GENERALIZED FACTORIZATION

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1 GENERALIZED FACTORIZATION GRANT LARSEN Abstract. Familiarly, in Z, we have unique factorization. We investigate the general ring and what conditions we can imose on it to necessitate analogs of unique factorization. The trivial ideal structure of a field, the extent to which rimary decomosition is unique, that a Noetherian ring necessarily has one, that a rincial ideal domain is a unique factorization domain, and that a Dedekind domain has unique rime decomosition, are all covered. The relationshis of quadratic recirocity and the class grou with the question of factorization are discussed, and Gauss s method of comuting the class number of quadratic fields and new work generalizing this to many other fields is briefly advertised. Contents 1. Introduction 1 2. Conventions 2 3. Prelimaries 2 4. Primary Decomosition 3 5. Noetherian Rings 5 6. Princial Ideal Domains 7 7. Dedekind Domains 8 8. Factoring in Extensions More Tools in Extensions Ramification and Quadratic Recirocity The Class Grou, Gauss, and Bhargava 17 References Introduction Factoring in Z is something we re taught to do in elementary school and is taken for granted. If one doesn t ursue a mathematics education, one rarely realizes that the existence of unique factorization is something that needs to be roven. Fortunately, in Z, the roof is simle number theory. When we generalize to more interesting rings than Z, unique factorization no longer necessarily holds: the clichéd examle is Z ( 5, in which 6 = 2 3 = ( ( 1 5. So what is to be done? Luckily, Kummer, Dedekind, Hilbert, and Noether have worked hard on this question and come u with the concet of ideals - sets of numbers where we can talk about factoring in all the domains that matter. The first half of this Date: SEPTEMBER 21, It is not immediate that these numbers are distinct rimes, but it is not hard to show. 1

2 2 GRANT LARSEN aer talks about how we can decomose ideals in which situations. The second half investigates the interesting questions of how factoring works in extensions, how quadratic recirocity is related to the answer, how ideal factoring and numerical factoring are related, and how to comute the extent to which the latter fails. 2. Conventions All rings are assumed to be commutative and have 1 0. The whole ring will in general be denoted R. The word unique always carries the caveat u to ordering. All conventions hold as far as is convenient; deviations will be exlicitly noted. 3. Prelimaries Definition 3.1. Let R be a ring. A non-emty subset a R that is closed under addition, and for which ra a for each r R and a a, is called an ideal of R. An ideal that is not equal to the ring is called a roer ideal. The ideal {0} is often denoted 0. Proosition 3.2. Let F be a ring. Then F is a field if and only if the only roer ideal of F is 0. Proof. Suose F is a field. Let a be a roer ideal, and a a. Then a 0 imlies that a 1 x F for every x F, so x a for every x F, so a = F, contradicting the roerness of a, so a = 0, since an ideal is defined to be non-emty. Suose now that the only roer ideal of F is 0. Let x F \ 0. Then xf is a non-zero ideal, since x xf. By assumtion, xf = F. Particularly, 1 xf, so there is a y F such that xy = 1. Therefore F is a field. Definition 3.3. Let A be a set. The smallest ideal that contains A can easily be shown to be { n } (3.4 a = a i r i n N, a i A, r i R, and is called the ideal generated by A, and A is its generating set or set of generators. We write a = (A. If a is generated by a finite set, say A = {a 1,..., a n }, we write a = (a 1,..., a n, and say it is finitely generated. Definition. Set and ideal oerations: (3.5 A + B = {a + b a A, b B}, A, B R. (3.6 ra = {ra a A}, A R, r R. { n } (3.7 ab = a i b i n N, a i a, b i b, a, b ideals of R. (3.8 a : b = {x R xb a b b}, a, b ideals of R. The following are direct consequences of the definitions. Proosition 3.9. Proerties of ideal oerations: (1 a b is an ideal,

3 GENERALIZED FACTORIZATION 3 (2 a + b is an ideal, (3 a : b is an ideal, (4 ab is an ideal, (5 a a : b, (6 (a : b b a, (7 (8 (a : b : c = a : (bc, and (9 (a (b = (ab. ( a i : b = (a i : b, i I i I Definition Let a be an ideal. Then its radical, Rad (a, is given by {x R n N such that x n a}. Proosition Let a be an ideal. Then Rad (a is an ideal. Proof. Given x, y Rad (a, there are m, n N such that x m, y n a. Also, there are α i, µ i, ν i N {0} such that (3.12 (x + y m+n = m+n i=0 α i x µi y νi. For each i, µ i +ν i = m+n, so either µ i m or ν i n. Therefore, x µi a or y νi a, so α i x µi y νi a for every i. Hence, (x + y m+n a, so x+y Rad (a, i.e. Rad (a is closed under addition. For any r R, (rx n = r n x n a, so rx Rad (a. Thus, Rad (a is an ideal of R. 4. Primary Decomosition Definition 4.1. A roer ideal is rime in R if for every ab, a or b. Proosition 4.2. If, i are all rime ideals and (4.3 i, then there is a k n such that k. Proof. Assume that for each i, i. Then for each i, there is a i i \, so (4.4 i i. But is rime, so there is a k n such that k, contradicting the method by which the i were selected. Therefore, there is a k n such that k. Definition 4.5. A rimary ideal is an ideal q such that ab q and a / q imly there exists an n N such that b n q. Proosition 4.6. Let q be a rimary ideal in R. Let = Rad (q. Then is a rime ideal, and for all rime ideals q, q.

