 1. This is a contradiction since m cannot be a unit. In fact, m is countably generated, namely m = X 0.1n n N. Now let 0 b m, a := bm, and consider the following factor ring S := R / a. It is easy to see that b + a is a non-zero element of S and that (b + a) 2 = b 2 + bm = 0 + bm. Therefore, S is not an integral domain. Let b := b + a and consider the divisional ideal 0 : b = { s S s b 0 } = { s S sb = 0 } = ann(b) = = { r + a S } { } (r + a)(b + a) = 0 = r + a S rb a = { = r + a R / } a r m = m/ a. Now assume that m/a is finitely generated, say since S is a chain ring m/a = m + a. Then, given any y m, we have y + a = rm + a for some r R. So y rm bm y rm = bn for some n m. This yields y = rm + bn and so y m, b, showing that m = m, b, a contradiction. This establishes that 0 : b is not finitely generated. Note 3.12: The ring of generalised power series, R, defined in the above example is a Prüfer domain since it is a chain ring (by Proposition 3.5.4) but not a Dedekind domain since it is not Noetherian (its unique maximal ideal m is not finitely generated). 59

63 3.5. PRÜFER DOMAINS Theorem Characterisation of Prüfer domains I Let R be an integral domain. Then R is a Prüfer domain if and only if R p is a chain ring for all 0 p prime ideals of R. We will show this by proving equivalence to part (b) of Theorem Assume that for every proper non-zero prime ideal of R, R p is a chain ring. Now consider ideals a p, b p, c p of R p. Then without loss of generality b p c p. This shows: a p (b p + c p ) = a p c p = a p b p + a p c p. Now consider ideals a, b, c of R (we make frequent use of Proposition 2.3.3): (a (b + c)) p = a p (b + c) p = a p (b p + c p ) = = a p b p + a p c p = (a b) p + (a c) p = = (a b + a c) p. By the Localisation Principle (Theorem 2.3.4), this shows that a (b + c) = a b + a c. Conversely, assume that (2) of Theorem holds. Let p be a proper non-zero prime ideal of R and let a, b R. Since a b + a b, we have a = a ( b + a b ) = ( a b ) + ( a a b ). Then a = t + c(a b) where t a b, c R and c(a b) a. Then cb a and (1 c)a = t cb b. If c p, then b ar p. If c p, then 1 c p and a br p. Proposition then shows that R p is a chain ring. This gives reason to a generalisation of Prüfer domains, the so-called arithmetical rings: Definition Arithmetical Ring A ring R is arithmetical if R m is a chain ring for all maximal ideals m of R. Theorem Characterisation of Arithmetical Rings In a commutative ring R, the following are equivalent: (a) R is arithmetical. (b) a : b + b : a = R for all finitely generated ideals a, b of R. 60

64 CHAPTER 3. EXTENSION RINGS AND FACTORISATION OF IDEALS (c) Rx : Ry + Ry : Rx = R for all x, y R. Consider the set of maximal ideals of R, denoted by {m λ } λ Λ. We will proceed in several steps: (1) Claim: ( ) c U = U mλ for any ideal U of R. λ Λ Since m λ is a maximal ideal, m λ R and so 1 m λ. Thus u 1 U, λ Λ and so U ( λ Λ U ) c. m λ U m λ u Let v λ Λ U m λ. Assume that v (u, 1) u U. Let v = u0 r 0 for some u 0 U, r 0 R\ { λ Λ m } λ = λ Λ R\m λ. But, by Proposition 2.2.2, there exists λ 0 Λ such that Rr 0 m λ0. Therefore, r 0 λ Λ R\m λ. Contradiction! Hence U ( λ Λ U ) c. m λ (2) By the Localisation Principle, Theorem 2.3.4, we know that for two ideals U 1, U 2 of R, U 1 = U 2 if and only if (U 1 ) mλ = (U 2 ) mλ λ Λ. (3) Consider the map Φ λ : { I Iλ U U mλ, λ Λ, where I are the non-zero ideals of R and I λ those of R mλ. Claim: For all λ Λ, Φ λ is surjective. Let I I λ. Then I c is an element of I and by Proposition 2.3.3, Φ λ (I c ) = (I c ) mλ = I. Therefore, Φ λ (I ) = I λ. (4) By Proposition we know that for a, b I Φ λ (a + b) = Φ λ (a) + Φ λ (b) Φ λ (ab) = Φ λ (a)φ λ (b) Φ λ (a b) = Φ λ (a) Φ λ (b) Φ λ (a : b) = Φ λ (a) : Φ λ (b), if b is finitely generated. 61

65 3.5. PRÜFER DOMAINS Now suppose that an equation (E 1 ) involving ideals and only the operations +,,, : is true for every local ring. If (E 1 ) holds for R, then, in particular, it is true for R mλ, λ Λ. Furthermore, if (E 1 ) (E 2 ) for some other equation (E 2 ) involving ideals and only the operations +,,, :, we hence note that (E 2 ) holds for R mλ, λ Λ. Then the Localisation Principle (Theorem 2.3.4) shows that (E 2 ) must also be true for R. This shows that we can assume that R is local. (a) = (b): Now, if R is arithmetical and local, R is a chain ring, since, if m is the unique maximal ideal of R, R = R m. Hence, if a, b are finitely generated ideals of R, we note that either a b or b a, yielding that either a : b = R or b : a = R. Therefore, a : b + b : a = R. (b) = (c): This is obvious. (c) = (a): Assume (c) holds. Let m be the unique maximal ideal of R. We know that not both Rx : Ry and Ry : Rx can be contained in m. Without loss of generality, Rx : Ry m. Hence, Rx : Ry must contain an element m m, which since m is unique must be a unit, yielding y Rx. Therefore, Ry Rx. Hence, by Proposition 3.5.6, R is a chain ring. Let a m b m be two ideals of R m such that a and b are ideals of R. Let a a. Then a 1 b m and so m R\m such that am b. Consider the ideal generated by m: m = Rm. Assume m R then, by Proposition 2.2.2, m m. Contradiction! Hence, m = R. Thus n R such that mn = 1 and so a = 1 a = mn a = n }{{} ma b. b This shows that R is arithmetical. Quite obviously, if R is also required to be an integral domain, Theorem yields even more equivalent definitions for Prüfer domains (since the proof of Theorem carries through when requiring R m to be a chain ring for non-zero maximal ideals of R only). The following theorem explains why Prüfer domains are of interest in the context of factorisation: they are merely a generalisation of Dedekind domains. 62

66 CHAPTER 3. EXTENSION RINGS AND FACTORISATION OF IDEALS Theorem Characterisation of Prüfer domains II In a commutative integral domain R, the following are equivalent: (a) All finitely generated non-zero ideals of R are invertible. (b) If ab = ac for a finitely generated, non-zero ideal a, b = c (the cancellation property for finitely generated ideals). (c) For all proper prime ideals p of R, R p is a chain ring. (d) (a + b) : c = a : c + b : c for all ideals a, b, c of R, c finitely generated. (e) Every non-zero ideal of R that is generated by two elements is invertible. (a) (b): If a is finitely generated, non-zero, and ab = ac, we see: b = Rb = a 1 ab = a 1 ac = Rc = c. (b) (c): Suppose (b) holds. Now let a, b, c be ideals of R, a finitely generated, such that ab ac. Then: ac = ab + ac = a(b + c) and so, by (b), c = b + c which gives b c Let p be a non-zero proper prime ideal of R. It is sufficient to show that for a s, b t R p, at least one of a s b t and b t a s holds (see Proposition 3.5.6). Since s, t p we see that 1 s and 1 t are units in R p and it therefore suffices to show that either ar p br p or br p ar p. This obviously is true for a = 0 or b = 0. Hence we can assume that both a 0 and b 0. Now consider the following: ab a, b = ab ( a + b ) = ab a + ab b = a 2 b, ab 2 a 2 b, ab 2, a 3, b 3 = a 3 + a 2 b + ab 2 + b 3 = = ( a 2 + b 2 )( a + b ) = a 2, b 2 a, b Therefore, by the above, ab a 2, b 2. Then ab = xa 2 + yb 2 for some x, y R. Now consider yb a, b = yab, yb 2 = y(xa 2 + yb 2 ), yb 2 = xya 2 + y 2 b 2, yb 2 = = xya 2, yb 2 a 2, yb 2 = a 2, xa 2 + yb 2 = a 2, ab = a a, b. This yields by the above yb a. Say yb = au for some u R. Then ab = xa 2 + uab and so xa 2 = ab(1 u). Thus xa = b(1 u). 63

67 3.5. PRÜFER DOMAINS Now, if u p, then a = b y u br p. If u p then 1 u p and therefore b = a x 1 u ar p. (c) (d): Assume (c) and let p be a non-zero proper prime ideal of R. Then (d) holds for ideals of R p since, for a p, b p, c p ideals of R p, without loss of generality we can assume that a p b p. Then (a p + b p ) : c p = b p : c p = a p : c p + b p : c p. Now let a, b, c be ideals of R, c finitely generated (we make frequent use of Proposition 2.3.3): ((a + b) : c) p = (a + b) p : c p = (a p + b p ) : c p = a p : c p + b p : c p = = (a : c) p + (b : c) p = (a : c + b : c) p Now, by the Localisation Principle (Theorem 2.3.4), (d) holds. (d) (e): Let 0 a, b R. If (d) holds, we see that R = a, b : a, b = a : a, b + b : a, b = a : b + b : a. Let 1 = x + y where xb a and ya b. Then xb b ab and ya a ab. Hence, a, b bx, ay ab. But ab = abx + aby, so ab = a, b bx, ay. Since a, b 0, the principal ideal ab is invertible. This shows that a, b is invertible. (e) (a): Let c = c 1,..., c n be a non-zero ideal of R. We will proceed by induction to show that c is invertible. The claim obviously is true for n = 1 and by (e) for n = 2. Now assume it is true for n 1, n > 2. Let Then a := c 1,..., c n 1, b := c 2,..., c n, d := c 1, c n, e := c 1 a 1 d 1 + c n b 1 d 1. ce = (a + c n ) c 1 a 1 d 1 + ( c 1 + b) c n b 1 d 1 = = c 1 d 1 + c n c 1 a 1 d 1 + c 1 c n b 1 d 1 + c n d 1 = = c 1 d 1 ( R + c n b 1) + c n d 1 ( R + c 1 a 1) ( ) = c 1 d 1 + c n d 1 = = c 1, c n d 1 = dd 1 = R. This shows that c is invertible. Note that the starred equality above holds because c n b 1 R and c 1 a 1 R. The following is an immediate result: 64