4 4 GRANT LARSEN Proof. is an ideal by Proosition If xy and x /, then there is an n N such that x n y n q. Since we assume x /, it follows that x n / q, so there is an m N such that y nm q, i.e. y. Thus, is rime. q trivially. For any rime ideal q, given x, there must be an n N such that x n q. is rime, so x, so. Definition 4.7. In the situation of the Proosition 4.6, q belongs to, and it is called -rimary. Corollary 4.8. Let q be -rimary. Then: (1 For any ab q such that a /, b q, (2 For any ideals ab q such that a, b q, and (3 a imlies q : a = q. Proof. (1 is an immediate consequence of the definitions, and (2 follows immediately from (1. For (3, we know that (q : a a q, so by (2, q : a q. We also know that q q : a, so q : a = q. Definition 4.9. Let a be an ideal. If (4.10 a = n q i, where each q i is rimary, a has a rimary decomosition, a is called decomosable, and the q i are called the rimary comonents of the decomosition. A decomosition in which no q j contains the intersection of the remaining q i is irredundant. An irredundant decomosition in which all the q i are distinct is a normal decomosition. Knowing these terms, we can claim a certain level of uniqueness to rimary decomosition, when it exists: Theorem If the ideal (4.12 a = m q i = n q j, where both are normal decomositions, q i is i -rimary for every i, and q j is j - rimary for every j, then m = n, and the comonents can be indexed such that i = i for every i. Proof. If a = R, then n = m = 1 and q 1 = q 1 = R. Otherwise, a is a roer ideal. From the finite set { i } i m { } j of rime j n ideals, we may select one which is not strictly a subset of any other. Assume without loss of generality that this is m. Assume that q m j for all j. Then by Corollary 4.8(3, q j : q m = q j, so (4.13 a : q m = n ( n q j : q m = q j = a. If m were 1, a would be q m, so by the above, a would be R, the case handled reviously, so in the current case, m must be greater than 1. By the selection of m, m i for every i < m, since the decomosition is normal. Therefore by

5 GENERALIZED FACTORIZATION 5 Proosition 4.6, q m i. This lets Corollary 4.8(3 be alied, giving q i : q m = q i. Since q m : q m = R, (4.14 a : q m = Combining (4.13 and (4.14, m (q i : q m = (4.15 a = m 1 This contradicts the normality of the decomosition in (4.12, so our assumtion must be false, i.e. there must exist a j m such that q m j. By Proosition 4.6, m j, so by the selection of m, m = j. Assume without loss of generality that m = n. By definition, q m q n belongs to m = n. By similar arguments to those above, (4.16 m 1 q i. q i = a : q = Since a : q is an ideal, we are in the same osition we started in, now with n 1 and m 1. Assume m < n. After m recursions of the above rocess, one would get (4.17 R = n m but the comosition was normal, so all the q j should have been roer, contradicting (4.17, so m n. A erfectly analogous argument shows n m, so m = n, and after m = n stes of this recursive rocess, all m = n airs of equal rime ideals have been identified. Definition The rime ideals roved to be unique in Theorem 4.11 are called the rime ideals belonging to a. q j, n 1 m 1 q j. q i. 5. Noetherian Rings In general we can t get any more unique than the above decomosition, and Theorem 4.11 rests on the condition that such a decomosition exists. We can make a ste in the right direction by demanding the following roerty: Definition 5.1. A ring R is Noetherian if, given an ascending chain a 1 a 2... of ideals, there is an m N such that for all n m, a n = a m. Proosition 5.2. Given a ring R, TFAE: (1 R is Noetherian; (2 Every non-emty set of ideals has a maximal element with resect to inclusion; (3 Every roer ideal is finitely generated.