68 CHAPTER 3. EXTENSION RINGS AND FACTORISATION OF IDEALS Theorem Dedekind Rings are Prüfer Let R be a Dedekind domain. Then R is a Prüfer domain. By definition, R is Noetherian, i.e. all its ideals are finitely generated. By Corollary 3.3.6, all these ideals are invertible. Then, by Theorem , the equivalence (a) (c) holds for R, whence, by Theorem 3.5.8, R is Prüfer. Our next goal is to find examples for Prüfer domains. We already have one: the ring of generalised power series (see Note 3.12) is a non-dedekind Prüfer domain. The next chapter forms the basis for our quest for Prüfer domains. 65

69 Chapter 4 An Introduction to Valuation Theory In this chapter, we will give some basic results about valuations. More detailed, in-depth material can either be found in [3], [28] or [38]. Note that this chapter in itself has little to do with the results obtained so far. Valuations are nothing but an important concept used quite heavily in chapter 5 to find more examples for Prüfer domains. Basically, there are two different approaches to valuations: it is possible to introduce them via a generalisation of the concept of absolute values (this is done in [3]). The other way is to directly enter into the theory of non-archimedian valuations (as in [12] or [38]). We endeavour to combine the two methods: section 4.1 is an introduction to absolute values, while section 4.2 explains the concept of places to motivate the introduction of valuations in section Absolute Values We begin with the concept of an ordered group: Definition Ordered Group Let G be a multiplicatively written group. We say that G is an ordered group if there exists a normal subsemigroup 1 S of G such that G = S {1} S 1. (4.1) 1 Let (G, ) be a (semi)group. A subset F G is called a sub(semi)group of G if e F (only if G is a group) and (F, F F ) is a (semi)group. The sub(semi)group is called normal if a F = F a for all a G. 66

70 CHAPTER 4. AN INTRODUCTION TO VALUATION THEORY Now it is possible to create a relation of order on G by means of the following result: Proposition Ordering an ordered group Let G be an ordered group. Then we can define a binary ordering relation < on G G by setting a < b : ab 1 S (4.2) for a, b G. We need to prove that < is indeed an ordering relation. Let a, b, c G. Then (a) either a < b, b < a or a = b, (b) if a < b then ac < bc and ca < cb, (c) if a < b and b < c then a < c. This is true, since: (a) From the decomposition in Equation 4.1 we have that either ab 1 S, then a < b, or ab 1 = 1, then a = b, or ab 1 S 1, then ba 1 S and hence b < a. (b) Since a < b, we have that ab 1 S and hence ab 1 = acc 1 b 1 = (ac)(bc) 1 S, yielding that ac < bc. On the other hand, if ab 1 S, then, since S is normal, b 1 a = a 1 ab 1 a S and if b 1 a S, then ab 1 = b(b 1 a)b 1 S. So it suffices to show that (ca) 1 (cb) S. This is true since (ca) 1 (cb) = a 1 c 1 cb = a 1 b S by the above. (c) If a < b and b < c, then ab 1 S and bc 1 S, yielding that ac 1 = ab 1 bc 1 S, showing that a < c. Note 4.1: If G is an ordered group, it is possible that an operation + is also defined on G G. But even if not, the relation a + b := max(a, b) satisfies the usual requirements (associativity, commutativity, distributivity) of an additive operation. This (somewhat unnatural) addition will be needed in Definition Example 4.2: Let p be a prime in Z and let (G, ) := ({ p n n Z }, ) (Q \ {0}, ) and G := G {0}. 67

71 4.1. ABSOLUTE VALUES Then G is an ordered group with the ordering relation p n < p m n < m since G = { p n } { } n N {1} p n n N ; furthermore we can use the addition p n + p m := p max(m,n). Now we can state the definition of an absolute value: Definition Absolute Values (a) Let K be a field and G be a (multiplicatively written) ordered Abelian group with addition. Assume that G is extended to G by adjoining an element 0 with the following properties: g 0 = 0 g = 0 0 = g = g + 0 = g, = 0 and 0 < g for all g G. Then an (archimedian) absolute value on K with values in G is a function : K G where (1) a = 0 a = 0 (2) a b = a b (3) a + b a + b. (b) If satisfies (3 ) a + b max( a, b ) instead of (3), it is called non-archimedian. Example 4.3: (a) Let G and G be as in Example 4.2. Then every non-zero element x Q can be written as x = u v pn 68

72 CHAPTER 4. AN INTRODUCTION TO VALUATION THEORY for some u, v Z \ {0} and n Z such that p u and p v. Then the function Q G { p : p n if x 0 x 0 if x = 0 is a non-archimedian absolute value on Q and is called the p-adic absolute value. (b) Let K be a number field with ring of integers O α and let p be a fractional prime ideal of O α in K. Then we can define an absolute value on K by K Q { p : N (p) r if x 0 x 0 if x = 0 where, for x 0, x p r but x p r+1. Furthermore, N is the absolute norm of p (see Definition ). p is called the p-adic absolute value. Before we move on to places, there is a nice result worth mentioning, all the more since we will make use of it later: Theorem Completion of a Field Let K be a field with absolute value. Then K := CS(K)/ NS(K) is a field where (a) there is a subfield k K such that k = K, (b) k is dense in K and (c) K is complete with respect to a valuation on K extending. Here, CS(K) is the set of all Cauchy sequences and NS(K) the set of all null sequences in K with respect to. Furthermore, CS(K) is a ring and NS(K) is a maximal ideal in CS(K). Since the proof is essentially the same as for the completion of Q with respect to the modulus-function, it shall not be given here but can be found in [3, p ]. 69

73 4.2. PLACES 4.2 Places Before the concept of valuations is introduced, a certain generalisation is helpful. The notion of places is a good foundation for understanding the main theorem of this chapter, Theorem To this end, some definitions and basic results. Definition Places Let K and L be fields and let R be a subring of K such that (1) x K \ R 1 x R. A ring homomorphism P : R L is called a place of K if (2) x K \ R P ( 1 x) = 0 and (3) P(x) 0 for some x R. Commonly, R is also denoted by K P. The place P can be extended to all of K by defining (4) P(x) := for all x K \ R. K P is called the valuation ring of P. If P is an isomorphism of K, we say that it is a trivial place. Some simple results: Proposition Let P and P be places on a field K into L and L, respectively. Then (a) P := P (K P ) is a subfield of L. It is called the residue field of P. (b) m P := kerp is the only maximal ideal of K P. It contains all non-units of K P. (c) Two places P and P with residue fields P and P, respectively, are said to be isomorphic if there exists an isomorphism ψ : P P such that P = ψ P. We write P P. P and P are isomorphic if and only if K P = K P. (a) Obvious. (b) m P obviously is an ideal. Now, let x 0 be a non-unit of K P. Then 1 x K\R and, by Definition 4.2.1, (2), P(x) = 0, i.e. x m P. On the other hand, if 70

74 CHAPTER 4. AN INTRODUCTION TO VALUATION THEORY n m P, it must contain at least one unit of K P, say a. But then 1 a K P and hence 1 = 1 a a n, establishing that n = K P. Hence, m P is maximal. Now assume that there is another maximal ideal m. Then, by the above, m cannot contain units of K P. This shows that m m P, a contradiction. (c) It is obvious that P P implies that K P = K P and, by (b), m P = m P. On the other hand, let φ be the canonical homomorphism of K P onto K P /m P. Then φ P 1 is an isomorphism of P onto K P /m P and, similarily, φ (P ) 1 is an isomorphism of P onto K P /m P. Hence, ( ψ := P P 1 = φ (P ) 1) 1 ( ) φ P 1 is an isomorphism of P onto P, establishing that P and P are indeed isomorphic. It turns out that every subring of a field satisfying Definition 4.2.1, (1) implies the existence of a place on this subring. Theorem Existence of Places Let K be a field and R a subring of K satisfying Definition 4.2.1, (1). Then there exists a place P of K such that K P = R. Let m be the set of all non-units of R. Let x, y m and a R. Then: ax cannot be invertible. Assume it were, with b being its inverse. Then 1 = b ax = (ba) x. However, this requires that ba = x 1, i.e. that x is invertible in R, a contradiction. If x = 0 or y = 0 it is obvious that, by the above, x y = x m or x y = 1 y m. Hence, we can assume that x, y 0. Now, without loss of generality, let y x R. Then x y = x ( 1 y ) x and since 1 y x R and x m we see, by the above, that x y m. This shows that m is an ideal in R. Furthermore, it is obvious that m is maximal. Now let P be the canonical homomorphism of R onto the field R/m. Then, obviously, P is a place with K P = R. By this theorem we see that, in fact, places are common and the existence of places is confirmed. Definition Finite Places Let K be a field and R be an integral domain contained in K. (a) We say that a place P of K is finite on R if R K P. 71

75 4.2. PLACES (b) Let P be a place of K, finite on R. Then the prime ideal c P,R := m P R of R is called the centre of P in R. An example: Example 4.4: Let A be a Dedekind Domain, K := quot (A). Let p be a nonzero prime ideal of A and let f be the homomorphic mapping of A onto A/p, i.e. f(a) = a+p. Extend f to g :A p quot (A/p) by setting g ( ) a b := a+p b+p = f(a) f(b). Then A p is a principal ideal domain (see Theorem 3.3.8). This yields that g is a place of K with K g = A p. We need to prove conditions (1) through (3) of Definition 4.2.1: (1) Let x K \ A p. Claim: We can select a, b A such that a p, b p and x = a b. Assume not. Then there is an x K \ A p such that x = a b where a, b p. By Corollary 3.3.9, there are units u 1, u 2 of A p, an m A p and positive integers n 1, n 2 such that a 1 = u 1m n1, b 1 = u 2m n2. But then u1 u 2 is a unit in A p and hence, Furthermore, since m A p, m = m1 m 2 K, a a b = 1 = v 1 m n1 v 2 m n2 b 1 1 mn2 2 1 mn1 2 Now, if n 1 n 2, we have that a b = v 1m n1 n2 1. v 2 m n1 n2 2. u 1 u 2 = v1 v 2 for some v 1, v 2 A \ p. for some m 1 A, m 2 A \ p. Then, in But since A \ p is a multiplicatively closed set, v 2 m n1 n2 2 A \ p and hence, x A p, a contradiction. On the other hand, if n 1 < n 2, we have a b = v 1m n2 n1 2. v 2 m n2 n1 1 Again, since A \ p is a multiplicatively closed set, v 1 m n2 n1 2 A \ p, a contradiction. But this yields that 1 x = b a A p. 72

76 CHAPTER 4. AN INTRODUCTION TO VALUATION THEORY (2) Let x K \ A p. Then, as above, g ( ) 1 x = b+p a+p = 0+p 1+p. (3) 1 1 A p. Then g ( ) 1 1 = 1+p 1+p 0. g is called the p-adic place of A and is denoted by P p. We now proceed to the main result of this section, a characterisation of finite places on the quotient fields of Dedekind domains. Theorem Finite Places on Dedekind Domains Let A be a Dedekind domain and let K := quot (A). If P is a non-trivial place of K, finite on A, then P is a p-adic place of A. Let P be such a place. Then the centre of P in A is a proper prime ideal p in A. It is obvious that K P A p. Since A p = K Pp, we only need to show by Proposition 4.2.2, (c) that K P A p. By Corollary we know that there exists an m A p such that for all x A p there is a unit u A p and an n N 0 such that x = um n. (4.3) Now, let x K. Then x = a p for some a A p and p p. Since p can be embedded into A p we see that both a and p can be written in the form of Equation 4.3, i.e. a = u 1 m n1, p = u 2 m n2, yielding x = u1 u 2 m n1 n2 = vm s where v is a unit in A p and s Z. Let S be a subring of K properly containing A p. Then there must be an element x S such that x = um r with u a unit in A p and r < 0. Then m 1 = (m r 1 u 1 )(um r ) S. This yields that m n S for all n N, establishing that for all s Z and all units v A p we have vm s S. Hence, S = K. This shows that K P cannot be a proper superring of A p and hence K P = A p, completing the proof. 4.3 Valuations In a very similar way to the absolute value, we can define valuations. Definition Valuations Let K be a field and Γ be an (additively written) ordered Abelian group. Assume that Γ is extended to Γ by adjoining an element with the following properties: γ + = + γ = + = for all γ Γ. 73

77 4.3. VALUATIONS Then a valuation on K with values in Γ is a function v : K Γ where (1) v(a) = a = 0 (2) v(a b) = v(a) + v(b) (3) v(a + b) min(v(a), v(b)). If Γ = Z we say that v is a discrete, if Γ R, that it is a real valuation. If v(x) = 0 for all x K we say that v is a trivial valuation. Note 4.5: We will now establish the connection between absolute values and valuations. Let K be a field. (a) Let be a non-archimedian absolute value on K with values on an ordered group G. Assume that G is order-isomorphic to an additive subgroup G of R by means of some φ : G G R. Then induces a discrete valuation v : K R by defining v(a) := log (φ( a )) for a K and v(0) :=. We need to show that v satisifies the conditions (1) through (3) of Definition Let a, b K. Then: (1) is obvious (2) multiplicativity: v(ab) = log (φ( ab )) = log (φ( a b )) = = [ log (φ( a )) + log (φ( b )) ] = v(a) + v(b) (3) additivity: v(a + b) = log (φ( a + b )) log (φ (max ( a, b ))) = = log (max(φ( a ), φ( b )) = = max (log φ( a ), log φ( b )) = = min [ log (φ ( a )), log (φ ( b )) ] = min (v(a), v(b)) 74

78 CHAPTER 4. AN INTRODUCTION TO VALUATION THEORY (b) Vice versa, if v is a valuation on K with values on an ordered group G, G being order-isomorphic to a multiplicative subgroup G of R by means of some ψ : G G R, v induces a non-archimedian absolute value : K R. To this end, select a c G, c > 1, and define a := c ψ(v(a)) for a K and 0 := 0. Then satisfies the conditions (1) through (3) of Definition a, b K. Then: Let (1) is obvious (2) multiplicativity: (3) additivity: ab = c ψ(v(ab)) = c ψ(v(a)+v(b)) = c ψ(v(a)) c ψ(v(b)) = a b a + b = c ψ(v(a+b)) c ψ(min(v(a),v(b))) = c min(ψ(v(a)),ψ(v(b))) ( max c ψ(v(a)), c ψ(v(b))) = max( a, b ). There are some immediate consequences of Definition 4.3.1: Corollary Let v be a valuation on K. Then we see that: (a) v(k ) =: Γ v is a subgroup of Γ. It is called the value group of v. (b) For all a K, v(a) = v( a) and v(1) = 0. (c) If v(a) < v(b), then v(a + b) = v(a) for a, b K. (d) O v := { a K } v(a) 0 is a subring of K and is called the valuation ring of v. (e) O v has exactly one maximal ideal, m v = { a K } v(a) > 0 0. (a), (b), (d) are obvious. (c): We know that v(a + b) min(v(a), v(b)) = v(a). Now assume that v(a + b) > v(a). 75

79 4.3. VALUATIONS Then v(a) = v(a + b + ( b) min(v(a + b), v( b)) = min(v(a + b), v(b)) = v(b), a contradiction. (e): Let m be as above. Assume m is not maximal. Then there is an ideal n such that O v n m. Since m contains all non-units, n must contain at least one unit. But this requires n = O v, a contradiction. On the other hand, 0 m O v since 1 m and v(k ) 0 by definition. Now assume that there is another maximal ideal, say m. Since m cannot contain units, we must have m m, a contradiction. This shows that O v is local. Example 4.6: Let K be a field and let R := K((X)) be the field of formal power series over K, i.e. K((X)) = a n X n a i K, n 0 Z, a n0 0. n n 0 Then we can define a valuation v on K((X)) by setting K((X)) Z { v := 0 if f = 0. f otherwise n 0 Then v is a discrete valuation and the valuation ring of v is K[[X]], the ring of formal power series with non-negative exponents over K. See [31, Ch. I, Ex. 2]. Note 4.7: It is not coincidental that O v is called valuation ring (cf. Definition 4.2.1): if x K, then either v(x) 0 or v(x) < 0. Now, if v(x) < 0, this shows that x O v but since 0 = v(1) = v ( ( x x) 1 = v(x) + v 1 x) we see that v ( ) 1 x = v(x) > 0, establishing that 1 x O v. Hence, every valuation v on K, by Theorem 4.2.3, gives birth to a whole equivalence class of places on K. In fact, the converse of this is true as well: Theorem Valuation induced by a place Let P be a place on a field K with valuation ring K P. Then there exists a valuation v on K such that O v = K P. The theorem is obvious if P is trivial: take the trivial valuation. Hence, let P be a non-trivial place on K. Let E := K P \ m P, the set of units in K P. Then E is a subgroup of (K, ). Now let v be the canonical homomorphism of K onto 76