6 6 GRANT LARSEN Proof. Suose R is Noetherian. Let I be a non-emty set of ideals of R. If I is finite, the existence of a maximal element is trivial. Otherwise, ick a 1 I. Given a i I, there is an a i I such that a i a i. Let a i+1 = a i. The assumtion that R is Noetherian imlies that this chain is eventually the same ideal over and over, and this ideal is necessarily maximal. Therefore, we have (1 (2. Suose now that (2 holds. Let a be an ideal of R, let A be the set of finitely generated ideals contained in a, and let m be a maximal element of A. Assume that there is an element x a \ m. Then m + xr is finitely generated, so m + xr A. But m + xr m, contradicting the maximality of m in A. Therefore m = a, i.e. a is finitely generated. This reveals that (2 (3. Suose now that every ideal of R is finitely generated. Let a 1 a 2... be an ascending chain of ideals of R. Let (5.3 A = i N a i. A is an ideal in R, so it is generated by a finite set, say {a 1,..., a n }, by assumtion. Then for every i N, a i (a 1,..., a n. For every j n, there is a k j such that a j a kj. Let k = max {k j } j n. Then for each j n, a j a k. Therefore a k (a 1,..., a n, so a k = (a 1,..., a n. Therefore, for all i k, a i = a k. This gives us (3 (1. In this case, the situation is slightly imroved, but in order to see why, we must introduce a new concet: Definition 5.4. An ideal a is irreducible if whenever a = b c, a = b or a = c. An ideal that is not irreducible is reducible. Lemma 5.5. If a is an ideal of a Noetherian ring, it is the intersection of a finite number of irreducible ideals. Proof. Let I be the set of ideals which cannot be written as finite intersections of irreducible ideals. By Proosition 5.2, if I is non-emty, it has a maximal element, m. There exist ideals a, b such that m = a b, but m cannot be irrdeucible, so m a and m b. Since m is maximal, a, b / I, i.e. they re both finite intersections of irreducible ideals. Therefore, so is m = a b, so m / I, a contradiction. Therefore I is emty. Lemma 5.6. In a Noetherian ring, all irreducible ideals are rimary. Proof. Let a be an ideal of a Noetherian ring that is not rimary, i.e. there are b and c such that bc a and b, c n / a for any n N. Thus, we find that a a : (c. By the roerties of ideal oerations, a : ( c k ( a : ( c k : (c = a : ( c k+1, so there is an ascending chain: (5.7 a a : (c a : ( c 2 a : ( c 3... Since the ring is Noetherian, there is an m N such that a : (c n = a : (c m for all n m. Let x (a : (c m (a + (c m. Since x a + (c m, there is an a a and an r R such that x = a + rc m. Since x a : (c m, we must have xc m = ac m + rc 2m a, meaning that rc 2m a, i.e. r a : ( c 2m = a : (c m by the selection of m. Therefore, rc m a, so x = a + rc m a. This means that (a : (c m (a + (c m a. It is clear that a a + (c m, and a a : (c m by (5.7.

7 GENERALIZED FACTORIZATION 7 Therefore a = (a : (c m (a + (c m, i.e. a is the intersection of two ideals that srtictly contain it, so it is reducible. Theorem 5.8. Every ideal of a Noetherian ring has a rimary decomosition. Proof. This is the direct combination of Lemma 5.5 and Lemma 5.6. To summarize, in Noetherian rings, we have existence of rimary decomosition in which the associated rimes are unique, though the decomosition itself it is not necessarily unique. 6. Princial Ideal Domains Of course, we can get unique factorization by imosing strict conditions. We know that the rational integers have unique factorization, so if we generalize just enough that the roof needn t change, then the claims still holds. Definition 6.1. If there is an a R such that a = (a, a is a rincial ideal. Definition 6.2. An integral domain where all ideals are rincial is a rincial ideal domain (P.I.D.. Unique factorization of ideals in a P.I.D. holds for exactly the same reasons as unique factorization of rational integers. To reeat that roof here would waste the time of a reader who knows it, and rob the reader that doesn t of a crucial exercise. It can also be found in [4], [6], and [8]. Theorem 6.3. Let R be a P.I.D. Then for any roer ideal a R, there exist roer rime ideals i R such that n (6.4 a = i. This rime decomosition is unique. Definition 6.5. An element with a multilicative inverse is a unit. Remark 6.6. The set of ideals of a P.I.D. is isomorhic to the P.I.D. itself in the obvious way, i.e. (6.7 R {a R}, a (a, where one can check that ab (a (b = (ab. Units ma to R. Definition 6.8. An element is irreducible if it is a nonzero non-unit r R such that r = ab imlies that a or b is a unit. Proosition 6.9. In an integral domain, any rime element is an irreducible element. Proof. Suose is a rime element and = ab. Without loss of generality, assume b, i.e. there is a c R such that c = b. Then ac = ab, so ac = 1, i.e. a is a unit. Definition A domain is a Unique Factorization Domain, U.F.D. for short, if any nonzero non-unit can be written uniquely as a roduct of irreducible elements.