80 CHAPTER 4. AN INTRODUCTION TO VALUATION THEORY the quotient group Γ := K /E. If we write the group operation on Γ additively, we see that Definition 4.3.1, (b) is satisfied for v. Now, let Γ + := v (m P \ {0}). If x Γ +, there exists an element u m P such that v(u) = x. If now w v 1 (x) we see that ue = we, hence w u E. Since m P is an ideal of K P and E K P we have w = w u u m P, establishing that v 1 (Γ + ) = m P \ {0}. Since m P is closed with respect to multiplication, Γ + is closed with respect to addition and so, because of the commutativity of K, is a normal subsemigroup of Γ. Now, let α Γ and x K such that v(x) = α. Then, if α Γ +, x m P and 1 x m P, whence α = 0 α = v(1) v(x) = v ( 1 x) Γ+. On the other hand, if α Γ +, α 0, we have that x m P and x E, whence x K P. And, lastly, if α = 0, we have that α Γ +, α m P but α E. This shows that Γ = Γ + {0} (Γ + ) 1, where (Γ + ) 1 = { α α Γ + }, i.e. Γ is an ordered group, see Definition It remains to show that v satisfies condition (c) of Definition It suffices to show that, for x, y K and v(x) v(y), v ( 1 + y x) 0 since this is equivalent to v(x + y) min(v(x), v(y)). But this is obvious since v(x) v(y) implies that y x K P, whence 1 + y x K P. Condition (1) of Definition can be satisfied by adjoining to Γ, setting v(0) := and properly extending the group operations. Since, due to the construction, K P = O v, we have that v indeed is a valuation. Definition Equivalence of two valuations Two valuations v and w are said to be equivalent if there is an order-preserving isomorphism φ between the value groups of v and w such that w = φ v. We write w v. If A is a subring of K, we say that v is non-negative on A if A O v. It is immediate that two equivalent valuations have the same valuation ring. The converse is true as well: Proposition Let v and v be two valuations on a field K with valuation ring R := O v = O v. Then v v. Let Γ and Γ be the value groups of v and v, respectively. Since v and v have the same kernel, namely the set of units of R, the map φ := v v 1 is an isomorphism of Γ onto Γ. Obviously, if v(a) > 0 for some a R, then v (a) > 0 as well since a must be a non-unit of R. If v(a) > v(b), then v ( ) a b > 0 and hence φ(v(a)) φ(v(b)) = φ ( v ( )) ( ) a b = v a b > 0, yielding that φ is an order-preserving 77

81 4.3. VALUATIONS isomorphism. This completes the proof. The following is a standard example: Example 4.8: (a) For any prime p Z and all 0 x Q there is an n Z and a a b Q with p and a, b relatively prime such that x = u p n. (4.4) Then we can define a valuation v p : { Q Z x n, v p(0) := on Q. This valuation is called the p-adic valuation on Q. It turns out that O vp = Z p, the localisation of Z with respect to the principal (prime) ideal generated by p. In fact, v p is the valuation induced by p from Example 4.3, (a), by means of log p. (b) If K is a number field, p from Example 4.3, (b), induces a valuation v p by means of log N (p), the p-adic valuation on K. If R is a Dedekind domain and K is its quotient field, we can extend the notion of the p-adic valuation on K. Let p be any non-zero prime ideal in R. We can now define a valuation v p on K. Let x K. Then, by Theorem 3.3.3, the ideal Rx has a unique factorisation Rx = q q nq(rx), where q runs through the prime ideals of R. Then define v p (x) := { np (Rx) if x 0 if x = 0. Then v p is a valuation (conditions (1) through (3) of Definition need to be checked): (1) This is obvious from the definition of v p (x). (2) v p (x y) = n p (Rxy) = n p (RxRy) = n p (Rx) + n p (Ry) = v p (x) + v p (y), by Corollary

82 CHAPTER 4. AN INTRODUCTION TO VALUATION THEORY (3) v p (x + y) = n p (R(x + y)) ( ) n p (Rx + Ry) = min(n p (Rx), n p (Ry)) = min(v p (x), v p (y)), by Corollary ( ) holds since R(x+y) Rx+Ry, whence, by Proposition 3.3.4, (Rx+ Ry) R(x + y), which, in turn, implies that n p (R(x + y)) n p (Rx + Ry). This shows that v p is indeed a valuation, it is called the p-adic valuation. Its valuation ring O vp obviously is R p. Furthermore, if K is a number field, the two definitions of the p-adic valuations are equivalent by Proposition Note 4.9: Since the modulus function : Q Q + 0 induce a valuation. is archimedian, it does not A nice classification for all non-negative valuations on the quotient fields of Dedekind domains is available: Theorem Valuations on the quotient field of a Dedekind domain Let R be a Dedekind domain and let K := quot (R). Then, if v is a valuation on K, non-negative on R, there exists a prime ideal p R such that v is equivalent to the p-adic valuation on K. Furthermore, the value group of v is order-isomorphic to the additive group of integers. Let v be a valuation on K, non-negative on R. Let O v be the valuation ring of v. Then, by Note 4.7 and Theorem 4.2.5, there is a prime ideal p R such that O v = R p, whence, by Proposition 4.3.5, v is equivalent to v p, completing the proof. Looking at the Dedekind domain Z and its field of quotients Q we see that all nonnegative, non-trivial valuations on Q must be p -adic (this is the same result as in Example 4.8, (a), obtained in a different way). It turns out that there is a not-obvious equivalent to a ring being a valuation domain. Theorem Valuation and Chain Rings Let R be an integral domain. Then R is a valuation ring if and only if it is a chain ring. : 79

83 4.3. VALUATIONS Let a, b be two ideals of R such that a b and b a. Then, there is an x a and an y b such that x Ry and y Rx. But since at least one of x y and y x must lie in R, this is a contradiction. Hence, the ideals in R form a chain with respect to inclusion. : Let K = quot (R) and let x y that either x Ry or y Rx, whence one of x y valuation ring. K. Then, either Rx Ry or Ry Rx, implying or y x must lie in R. Thus R is a Finally, we can extend the concept of completion of a field with respect to an absolute value to valuations. Definition Completion with respect to a valuation Let v be a valuation on some field K. Then the completion K of K with respect to v is defined as the completion of K with respect to the absolute value induced by v according to Note 4.5, (b). If v v p, we also write K p and say that K p is the p-adic completion of K. Sometimes it is necessary to look at p-adic completions of subrings of K only. That this is indeed possible shows the next Theorem. Theorem Completion of a Dedekind Domain Let R be a Dedekind domain with quotient field K, let v be a valuation on K, non-negative on R. Then there is a prime ideal p of R such that the set R p := CS(R)/ NS(R) is complete with respect to v. R p is called the p-adic completion of R. By Theorem there must be a prime ideal p of R such that v v p. Then R p is a subset of K and, by an argument similar to that of Theorem 4.1.4, is complete with respect to the absolute value p induced by v p and contains a dense subring isomorphic to R. Hence, by Definition 4.3.8, R p is complete with respect to v. Here, CS(R) and NS(R) are the Cauchy and null sequences on R with respect to p. For some purposes, an equivalent description of the p-adic closure is more convenient. To this end, we require a definition. 80

84 CHAPTER 4. AN INTRODUCTION TO VALUATION THEORY Definition Inverse Systems and Projective Limit Let (I, ) be a poset (see Definition A.1), let (A i ) i I be a family of groups (or sets, rings, modules over a fixed ring, algebras over a fixed field, etc.) and let (f ij : A j A i ) i j, i,j I be a family of homomorphisms such that f ii = id Ai and f ik = f ij f jk for all i j k. Then the set of pairs (A i, f ij ) is called an inverse system over I. Then the inverse limit or projective limit is defined as { } lim A i := (a i ) i I A i a i = f ij (a j ) for all i j Obviously, lim A i is an object of the same category as the A i s and a subgroup (-set, -ring, -module, -algebra, etc.) of the direct product of the A i s. For further information on inverse systems and projective limits, see [2, Ch. 10].. This now allows us to show the aforementioned equivalent characterisation. Theorem Completion of a Dedekind domain II Let R be a Dedekind domain with quotient field K, let v be a valuation on K, non-negative on R. Then there is a prime ideal p of R such that v v p and / R p = lim R p n. The existence of p and of R p follows from Theorem Now, let CS(R) be the set of Cauchy sequences and NS(R) be the set of null sequences on R with respect to p, the p-adic absolute value on R induced by v p. Then we note that the ideals p n, n N, can be seen as p n = { x R vp (x) n }. Now, if (x n ) n N is a Cauchy sequence in R, we have that for all ε > 0 there is an index N(ε) N such that x n1 x n2 p < ε for all n 1, n 2 N(ε), which is equivalent to there being an M(m) N for all m N such that v p (x n1 x n2 ) m for all n 1, n 2 M(m), which, in turn, is equivalent to x n1 x n2 p m for all n 1, n 2 M(m). 81

85 4.3. VALUATIONS This shows that for all m N, x n x M(m) (mod p m ) for all n M(m), establishing that the sequences (x n + p m ) n N, m N, eventually become stationary (at M(m) or earlier). Furthermore, it is clear that ( / ) R p m, f mn, m, n N, m n, is an inverse system for R / f mn : p n R/ p m. x + p n x + p m Now we can define a map CS(R) R / p Φ : m m N (x n ) n N ( x M(1) + p, x M(2) + p 2, x M(3) + p 3,... ). It is clear that Φ is a ring homomorphism, that the image of Φ is the projective limit of R/p m and that the kernel of Φ is NS(R), i.e. / Im(Φ) = lim R m p and Ker(Φ) = NS(R). This, finally, shows that R p = CS(R)/ / NS(R) = lim R p m, completing the proof. It would now be interesting to actually determine which elements lie in R p. From the definition it is clear that R R p. But, and that is more noteworthy, it turns out that R p R p. To see this, let b R \ p, i.e. 1 b R p. Then we want to construct a sequence (s i ) i N0 such that ( v p s i 1 ) for i. (4.5) b Now consider the polynomial q(x) = bx 1. It is clear that a sequence (s i ) satisfies Equation 4.5 if and only if it satisfies v p (q(s i )) for i. (4.6) Now, we construct such a sequence. Since b p and R is Dedekind, R/p is a field and b + p is invertible in R/p. Let c + p be the inverse of b + p. Then q(c) bc (mod p), 82

86 CHAPTER 4. AN INTRODUCTION TO VALUATION THEORY establishing that q(c) p. Then, by Note 3.6, there is a unit ε R p and an element t p, t p 2, such that q(c) = εt j for some j N. This shows that v p (q(c)) = j. Let s 0 := c. If we now define s i := s i 1 (1 q (s i 1 )), i 1, we see that q(s i ) = bs i 1 (1 q (s i 1 )) 1 = bs i 1 1 bs i 1 q(s i 1 ) = = q(s i 1 (1 bs i 1 ) = q(s i 1 ) 2 = q(s 0 ) 2i = ε 2i t 2ij, establishing that v p (q(s i )) = 2 i j, proving that the sequence we just constructed indeed satisfies Equation 4.6. Hence, R p R p. This why the notation R p for R p is used quite often. As an example, Z p contains all elements a b from Q such that p b. If p = 5, a = 1 and b = 3, the first few elements of (s i ) are: s 0 = 2 s 1 = 8 s 2 = 208 s 3 = s 4 = s 5 = s 6 = But this is not all: it can be shown that several elements which are algebraic over Q are contained in Z p, an example is e 2πi/(p 1) for p > 2. For p = 2, it is known that a Z p if and only if 2 a, 4 a and a 1 (mod 8). As an example, 17 Z 2. Similarly, it can be shown that R p is properly contained in R p. More information can be found in [3] and [38]. 83

87 Chapter 5 Skolem Rings We return to the quest for Prüfer rings. To this end, the class of rings found by T. Skolem in 1936 (see [32]) proves to be an interesting source of examples. In this chapter, we bring together the results Skolem found in 1936, include generalisations by D. Brizolis (see [6], [7] and [8]), move to the core result (Theorem 5.3.8) of this chapter, also by Brizolis, and finish by mentioning recent results in this field by Cahen, Chabert and Loper (see [9] and [21]). We assume all rings to be commutative. 5.1 A-Valued Polynomials The whole concept of Skolem rings is based on polynomial rings with a certain invariance property, as the next definition shows. Definition A-valued polynomials Let A be a ring and let K := quot (A) be its total ring of quotients. Then we say that a polynomial f K [X 1,..., X n ] is A-valued if f(x) A for all x A n. The ring of all A-valued polynomials in n variables is denoted by Int (A, n). Of course, it is clear that the ring of integer-valued polynomials, Int (Z, 1), must be a first example. And, in fact, it is already an important one. As far back as the 17th century, integer-valued polynomials have been used for interpolation purposes. We find that, of course, Z[X] Int (Z, 1). But then? Is there anything else? As it turns out, there is quite a bit! It is clear that the binomial coefficient ( n k) is an integer for all n and k in N. We define the polynomials f n (X) := ( ) X := n X(X 1) (X n + 1), n N; f 0 (X) := n! ( ) X :=

88 CHAPTER 5. SKOLEM RINGS Then, obviously, they do not have integer coefficients. We will now show that they are integer-valued. Let n N, k Z. Then there are three cases: 0 k < n: n k: k < 0 : f n (k) = 0 since the factor X k occurs in the numerator of f n. f n (k) = ( k n) N. f n (k) = k(k 1) (k n+1) n! = = ( 1) n ( k+(n 1))( k+(n 1) 1) ( k+1)( k) n! = = ( 1) n f n ( k + (n 1)), hence f n (k) Z since k + (n 1) n. It is immediate that the polynomials ( ) X, n N 0, n form a basis of the Q-module Q[X] since there is exactly one such polynomial of each degree. Furthermore, we see that they also form a basis for the Z-module Int (Z, 1): let f Int (Z, 1). Then there are rational numbers a 0, a 1,..., a n such that f(x) = a 0 + a 1 ( X 1 ) ( ) X a n. n Obviously, a 0 = f(0) Z. We proceed by induction. Now, assume that for i = 0, 1,..., k, k < n, the coefficients a i are integers. Then the polynomial ( ) ( ) ( ) ( ) X X X X g(x) = f(x) a 0 a 1... a k = a k a n 1 k k + 1 n is integer-valued as well. Hence, a k+1 = g(k + 1) Z, establishing that all coefficients of f are integers, whence Int (Z, 1) = ( ) X Z. n n N 0 This result can, in a natural way, be extended to the ring Int (Z, n). Proposition Bases for the Rings of integer-valued Polynomials The polynomials ( ) X1 n 1 ( Xk n k ), n 1,..., n k N 0, form a basis for the ring Int (Z, k). Furthermore, Int (Z, k) = ) ( ) Xk. n 1 N 0 ( X1 Z n 1 n k N 0 85 n k

89 5.1. A-VALUED POLYNOMIALS See [9, Th. XI.1.12]. It is interesting to note that although Z is Noetherian, Int (Z, 1) is not: Proposition The Ring Int (Z, 1) is not Noetherian. Assume it were. Consider the ideal m := {f Int (Z, 1) f(0) 2Z}. Assume that m is finitely generated, say by the polynomials g 1,..., g s. Select k such that 2 k+1 does not divide any of the denominators of any of the coefficients of the g i. Then, obviously, 2 k g i Z 2 [X] for all i = 1,..., s. We thus can write 2 k g i (X) = n i j=0 a ij b ij X j, a ij Z, b ij Z \ 2. If we define B i := b i1... b ini and B ij := B i /b ij for all i, j, we see that g i (2 k+1 ) = g i (0) + 1 B i n i j=1 a ij B ij 2 j(k+1) k. Then, since g i Int (Z, 1), B i must divide the sum. But since all summands are even and 2 B i, this quotient must be even, too. And since g i m we have that g i (0) is even. This establishes that g i (2 k+1 ) is even for all i = 1,..., s. Now, let f m. Then there are polynomials a 1,..., a s Int (Z, 1) such that f = a 1 g a s g s. Therefore, f(2 k+1 ) is even. It is clear that ( ) ( ) X 2 k+1 2 k+1 m and 2 k+1 = 1. By the above, this is a contradiction, showing that Int (Z, 1) is not Noetherian. Again, this result can be stated in a more general way. Theorem Noetherian properties Let A be an infinite integral domain and B be its integral closure in quot (A). Then, if Int (A, 1) is Noetherian, Int (A, 1) B[X]. If A is a normal Noetherian domain, the following are equivalent: 86

90 CHAPTER 5. SKOLEM RINGS (a) Int (A, 1) is Noetherian (b) A/p is infinite for all prime ideals p of A which are maximal (c) Int (A, 1) = A[X]. See [9, Prop. VI.2.4] and [9, Cor. VI.2.6]. But, as it turns out, at least a little of the Noetherianness of Z can be transferred to Int (Z, 1). Proposition ACCP on Int (A, 1). Let A be an infinite commutative ring. Then A has the ACCP if and only if Int (A, 1) has the ACCP. See [9, Prop. VI.2.9]. We can readily proceed to a first result, providing some information about the structure of (certain) maximal ideals in rings of A-valued polynomials: Proposition Let A be a Dedekind domain and A R Int (A, n). Then, if p is a non-zero prime ideal of A, m(r, α, p) := {f R such that v p (f(α)) > 0} ( is a maximal ideal of R for all α A p) n. First, we note some facts: (a) A is dense in A p. (b) For all f R, f is continuous on A n. (c) For all f R, f (A n ) A A p. ( (A p) ) n Hence, f A p. Let m := { } β A p such that v p (β) > 0 87

91 5.2. SKOLEM S THEOREM be the unique maximal ideal of A p (in fact, m = m vp A p ). Then the homomorphism R A p/ φ : m f f(α) (mod m) is onto and well-defined with kernel m(r, α, p). Since m is maximal, Im(φ) is a field and hence m(r, α, p) must be a maximal ideal of R. Note that in the proof of [7, Th ], Brizolis shows that for α β, the ideals m(r, α, p) and m(r, β, p) are distinct. Finally, as a fundament for further results to be shown later, we proceed to a proposition. Proposition Let A be a Dedekind domain and p a prime ideal of A. Then, for V := {f Int (A, n) such that f (A n ) p}, there is some π p 1 such that v p (π) < 0 and πf Int (A, n) for all f V. We proceed in several steps. First, note that p 1 A: assume it were. Then we would have A = pp 1 = pa p, a contradiction. Now let π p 1 \ A. Then v p (π) 0, since for v p (π) > 0 we see that π p A, a contradiction. Now, if v p (π) = 0, v p (tπ) = v p (t) > 0 for all t p. Hence, tπ p for all t p. But this shows that πp p and subsequently that π pp 1 = A, a contradiction. So v p (π) < 0. Now, for any f V, the relation f (A n ) p shows that p 1 f (A n ) A, establishing that (tf) (A n ) A for all t p 1. But, since tf K[X], we have that tf Int (A, n) for all t p 1 and so πf Int (A, n), completing the proof. 5.2 Skolem s Theorem In this section, we introduce an intriguing, yet not widely known, fundamental result by Thoralf Skolem. It was first published in 1936 in [32]. Theorem Skolem s Theorem Let f 1,..., f n be Z-valued polynomials in m variables. If gcd (f 1 (x),..., f n (x)) = 1 for all x Z m, (5.1) 88

92 CHAPTER 5. SKOLEM RINGS there exist Z-valued polynomials F 1,..., F n such that f 1 F f n F n = 1 holds. This theorem can be re-phrased in terms of ideal theory: Corollary If a is a finitely generated ideal in Int (Z, m) where a(x) := {f(x) f a} = Z for all x Z m, then a = Int (Z, m). Here, it is clear that a(x) is an ideal of Z. It is called the ideal of values of a. Before we are able to prove this theorem, we need some intermediate results. Lemma Let p be a rational prime and let f 1,..., f n Int (Z, m) such that Equation 5.1 holds. Let j 1 A 1 := 1, A j := s=1 for j = 2,..., n and let A := ( f p 1 j ) fs p 1 n f j A j. (5.2) j=1 Then we see that A(x) 0 (mod p). Let x = (x 1,..., x m ) Z m. There are three cases: (a) f j (x) 0 (mod p) for all j Then A j (x) 0 (mod p) for all j > 1 and A 1 (x) 1 (mod p). Hence, A(x) f 1 (x) 0 (mod p). (b) There is a k N, k < n, such that k of the f j (x) are congruent to 0 (mod p). Let V := {ν 1,..., ν k } {1,..., n} such that f j (x) 0 (mod p) for all j V. Then, setting j 0 := min ({1,..., n} \ V ), we have that f j (x)a j (x) 0 (mod p) for all j V 89