8 8 GRANT LARSEN Corollary Any P.I.D. is a U.F.D. Proof. This is immediate from Theorem 6.3, Remark 6.6, and Proosition 6.9. Also, the language of rincial ideals lets us rove a conclusion about the elements of a Noetherian ring: Proosition If R is a Noetherian integral domain, every nonzero non-unit can be written as the roduct of finitely many irreducible elements. Proof. Analogously to the roof of Lemma 5.5, let I be the set of rincial ideals generated by elements of R that cannot be written as the roduct of finitely many irreducibles. If I was nonemty, there would be a maximal element (m I, and since m could not be reducible, there would be non-units a, b R such that m = ab. Therefore (m (a and (m (b, so by the maximality of (m in I, (a, (b / I, so a and b would both be roducts of finitely many irreducibles, so ab = m would be as well. Therefore I is necessarily emty. 7. Dedekind Domains The condition of being a rincial ideal domain is very strict, so it is natural to inquire if there is a more general case in which unique factorization holds. This generalization is what is known as a Dedekind domain, the conditions of which we need to define. Definition 7.1. An ideal is maximal if it is not strictly contained in any roer ideal. Proosition 7.2. A maximal ideal is rime. Proof. Let a be a non-rime roer nonzero ideal, i.e. let x, y / a such that xy a. Then x (a : (y \ a and 1 / a : (y, so a a : (y R, i.e. a is not maximal. Proosition 7.3. An ideal m is maximal if and only if R/m is a field. Proof. The Lattice Isomorhism Theorem for rings states that the ideal structure of R/m is the same as the ideal structure of R, limited to those ideals containing m. Therefore, this claim is a direct consequence of Proosition 3.2. Definition 7.4. R S is an extension of rings/fields if both sets are rings/fields, and the notions of addition and multilication agree on R. In the case of fields, one writes S R. Definition 7.5. Let R S be an extension of rings. An element s S that satisfies a monic olynomial with coefficients in R is said to be integral over R. If s is integral over R for all s S, S is an integral extension of, or integral over, R. The subset of S consisting of all elements integral over R is the integral closure of R in S. A ring R that equals its integral closure in S is integrally closed in S. The integral closure of R in its field of fractions is its normalization. A ring R that is its own normalization is normal or integrally closed. The field of fractions takes a bit of effort to construct rigorously, but is essentially as it sounds: fractions where the numerator and denominator come fom the domain in question. Q is the field of fractions of Z. For more, see [4], [7], or [10].

9 GENERALIZED FACTORIZATION 9 Definition 7.6. For a 1,..., a m S R, R adjoin a 1,..., a m is n m R [a 1,..., a m ] = r j a j i r j R, n N S. R [x] will be used in this aer to denote the similar but distinct construct, the olynomial ring over R. Definition 7.7. An R-module M is finitely generated if there is a finite set A M such that for every m M, there are elements r a R such that (7.8 m = a A r a a. Proosition 7.9. The elements s 1,..., s m S are integral over R if and only if R [s 1,..., s m ] S is a finitely generated R-module. Proof. Suose s S is integral over R. Let f (x R [x] be a monic olynomial of degree n, such that f (s = 0, and g (x an arbitrary olynomial in R [x]. R [x] is Euclidean, i.e. there are q (x, ρ (x R [x] such that g (x = q (x f (x + ρ (x and degρ (x < n. Then g (s = ρ (s, a sum of finitely many roducts of elements of R and owers of s. Therefore, R [s] is a finitely generated R-module. Suose now that R [s] is an R-module generated by {α 1,..., α k }. Then for every element t R [s], there are r ij R such that (7.10 tα i = k r ij α j, i. We know from linear algebra that det (t1 k (r ij α i = 0 for every i. That is, for every t R [s], there is an R-linear transformation (r ij such that t is an eigenvalue of the transformation. det (t1 k (r ij = 0 is a monic olynomial for s with coefficients in R, so s is integral over R. Observe that R [s 1 ] [s 2 ] = R [s 1, s 2 ] and induct. Corollary If R S T are ring extensions, where T is integral over S, and S is integral over R, then T is integral over R. Proof. Let t T. Then there is an n N and s i S such that n (7.12 t n + s i t n i = 0. Let M = R [s 1,..., s n ]. By Proosition 7.9, M is finitely generated over R, and R [t] is finitely genereated over R, so t is integral over R, so T is integral over R. Corollary The integral closure of a ring in another ring is integrally closed. Proof. This is immediate from Corollary Definition A Noetherian integral domain that is integrally closed, in which every nonzero rime ideal is maximal, is a Dedekind domain. A module over a ring is the generalization of a vector sace over a field. For more, see [4]. 1 k denotes the k k identity matrix, (δ ij in Kronecker delta notation.

10 10 GRANT LARSEN In order to get where we want, we ll need a coule of lemmas concerning rime ideals in Dedekind domains: Lemma Any nonzero ideal of a Dedekind domain D contains the roduct of finitely many rime ideals. Proof. Let I be the set of ideals which do not contain the roduct of finitely many rime ideals. By Proosition 5.2, there is a maximal element m I. m cannot be rime, so there must be a, b D such that ab m and a, b / m. Letting a = m+(a and b = m + (b, it is clear that m a and m b. By the maximality of m, a and b are suersets of finite roducts of rime ideals, but a quick check shows that ab m, so m is as well. Therefore I is emty. Definition Let be a rime ideal in the Dedekind domain D with field of fractions K. Then 1 = {k K k D}. Definition A fractional ideal of K is a nonzero finitely generated D- submodule of K. Definition A rincial fractional ideal is a fractional ideal of the form kd for some k K. It will be denoted (k. Remark is a fractional ideal, and the same oerations that aly to ideals aly to fractional ideals. Lemma For a rime ideal of a Dedekind domain D, 1 D. Proof. By Definition 7.16 and the definition of ideal, 1 D. Let \ 0. By Lemma 7.15, there must be nonzero rime ideals i such that m (7.21 i (, where m is as small as ossible. By Proosition 4.2, there is a k m such that k. Assume without loss of generality that k = m. Since D is a Dedekind domain, m is maximal, so m =. By the minimality of m, (7.22 m 1 i (, i. e. x ( m 1 i \ D, i. e. 1 x K \ D. By (7.21, x (, so 1 x D, so 1 x 1, so 1 D by (7.22. Lemma For a rime ideal of D, a 1 a for all nonzero ideals a. Proof. By Proosition 5.2, there must exist α 1,..., α n that generate a. Assume a 1 = a, so for any x 1, there are a ij D such that n (7.24 xα i = a ij α j. Let A = (x1 n (a ij. Then A (α 1,..., α n t = 0. We know that det (A α i = 0 for every i, so det (A = 0, so x is integral over D. Since D is a Dedekind domain, this imlies that x D, so 1 D, contradicting the revious lemma. Therefore a 1 a. Theorem Every nonzero, roer ideal of a Dedekind domain can be uniquely factored into nonzero rime ideals.