93 5.2. SKOLEM S THEOREM and, for j V, that A j (x) j 1 f j(x) p 1 f s (x) p 1 }{{}}{{} s=1 1 ( ) 0 s V 1 s V { 1 for j = j0 (mod p). 0 for j > j 0 whence A(x) f j0 (x) 0 (mod p). Note that in ( ), we make use of Fermat s Little Theorem, see Theorem (c) f j (x) 0 (mod p) for all 1 j n This is not possible since it would be a contradiction to Equation 5.1. This shows that the assertion holds for all x Z m, completing the proof. Lemma Let f 1,..., f n Int (Z, m). If there are algebraic numbers ξ 1,..., ξ m over Q, ξ := (ξ 1,..., ξ m ), such that f 1 (ξ) = = f n (ξ) = 0, then there are rational integers x 1,..., x m, x = (x 1,..., x m ) and infinitely many rational primes p such that f 1 (x) f n (x) 0 (mod p). Let Q(α) be the smallest algebraic field extension containing the ξ 1,..., ξ m and let O α be the Number Ring of Q(α). Then there are infinitely many prime ideals of degree 1 in O α (see Theorem 3.4.3). Select such an ideal p with the additional condition that the denominators if they are not 1 of the ξ i and the f j are coprime to p. This rules out only finitely many such ideals. Then by Theorem there exists exactly one rational prime p such that O α / p = Z p. (5.3) For all 1 i m there are a i, b i O α, b i 0, such that ξ i = ai b i. Then, by the above, there are non-zero x 1,..., x m Z such that a i b i x i (mod p). And, since the b i were required to be coprime to p, they are invertible in Z p and hence we can say that ξ i x i (mod p). Now let c j be the lcm of the denominators of f j. Again, by definition, the c j are coprime to p. If we now define f j := c j f j, we see that c j f j (x 1,..., x m ) = f j (x 1,..., x m ) f j (ξ 1,..., ξ m ) (mod p) for all j = 1,..., n. 90

94 CHAPTER 5. SKOLEM RINGS But since f j (ξ 1,..., ξ m ) = c j f j (ξ 1,..., ξ m ) = 0, we find that f j (x 1,..., x m ) 0 (mod p). Since the f j are Z-valued polynomials, we also see that f j (x 1,..., x m ) = c j e j = d j 0 (mod p) for some rational integers e j and d j. And because of Equation 5.3 we see that c j e j 0 (mod p) for all j, yielding, since the c j are coprime to p and hence must be invertible in Z p, that the e j 0 (mod p), completing the proof. Now we can proceed to the proof of Theorem 5.2.1: Assume that there exist algebraic numbers ξ 1,..., ξ m being simultaneous roots of the polynomials f 1,..., f n. Then, by Lemma 5.2.4, there must be rational integers x 1,..., x m and a prime p such that p gcd (f 1 (x),..., f n (x)). Since we required that gcd (f 1 (x),..., f n (x)) = 1 for all x Z m, this cannot be. Now, by a special case (see Corollary ) of Hilbert s Nullstellensatz (see Theorem ), there are polynomials P 1,..., P n Z[X 1,..., X m ] and a non-zero rational integer M such that f 1 P f n P n = M. If we assume that M = p α1 1 pαν ν where the p i are distinct rational primes and we define j 1 A i1 := 1, A ij := s=1 Lemma yields that ( f pi 1 j ) fs pi 1, A i := n f j A ij, i = 1,..., ν, j=1 A i (x) 0 (mod p i ) for all i = 1,..., ν and all x Z m. Hence, for M := p 1 p ν, the values g(x) of the polynomial g := M p 1 A M p ν A ν are relatively prime to M for all x Z m. Thus, by Theorem 3.4.9, g(x) φ(m) 1 0 (mod M) for all x, and therefore the polynomial h := gφ(m) 1 M, where φ is Euler s function, is Z-valued in m variables. If we now define ( M F r := A 1r M ) A νr g φ(m) 1 P r h, r = 1,..., n, p 1 p ν 91

95 5.3. GENERALISATIONS AND CONSEQUENCES we see that n f r F r = r=1 [ ( n ν f r M r=1 k=1 A kr p k = hm + g φ(m) 1 n = hm + g φ(m) 1 ν ) r=1 k=1 k=1 = g φ(m) g φ(m) 1 ν g φ(m) 1 f r P r h ] ν M f r A kr = p k ( n ) M f r A kr = p k k=1 r=1 = M p k A k = 1 g φ(m) + g φ(m) 1 g = 1, completing the proof. As an example of this, consider the two polynomials f = X 2 + 1, g = 3. It is clear that for all x Z, gcd(f(x), g(x)) = 1. Now, assume that Z[X] satisfies Theorem Then there must be polynomials F, G Z[X] such that F f + Gg = F (X 2 + 1) + 3G = 1. Consider (F f + Gg)(i) = 3G(i). This yields that that 3G(i) = 1, i.e. that 3 is a unit in Z[i], which is a contradiction. On the other hand, it is clear that ( ) X (X 2 6X + 10) (X 2 + 1) 8 3 = 1, 4 a consequence of Theorem This example comes from [32]. 5.3 Generalisations and Consequences The majority of this section is based on the work of Demetrios Brizolis, of UCLA. In his dissertation (which has partially been published, see [7]), he managed to generalise Skolem s result to certain other domains which will be introduced shortly. In one of his earlier publications, [6], he laid the foundation for the two core theorems, both stated in [8], one of which we will prove in due course. We endeavour to be as thorough as possible in our approach: only references to certain basic results from algebraic number theory shall be made and, wherever necessary, explained. The reader will note, though, that it is not possible to have the remainder of this thesis be self-contained. Due to the complexity of the matter we have to assume a certain familiarity with algebraic concepts. Definition D*-domains Let A be a ring. Then A is said to be a D*-domain if 92

96 CHAPTER 5. SKOLEM RINGS (a) A is a Dedekind domain, (b) the quotient field K of A has characteristic 0, (c) A/ p is finite1 for all non-zero prime ideals p in A, (d) for all f A[X], f(x) 0 (mod p) has a solution for infinitely many prime ideals p in A. Example 5.1: It is clear that Z is D*. Furthermore, for any algebraic number field K the ring O K is D* as well, see [7, Cor ]. But, from [7, p. 52] we also know that the question of characterising D*-domains remains unsolved. Furthermore, Brizolis shows that Q[X] X is a Dedekind domain which is not D*. Now, we rephrase the assertion of Theorem to Definition Skolem Rings Let A be a ring. We say that R := Int (A, m) is a Skolem Ring if for all finitely generated ideals i of R the relation i(x) = A for all x A m implies that i = R. This leads to the following theorem: Theorem Generalised Skolem s Theorem Let A be a D*-domain. Then Int (A, m) is a Skolem Ring. Before we can prove this result, we require an intermediate lemma. Lemma Let A be a Dedekind domain with property (d) of Definition Let K be the field of quotients of A, L a finite separable extension of K and B the integral closure of A in K. Then there are infinitely many prime ideals P of B such that [ / B P : A/ ] P A = 1. By Lemma there is an element α B such that L = K(α). Then, by [18, Ch. I, 4, Remark 2] and [18, Ch. III, 3, Prop. 16], B p = A p [α] for all but a finite number of prime ideals of A. Now let f be the minimal polynomial of α over A. Then, by hypothesis, there are infinitely many prime ideals of A such that 1 If m is a maximal ideal of A, A/m is called a residue field of A. 93

97 5.3. GENERALISATIONS AND CONSEQUENCES f(x) 0 (mod p) has a solution in A. This shows that there are infinitely many prime ideals satisfying both conditions. Now let p be such an ideal and let a be a solution of f(x) 0 (mod p). Then there must be an e 1 N and a polynomial P A[X], P (a) 0 (mod p) such that f(x) = (x a) e1 P (x), where denotes the reduction of elements modulo p. Then, if m p is the unique maximal ideal of A p, we have (by Note 2.8) that m p A = p. Furthermore, we can regard f and P as elements of A p [X] by means of the canonical embedding of A into A p. Then, reducing modulo m p, we see that P (a) 0 (mod m p ) and hence f(x) = (x a) e P (x) = (x a) e1 P 2 (x) e2 P r (x) er (5.4) where the P i, i = 2,..., r, are the irreducible factors of P in A p /m p. Let P 1 (x) := x a. It is clear that since B is integral over A, B p must be integral over A p. Now, let P be a prime ideal of B p. Then A p P must be a prime ideal of A p which, since A p is a Dedekind domain (see Theorem 3.3.8), must be maximal. Then, B p /P is a ring integral over the field A p /m p. From elementary field theory (see [11]) it is clear that B p /P must also be a field, yielding that P is a maximal ideal of B p. We furthermore note that quot (A p ) = quot (A) = K and that B p is the integral closure of A p in L. Then, by Equation 5.4 and [18, Ch. I, 8, Prop. 25] we see that m p B p = P e1 1 Pe2 2 Per r where the P i, i = 1,..., r, are by the above maximal ideals of B p. Furthermore, in the course of the proof of [18, loc. cit.] it is shown that [ / Bp P : A ] p i /m = deg P i, p i.e. [Bp/ P : A ] p 1 /m = 1. p Consequently, since A p /m p = A/p and Bp /P 1 = B/(B P1 ) by [15, Cor. 5.9], we see that [ / B B P : A/ ] 1 p = 1. It is clear that B P 1 is a prime ideal of B and lies above p: (B P 1 ) A = P 1 A = P 1 A p A = m p A = p This completes the proof. Now we can proceed to the proof of Theorem 5.3.3: 94

98 CHAPTER 5. SKOLEM RINGS Let f 1,..., f n Int (A, m) such that f 1 (a),..., f n (a) = A for all a A m. Assume that there are elements β 1,..., β m algebraic over K such that f 1 (β) = = f n (β) for β := (β 1,..., β m ). Let L := K(β 1,..., β m ) Then, by the Primitive Element Theorem (see Theorem 3.2.5), there is an α algebraic over K such that K(α) = K(β 1,..., β m ). Then, by Note 3.7, there are elements c i A, c i 0, such that the c i β i are integral over A, i = 1,..., m. Let B be the integral closure of A in L. By Lemma there exist infinitely many prime ideals P of B such that [ / B P : A/ ] P A = 1. (5.5) Now choose such a prime ideal P so that P is coprime to the c i and the denominators of the coefficients of the f i s. Since P A is prime in A, A/(P A) is finite. Now, if S = {r 1,..., r N } is a complete system of residues of A, we see that, by Equation 5.5 and since S A B, S also is a CSR of B. This shows that there are ν 1,..., ν m {1,..., N} such that r νi β i (mod P), i = 1,..., m. This shows that in turn 0 = f i (β) f i (r ν1,..., r νm ) (mod P) and so 0 f(r ν1,..., r νm ) (mod P A) for all i {1,..., n}, a contradiction. Now, by Corollary , we can find polynomials g 1,..., g n K[X 1,..., X m ] such that f 1 g f n g n = 1. Let M be the product of the denominators of the coefficients of the g i s. Then M A and M = p e1 1 pe pe k k for some distinct nonzero prime ideals p j, j = 1,..., k, of A. Furthermore, h i := Mg i A[X 1,..., X m ]. Now, if A/p j has λ j elements, by Theorem , b λj 1 1 (mod p j ) for all j = 1,..., k if A b 0 (mod p j ). Define the polynomials j 1 A i1 := 1, A ij := s=1 ( f λi 1 j ) fs λi 1 for i = 1,..., k and j = 1,..., n and let A i := n f j A ij. j=1 Then, by a similar argument to Lemma 5.2.3, we see that A i (x) 0 (mod p i ) for all x A m. By the Chinese Remainder Theorem, there are numbers m i, i = 1,..., n, such that m i 0 (mod p i ) and m i 1 (mod p j ) for i j. Define M to be m 1 m n. Then M m i A i (x) 0 (mod p i ) and M m j A j (x) 0 (mod p i ) for all x A m. Then, if we define g := M m 1 A M m k A k, we have that g(x) 0 (mod p i ) for all x A m, i = 1,..., k, 95

99 5.3. GENERALISATIONS AND CONSEQUENCES yielding that g(x) 0 (mod M ) for all x A m. Since, by hypothesis, A/p ei i Theorem 3.4.8, is finite for each i, A/ M must be finite. Hence, by g(x) ord((a/ M )\{0}) 1 (mod M ) for all x A. Let T := ord((a/ M ) \ {0}). Let ĝ := gt 1 M. Then we see that both g and ĝ are A-valued on A m by construction. Now, taking ( k ) M F j := A ij g T 1 + ĝh j, m i i=1 we have that the F j Int (A, m) for all j = 1,..., n and that n F j f j = j=1 = [( n k j=1 ( k i=1 i=1 M m i ) M A ij f j m i ] g T 1 + gt 1 M Mg jf j = ) n f j A ij g T 1 + ( g T 1 ) n g j f j = j=1 }{{} =A i } {{ } =g = gg T 1 + g T 1 = 1, j=1 } {{ } =1 completing the proof. Now, we wish to find out more about the properties of Int (A, 1). In turn, we require two more lemmas. Lemma Let A be a D*-domain, K its quotient field, L a finite separable extension of K and B the integral closure of A in L. Then there are infinitely many primes P in B such that A P A = B P. By Lemma we know that there are infinitely many primes such that B/P = A/(P A). Now, by Theorem 3.4.6, all but finitely many of these prime ideals are unramified over A, whence for all those primes A/(P A) i = B/P i and hence / / lim A (P A) i = lim B P i. 96

100 CHAPTER 5. SKOLEM RINGS But this shows (see Theorem ) that A P A = B P, as claimed. Lemma Let A be a D*-Domain. Let A R Int (A, n). Define { V a,p := α ( A p) } n vp (f(α)) > 0 f a = {α ( A p) } n a m(a, α, p) and V a,0 := { α A n f(α) = 0 for all f a } for any non-zero prime ideal p of A and a an ideal of R. If a A = 0 then there are infinitely many prime ideals p in A such that V a,p. Let K, L and B as before. Since a A = 0, ak is a proper ideal of K[X 1,..., X n ]. Let K be the algebraic closure of K. Then, by the weak Nullstellensatz (see Corollary ), there are α 1,..., α n K such that f(α 1,..., α n ) = 0 for all f ak. Let L = K(α 1,..., α n ). Since A is D*, K is perfect and hence L is a finite separable extension of K. Then, by Lemma 5.3.5, there are infinitely many prime ideals P of B such that A P A = B P. Now, by Note 3.7, there are elements c i A such that c i α i B for i = 1,..., n. Since B is Dedekind, for almost all prime ideals of B the elements c i P. Hence, ciαi c i B P and thus α := (α 1,..., α n ) (B P) n for almost all prime ideals P of B. By the above, there is an element α (A P A) n corresponding to α. Thus, f(α ) = 0, whence v P A (f(α )) = > 0, thus showing that α V a,p A, completing the proof. Now we can proceed to a characterisation of maximal ideals of Skolem rings. Theorem Maximal Ideals of Skolem Rings Let A be a D*-domain and let A R Int (A, n) be a Skolem ring. Then if m is a maximal ideal of R, there is a prime ideal p of A and an α ( A p) n such that m = m(r, α, p). First, note that by Proposition m(r, α, p) is a maximal ideal of R for all α and p. 97

101 5.3. GENERALISATIONS AND CONSEQUENCES Now, Let m be a maximal ideal of R. Now assume that m A = 0. Then, by Lemma 5.3.6, there is a non-zero prime ideal in p and an α ( A p) n such that α V m,p. Recall that, for p 0, { V m,p = α ( A p) } n vp (f(α)) > 0 for all f m and that m(r, α, p) = { f R v p (f(α)) > 0 }. Hence, m m(r, α, p). But since m is maximal, equality holds. Now note that for f A, v p (f) > 0 if and only if f p, whence m A = m(r, α, p) A = p 0. A contradiction. Let m A = p. Now, for all f R, f : ( A p) n p A is continuous in the p-adic topology. Let m be the unique maximal ideal of A p and { M f := α ( A p) } n vp (f(α)) > 0. Recall that { } m = β A p vp (β) > 0. This shows that we have f 1 (m) = M f. Since m is closed in A p, M f must be closed in ( A p) n. Now let f1,..., f r m. Since A is D*, it is Noetherian and hence p = p 1,..., p s for some elements p i A. Then a := f 1,..., f r, p 1,..., p s m R. And since R is Skolem, there must be an a A n such that p f 1 (a),..., f r (a), p 1,..., p s A. And hence, since A is Dedekind, p = f 1 (a),..., f r (a), p 1,..., p s, establishing that v p (f i (a)) > 0 for i = 1,..., r. Hence, a M fi for i = 1,..., r and thus M f1 M fr. This shows that, for f m, the sets M f have the finite intersection property. And since A/p is finite, A p = lim A/p i is compact (by Tychonoff s Theorem, see [34]), establishing that M := M f. f m Now let α M. Then v p (f(α)) > 0 for all f m, showing that m m(r, α, p). Again, since m is maximal, equality holds. This completes the proof. We now proceed to the core result of this section. It was shown by Brizolis in [8]. Theorem Int (A, 1) is Prüfer Let A be a D*-domain. Then R := Int (A, 1) is Prüfer. 98

102 CHAPTER 5. SKOLEM RINGS By Theorem and Theorem it suffices to show that R m is a valuation domain for all maximal ideals m of R. Since, by Theorem 5.3.3, R is Skolem, the maximal ideals of R are, by Theorem 5.3.7, of the form m(α, p) := m(r, α, p) for some prime ideal p of A and some α A p. Note that m(α, p) A = p. We need to show that for any f, g R either f/g R m(α,p) or g/f R m(α,p). Since R m(α,p) is a subring of quot (A[X]), we can write f/g = f/ g for some f, g A[X]. Now, it is possible that both f(α) and g(α) = 0. Then let h A[X] a polynomial of minimal degree such that h(α) = 0. Then h divides both f and g, whence there are positive integers r and s such that f g = h r f h s g for some f, g quot (A) [X] where f(α) 0 g(α). Then there is an a A such that af, ag A[X]. Without loss of generality we assume that r s. Then f g = h r s af ag where ˆf := h r s af and ĝ := ag lie both in A[X] and ĝ(α) 0. Note that the argument of this paragraph does not work for more than one variable. Now, if either ˆf(α) p = 1 or ĝ(α) p = 1, either ˆf or ĝ does not lie in m(α, p) whence either ˆf/ĝ or ĝ/ ˆf belongs to R m(α,p). Then we would be done. Thus we assume that both ˆf(α) p < 1 and ĝ(α) p < 1. Let N := #A/p. Then, since A is D*, N <. Hence, for all x A, x N x 0 (mod p) by Theorem This shows that ( ˆf N ˆf) (A) p and ( ĝ N ĝ ) (A) p. Then, by Proposition 5.1.7, there is an element π p 1 such that π p > 1 and ( π ˆf N ˆf ), π ( ĝ N ĝ ) R. We now write ( ˆf) ˆf π ˆf (ĝn 1 N 1 ) ĝ = ( ). π (ĝ N ĝ) ˆf N 1 1 Note that ˆf N 1 (α) = ˆf(α) p and that N 1 p ĝn (α) p = ĝ(α) N p < ĝ(α) p. < ˆf(α) < 1 p 99