11 GENERALIZED FACTORIZATION 11 Proof. Let I be the set of nonzero roer ideals of the Dedekind domain D that do not factor into rime ideals. Assume I is non-emty. By Proosition 5.2, there is a maximal element m I. Alying Proosition 5.2 to the suersets of m, we find that it is contained in a maximal ideal. By Lemma 7.20, m m 1 1 D, so m 1 is an ideal of D. By Lemma 7.23, m m 1. It s simle to show that 1 is an ideal, so since is maximal, 1 = D. We know is maximal, hence rime by Proosition 7.2, but m cannot be rime, so m, so m 1 1, so m 1 is a roer ideal containing m. Since m is a maximal element of I, m 1 has a rime factorization. But m = m 1, so m also has a rime factorization, so I is necessarily emty. Therefore all ideals in a Dedekind domain can be factored into rimes. Let the following be two rime factorizations of the same ideal: m (7.26 a = i = Since m is rime, there must be a j such that m j, but rime ideals are maximal in a Dedekind domain, so m = j. Assume without loss of generality that j = n. Since 1 = D, by multilying by 1 m, we find that (7.27 m 1 i = Using this recursive rocess, and remembering that the ideals are rime, we find that m = n and air u all m = n airs of equal rime ideals. Thus, the factorization of ideals in a Dedekind domain into rime ideals is unique. n 1 j j. 8. Factoring in Extensions But what about factoring in extensions of Dedekind domains? Remark 8.1. In the situation of a field extension L K, L is a K-vector sace. Definition 8.2. The degree of the extension L K is [L : K] = dim K L. If this is finite, L is called a finite extension of K. Definition 8.3. For a field extension L K and a set A L, K adjoin A, denoted K (A, is the inersection of all subfields of L containing K and A. If A = {a 1,..., a n }, we write K (a 1,..., a n. Definition 8.4. A olynomial is searable over K if all of its irreducible factors have distinct roots in the algebraic closure of K. A field extension L K is searable if there is a set of roots of searable olynomials over K that, when adjoined to K, give L. Definition 8.5. The discriminant of a basis α 1,..., α n of a searable extension L K is d (α 1,..., α n = det ( T r L K (α i α j. Remark 8.6. It can be shown that if L K is searable, then the discriminant of any basis is nonzero. See [9] for details. Actually, in any ring, any roer ideal is contained in a maximal ideal, but the general roof of this requires Zorn s Lemma. See [4].

12 12 GRANT LARSEN Henceforth, D denotes a Dedekind domain with field of fractions K, L K an extension of fields, and O the integral closure of D in L. Lemma 8.7. If O contains α 1,..., α n, a basis for L K, and d = d (α 1,..., α n, then n (8.8 O Dα i /d. Proof. If α O and (8.9 α = Then n a i α i O, a i K. (8.10 T r L K (α i α = n T r L K (α i α j a j. Since T r L K (α i α D, det ( T r L K (α i α j a j D. But det ( T r L K (α i α j = d, so n n (8.11 α Dα j /d, i. e. O Dα j /d. Proosition If the extension L K is searable, then O is a Dedekind domain. Proof. The domain O is integrally closed by assumtion. Let P be a nonzero rime ideal in O. Then for a y P \ 0, there are m N, x i D such that x m 0 and m (8.13 y m + x i y m i = 0, so x m P D, i.e. P D 0. P D is clearly a rime, and therefore maximal, D ideal in D, so Proosition 7.3 shows P D to be a field. By Proosition 3.2, it has no roer nonzero ideals, so neither can O/P, or else the intersection of such an D ideal with P D would yield a roer nonzero ideal in D P D. By Proosition 3.2, O/P is a field, so by Proosition 7.3, P is maximal in O. Since L K is searable, let α 1,..., α n be a basis of L K contained in O of nonzero discriminant d. By Lemma 8.7, O is contained in a finitely generated D-module, so every ideal of O is as well, making every ideal of O a finitely generated D-module. Since D O, an ideal of O is a finitely generated O-module, i.e. O is Noetherian. Since O is integrally closed, has every rime ideal maximal, and is Noetherian, it is a Dedekind domain. Proosition If is a rime ideal of D, then O is a roer ideal of O. The searable condition here is actually weaker than necessary, but the roof is trickier and requires more machinery if one wants to rove this for finite extensions in general. See [1], [7], or [9] for details. Also, most extensions considered here are searable.