103 5.3. GENERALISATIONS AND CONSEQUENCES This shows (by Corollary 4.3.2) that ˆf N 1 (α) 1 = 1 and ĝn (α) ĝ(α) p p = ĝ(α) p. Now, if we let g 1 := π ( ĝ N ĝ ) ( ) ˆf N 1 1 we see that g 1 R and that g 1 (α) p = π p ĝ(α) p > ĝ(α) p since π p > 1. Now let f 1 := π ( ˆf N ˆf) (ĝn 1 1 ), showing that f 1 R and that ˆf/ĝ = f 1 /g 1 where g 1 (α) p > ĝ(α) p. By this procedure we are able to produce sequences of polynomials f 1, f 2,... and g 1, g 2,... in R such that and ˆf ĝ = f 1 g 1 = f 2 g 2 =... ĝ(α) p < g 1 (α) p < g 2 (α) p <. We know that 1 g i (α) p = N (p) ci where the c i are non-negative integers and N (p) > 1. This shows that the c i must be a decreasing sequence and thus there must be a k such that c k = 0, establishing that g k (α) p = 1 and so that g k m(α, p), whence f/g = ˆf/ĝ = f k /g k R m(α,p). Similarly, if f j (α) p = 1 for some j k, then g/f = ĝ/ ˆf = g j /f j R m(α,p). In either case, either f/g R m(α,p) or g/f R m(α,p), showing that R m(α,p) is a valuation domain. This completes the proof. Note that this result cannot be generalised to Int (A, m) for m > 1: it is known (see [15, 22, Ex. 15]) that, for K := quot (A), K[X 1,..., X m ] is Prüfer if and only if m = 1. Furthermore, by [15, Th. 26.1], if an integral domain R is Prüfer and R R quot (R) for some overring R of R, then R is Prüfer as well. Hence, since quot (Int (A, m)) = quot (K[X 1,..., X m ]) and Int (A, m) K[X 1,..., X m ], we have that Int (A, m) can never be Prüfer for m > 1. This was also stated by Brizolis in [8]. Example 5.2: This theorem is an important result: it establishes that Int (O K, 1) is Prüfer for all number rings O K. A lot of questions are still open: how can Skolem rings be characterised? We have shown sufficient conditions for some rings to be Skolem but this is far from being a complete characterisation. In fact, some more is known. In his dissertation Brizolis 100

104 CHAPTER 5. SKOLEM RINGS shows that the rings Int(A, m, k), k N { }, of A-valued polynomials whose derivatives up to f (k) are A-valued as well, are non-prüfer Skolem rings. But what comes beyond? So far, no good solution has been found. It is immediate from the definition of Skolem rings to ask the following question: Let R be a subring of Int (A, 1) for some ring A. Let a and b be ideals of R such that a(a) = b(a) for all a A. Is it then true that a = b? In general, this question is to be answered in the negative. But in certain cases it is true: such rings are then said to satisfy the strong Skolem property. An example of this is, again, the ring Int (Z, 1). Consider the ideals X(X 1) a := 2, X, and b := 2, X(X2 + 1). 2 2 It is clear that a(a) = b(a) = { 2 if a 0 (mod 4) or a 2 (mod 4) 1 if a 1 (mod 4) or a 3 (mod 4). And yes, here we have a = b: X(X 1) 2 X(X 2 + 1) 2 X = 2 X2 (X 1) 2 4 = 2 X3 (X + 1) 3 8 = X X + X(X 1) 2 + X(X2 + 1) 2 + X(X2 + 1) 2 (X 1) (2 X) ( X 2 (X + 3) 2 This is an example from [8], where Brizolis also shows the following: ) + X 1 Let A be a D*-domain and a, b be finitely generated ideals of Int (A, 1) such that a(a) = b(a) for all a A. Then a = b. Furthermore, Cahen and Chabert give an exhaustive survey of rings with this property alongside a lot of material on A-valued polynomials in general in [9], to which the reader is referred for more information. Going away from Skolemness we find that it is not necessary that A be a D*-domain for Int (A, 1) to be Prüfer: if K is a field, K is not D* but Int (K, 1) = K[X] is Prüfer. In 1996, Cahen and Chabert succeeded in giving us some more information: (a) Let A be an infinite Noetherian ring which is not a field. Then Int (A, 1) is Prüfer if and only if A is Dedekind and A/p is finite for all prime ideals p of A. See [9, Th. VI.1.7]. (b) If A is not a field and Int (A, 1) is Prüfer, A must be almost Dedekind (i.e. A m must be a Noetherian valuation ring for all maximal ideals m of A) with only finite residue fields. See [9, Prop. VI.1.5]. 101

105 5.3. GENERALISATIONS AND CONSEQUENCES It turns out that the characterisation in (b) is not sufficient: Chabert has constructed an example of an almost Dedekind domain A with all its residue fields finite and the ring Int (A, 1) still not being Prüfer. Finally, in 1998, K. Alan Loper provided us in [21] with this characterisation: (c) Let A be a non-noetherian, almost Dedekind domain with only finite residue fields. Then Int (A, 1) is Prüfer if and only if A is doubly bounded. See [21, Th. 2.5]. Loper defines a a non-noetherian, almost Dedekind domain with only finite residue fields to be doubly bounded if the sets and F p = { A/P p P} E p = { v (N) P (p) p P } are bounded for all primes p which are nonunits in A. Now we are there: characterisations (a) and (c) provide necessary and sufficient conditions in both the Noetherian and non-noetherian cases such that Int (A, 1) is a Prüfer domain. 102

106 Appendix A Zorn s Lemma A well-known yet fairly hard to grasp result is Zorn s famous lemma. Since we used it in chapter 3, it shall be stated here. Definition A.1 Partial Ordering Let X be some set. We say that X is partially ordered with respect to some binary relation defined on X X if for any a, b, c X the following holds: a a (reflexivity) If a b and b a, then a = b (antisymmetry) If a b and b c, then a c (transitivity) Then (X, ) is called a partially odered set. Definition A.2 Properties of partially ordered sets Let (P, ) be a partially ordered set, also called poset. Given a subset X P, an element u P is called an upper bound of X, if x u x X. An element m X is called maximal if there exists no y P such that m < y. A subset C P is called a chain in (P, ) if each pair of elements c, d C is comparable, i.e. c d or d c Theorem A.3 Zorn s Lemma If (P, ) is a poset such that every chain has an upper bound in P, then P has a maximal element. 103

107 Appendix B The Chinese Remainder Theorem The number-theoretical version of the Chinese Remainder Theorem (the ring theoretic equivalent was introduced in Definition 3.5.1) was first stated by the Chinese mathematician Qin Jiushao in Theorem B.1 Chinese Remainder Theorem Suppose that n 1,..., n k are pairwise coprime integers for some k N. Then, for any given integers a 1,..., a k, there exists a solution x Z to the system of simultaneous congruences x a i (mod n i ) for all i {1,..., k}. Furthermore, all those solutions are congruent modulo n := n 1 n 2 n k. The solution in question can be found using the Extended Euclidean Algorithm. For further information, see [10], [13] and [24]. 104

108 Appendix C Notation Throughout the entire project, the following symbols were used: Definition C.1 Notation denotes containment, proper containment := is used as a defining operation: a := 1 means define a to be 1 : is used as a defining operation: (1) : (2) means define (1) to be equivalent to (2) and are used as division indicators, i.e. if a b for some a, b R, R a commutative ring, this means there is an element c R such that ac = b. Analogously, a b means that for all c R, ac b. N = {1, 2, 3,...} N 0 = N {0} Z = {..., 2, 1, 0, 1, 2,...} Z + = N, Z + 0 = N 0, Z = Z\N 0, Z 0 = Z\N { } Q = ± p q p N 0, q N M(n, R) (or M n (R)) is the ring of all n n matrices where the entries stem from R In the context of groups, denotes a subgroup: (H, +) (G, +) means H is an additive subgroup of the additive group G Let S be some set and let x, y be some elements such that x s and s y are defined for all s S. Then xs := { x s s S } If x s y is defined for all s S, we set xsy := { x s y s S }. and Sy := { s y s S }. 105

109 Let S be a subset of a multiplicative group. Then S 1 := { s 1 s S }. A B denotes the disjoint union of the sets A and B, i.e. A B =. #A denotes the size of the set A, i.e. #A = a A 1. If R 1,..., R n are R-modules, n i=1 R i denotes the direct sum of these modules. 106

110 Appendix D Noetherian Domains In this thesis, a lot of definitions of different types of rings and domains were given. To clarify things, Figure D.1 shows the relations between the most important (Noetherian) domains. Data for this figure comes primarily from [17], although most of the results are shown in this thesis. Figure D.1: Noetherian Domains 107