13 GENERALIZED FACTORIZATION 13 Proof. The statement is trivial if = 0. Otherwise, let \ 2, so there is some ideal a of D such that a = D and a. In this case, + a = D. Therefore, there exist q and a a such that q + a = 1. Then a cannot be in and a a = D. Assume O = O. Then ao = ao O. This means that there is an x O K such that x = a, so a, contradicting our earlier conclusion. Hence, O is a roer ideal of O. Corollary A nonzero rime ideal of D factors uniquely into rime ideals in O, i.e. there exist unique rime ideals P i O and natural numbers m, ν i N such that (8.16 O = m P νi i. Proof. Proosition 8.12 and Proosition 8.14 allow us to aly Theorem More Tools in Extensions In order to understand the intriguing conclusions of the next section, some vocabulary and machinery must be develoed. Definition 9.1. In the situation of Corollary 8.15, ν i [ is the ramification ] index of P i over. The inertia degree of P i over is f i =. Proosition 9.2. If L K is searable, then the above quantities are related to the degree of the extension L K in the fundamental identity: m ν i f i = [L : K]. Proof. Let ϕ : D [x] D [x] be the natural homomorhism. Let ϕ (ω 1,..., ϕ (ω n O be a basis of O over D/. Assume the ω i are linearly deendent in K. Then they are linearly deendent in D. Then there are a i D and k n such that a k 0 and n (9.3 a i ω i = 0. This means we can generate a nonzero ideal a = (a 1,..., a n of D. We can find an a a 1 such that a / a 1, so aa. This means that {aa i i n} D, but {aa i i n}. Therefore, since ( n (9.4 ϕ aa i ω i = ϕ (0 = 0, the ϕ (ω i are linearly deendent over D/, contradicting the fact that they form a basis. Therefore the ω i are linearly indeendent in K. Let n (9.5 M = Dω i and N = O/M. Then O = M + O, so N = N. L K is searable, so O is a finitely generated D-module, as shown in the roof of Proosition 8.12, so N must O P i : D

14 14 GRANT LARSEN be as well. Therefore, we can find generators α i,..., α m of N. Then for every i, j there must be ρ ij such that m (9.6 α i = ρ ij α j. Let A be the matrix (ρ ij 1 m, let the minor B ij be the determinant of the (m 1 (m 1 matrix formed by deleting the ith row and jth column of A, and let adj (A be the classical adjoint of A, meaning adj (A ij = ( 1 i+j B ji. Then we have A (α 1,..., α m t = 0 and adj (A A = det (A 1 m. Therefore, adj (A A (α 1,..., α m t = (det (A α 1,..., det (A α m t = 0, so det (A N = 0, so det (A O M. Since ρ ij, we find that ϕ (det (A = ( 1 m. Hence, we can find that n (9.7 L = det (A L = Kω i. Thus, we find that ω 1,..., ω m is a basis of L K, so dim D/ ( O O = [L : K]. Now fix i n. Let Q µ i = P µ i /Pνi i for each non-negative integer µ ν i. Then we have O/P νi i = Q 0 i Q 1 i... Q νi i = 0, all of which are D -vector saces. We also have that Q µ i /Qµ+1 i = P µ i /Pµ+1 i for all µ < ν i. Let q P µ i /Pµ+1 i. Let ψ q : O P µ i /Pµ+1 i, ψ q (x = xq. ψ q is a homomorhism, ker ψ q = P i, and ψ q is surjective, because P µ i = P µ+1 i + qo. Therefore, ( Q µ i /Qµ+1 i = P µ i /Pµ+1 i = O/P i. By the definition of inertia degree, f i = dim D/ Q µ i /Qµ+1, so νi 1 (9.8 dim D/ (O/P νi i = ( dim D/ µ=0 The Chinese Remainder Theorem tells us that O (9.9 O m = O/P νi i, so ( O (9.10 dim D/ O by (8.24. = i Q µ i /Qµ+1 i = ν i f i. m m dim D/ (O/P νi i, i. e. [L : K] = ν i f i, Definition Suose (x D [x], and θ L such that (θ = 0 and L = K (θ. The conductor of the ring D [θ] is the largest ideal of O, the integral closure of D in L, contained in D [θ], exlicitly given by F = {α O αo D [θ]}. Proosition Let a rime ideal of D such that O + F = O, and let m (9.13 ϕ ( (x = ϕ ( i (x νi be the factorization of ϕ ( (x into irreducibles such that i (x is monic for all i. Then one finds the rime ideals over of (8.16 by P i = O + i (θ O. The inertia

15 GENERALIZED FACTORIZATION 15 degree of P i is the degree of ϕ ( i (x. Also, m (9.14 O = P νi i. Proof. By assumtion, O+F = O, so since F D [θ], O = O+D [θ], so the natural homomorhism from D [θ] to O O is surjective. Its kernel is O D [θ] = D [θ]. Since + F D = D, O D [θ] = ( + F (O D [θ] D [θ]. Therefore O O = D[θ] D[θ]. Let R = ϕ (D [x] / (ϕ ( (x. Let φ : D [x] R be the natural homomorhism. It is surjective and ker φ is generated by and (x. Since D [θ] = D [x] / ( (x, we must have D[θ] D[θ] = R, so R = O O. Since m (9.15 ϕ ( (x = ϕ ( i (x νi, the Chinese Remainder Theorem gives us (9.16 R m = ϕ (D [x] / (ϕ ( i (x νi. This shows that the rime ideals of R are the rincial ideals (φ ( i, that [R/ (φ ( i : D/] = deg ϕ ( i (x, and that m ( = φ ( = (φ ( i νi. Let Φ : O O O be the natural homomorhism. It is surjective, i.e. Φ (O = O O. Since R = O O, the conclusions above hold in O O. That is, the rime ideals Φ (P i of Φ (O corresond to the rime ideals (φ ( i and are (Φ ( i (θ, [Φ (O /Φ (P i : D/] = deg ϕ ( i (x, and m ( = Φ (P i νi. Thus, the reimage of Φ (P i, i. e. P i, is O = i (θ O, and by definition, f i = [Φ (O /Φ{(P i : D/] = deg} ϕ ( i (x, as desired. Since ν i = Φ (P k k N, P νi i = Φ 1 (Φ (P i νi. Also, m m (9.19 O P νi i, O P νi i, so by the fundamental identity (8.18, (9.20 O = m P νi i. We want to know more about how a rime ideal factors in an extension, but we lack the vocabulary to ask about it, hence some definitions: Definition A rime ideal of D slits comletely, or is totally slit, in L if the unique factorization of O (8.16 has m = [L : K], so ν i = f i = 1 for all i. is nonslit, or indecomosed, if n = 1.

16 16 GRANT LARSEN Definition A rime ideal P i of O from (8.16 is unramified over D or K if ν i = 1 and O P i D is searable. Otherwise, it is ramified. It is totally ramified if it is ramified and f i = 1. The rime ideal is unramified if all the P i are unramified; otherwise it is ramified. The extension L K is unramified if all rime ideals of K are unramified in L. 10. Ramification and Quadratic Recirocity Now, the fun art: What s the easiest way to figure out just when this total slitting haens in, say, a quadratic number field, Q ( a? Definition Let a, b Z. Then a is a quadratic residue of b if there is an integer k such that k 2 a mod b. ( Definition The Legendre symbol,, read a on, is defined for any a Z and odd rime N to be 1 if a is a quadratic residue ( modulo, 0 if a, a and 1 otherwise. It can be shown that this is equivalent to a 1 2 mod. ( ( ( a b ab Proosition =. Proof. This is an immediate consequence of the alternate definition mentioned above. This tool lets us aly Proosition 8.28 rather succinctly in the case of quadratic fields: ( Corollary For square-free a and odd rime a, we find that = 1 if and only if ( is totally slit in Q ( a. Proof. It is easily verifiable that Z is a Dedekind domain, so let D = Z, K = Q, θ = a, (x = x 2 a, and L = Q ( a. What is less trivial, but not difficult to show, is that F (2. See [4] or [9]. Since is odd, this means that we can aly Proosition 8.28 to (. For square-free a, a ( a a = 1 is equivalent to the existence of some α Z such that x 2 a (x + α (x α mod. Hence, the decomosition given ( by (9.13 has all linear factors, i.e. the inertial degrees are all 1, if and only if = 1. a is square-free, so all the ramification indices are 1. a Hence, ( by the fundamental identity, m = [L : K], i. e. ( slits comletely, if and only if = 1. a If you re wondering how this makes life any easier, I have the leasure of introducing you to one of the truest gems of number theory, Gauss s famous Quadratic Recirocity Law: Theorem For two distinct odd rimes, q N, ( q ( q = ( 1 (q 1( 1 4. This, along with the two simle sulementary statements below that handle 1 and 2, makes comutation of a Legendre symbol as easy as a few stes of modular arithmetic. Theorem For an odd rime N, ( 1 = ( and ( 2 = (

17 GENERALIZED FACTORIZATION 17 There are literally hundreds of distinct roofs of this law, with 8 by Gauss himself. The most enlightening, i.e. the one that best answers the question, But, why? that I have encountered is in the section on cyclotomic fields in [9]. There are other roofs in [5], [6], [7], [8], and [9]. Corollary Let q and be distinct odd rimes. Then if ( 1 (q 1 /4 is odd, (q slits comletely in Q ( if and only if ( does not slit comletely in Q ( q. If ( 1 (q 1 /4 is even, (q slits comletely in Q ( if and only if ( also slits comletely in Q ( q. Proof. This is a restatement of Theorem 10.5 in terms of Corollary The Class Grou, Gauss, and Bhargava Previously, we used our understanding of ideal structure to come to conclusions about factorization of elements. It is natural to ask if there is there some analogous statement that can be made for Dedekind domains. Henceforth, unless otherwise stated, R S is the notation for the localization of R at S, where S must be a multilicative subset of R. Proosition If a b are two ideals in a Dedekind domain D, then there is an a D such that a = b + (a. Proof. By Theorem 7.25, there are unique natural numbers ν i N and rime ideals i of D such that (11.2 b = νi i. Because a b, there are integers µ i such that 0 µ i ν i and (11.3 a = µi i. \ µi+1 i. By the Chinese Remainder mod νi i for all i n. Because of this, every element of the fractional ideal generated by b + (a in D i is congruent to an element of the fractional ideal generated by a in D i, and vice versa, i.e. the two fractional ideals are equal, for every i. In the localization of D at any other rime ideal, the fractional ideals generated by both are trivially the same. Since For each i N such that i n, let a i µi i Theorem, there is an a D such that a a i the two ideals generate the same fractional ideals in the localization of D at any rime ideal of D, b + (a = a. Definition Two nonzero roer ideals a and b are relatively rime if they have no rimes in common in their unique factorizations as given by Theorem This can be shown to be equivalent to a + b = R. Corollary Given any nonzero ideals a, b of a Dedekind domain D and b a, there exist nonzero ideals a b and a b of D such that aa b is rincial, a b is relatively rime to b, and aa b = (b. e.g. Corollary 6.11 and Proosition 6.12 Good resources for learning about localization are [4], [5], [7], and [9].

18 18 GRANT LARSEN Proof. The claim for b a is trivial if b = 0. Otherwise, (b a is nonzero, so by Theorem 7.25, as in the revious roof, there exist unique rime ideals i of D, naturals n and ν i, and integers µ i such that 0 µ i ν i, (11.6 a = Let (11.7 a b = µi i, and (b = νi µi i. Then aa b = (b. We know ab a, so by Proosition 9.1, there is an a D such that a = ab + (a. Therefore, a = ab + aa a, so D = b + a a. Set a b to be a a. νi i. Theorem A Dedekind domain is a U.F.D. if and only if it is a P.I.D. Proof. A P.I.D. is necessarily a U.F.D. Suose then, that D is a Dedekind U.F.D. Let be a rime nonzero ideal of D. There is a nonzero element in, and by Proosition 6.9, it has an irreducible factor. By Corollary 11.5, there are elements a, b D and nonzero ideals, a, ( a b of D such that = (, a = (a, + a = D, a ( a b = (b, and ( a b + = D. We can see that (b = a ( a b = (a ( a b. Therefore a b, i.e. there exists a c D such that ac = b. Because D is a U.F.D., a or c. Assume that c. Then there is a d D such that d = c, so d = ( a b, contradicting their selection as relatively rime. Hence, a, i.e. there is an e D such that e = a, so e = a. This means that any ideal that was a factor of would also be a factor of a, which cannot be because they are relatively rime. Thus, = D. Since = ( by construction, this argument tells us that = (. Our hoe seems to have been naïve - in reality, to talk about unique rime factorization of elements in a Dedekind domain, we necessarily must require it to be a P.I.D. However, with a new tool, we can measure how far away we are. Proosition The set of fractional ideals form an abelian grou under multilication, with 1 = D and a 1 = {k K ka D}. Proof. Associativity, commutativity, and identity are all trivial to show. Lemma 7.23 says that for a rime ideal, 1, and since D is Dedekind, this means that 1 = D. Therefore, if a is an ideal and (11.10 a = i, b = will give ab = D. By this, b a 1. If b a 1, bab b, so b b, remembering that ab = D. Therefore a 1 = b. Were a a fractional ideal and not an ideal, we could ick an a a such that aa is an ideal of D, with an inverse that is easily shown to be a 1 a 1, so aa 1 = D. Definition The grou in Proosition 11.9 is the ideal grou of K, denoted J K. 1 i

19 GENERALIZED FACTORIZATION 19 Corollary Every fractional ideal a J K has unique rime factorization (11.13 a = νi i, ν i Z. Proof. After realizing that there are ideals b, c such that a = b/c for any fractional ideal a, this is a direct consequence of Theorem Definition It is trivial to show that P K = {kd k K} is a subgrou of J K. Cl K = J K /P K is the class grou of K. Cl K is the class number of K. Remark It is elementary to show that a domain is a P.I.D. if and only if its class number is 1. In a very hands-waving manner, let me say that the class number indicates to what extent the ideals can vary from being rincial, i.e. to what extent unique rime factorization can fail for elements desite holding for ideals, in a Dedekind domain. This makes a little more sense if Theorem 11.8 and Remark are ket in mind. The class number is in fact necessarily finite, but the roof of this is outside the scoe of this aer. Different roofs can be found in [5], [7], and [9]. Calculating the class number of a domain is tricky. One way to do it is using Minkowski s Lemma. That is not the subject of this aer, but more about it can be found in [7] and [9]. One of the more common settings where this is an issue is the quadratic field, as addressed in the last section. It turns out that Gauss develoed an exlicit isomorhism between ideal classes in quadratic rings of a given discriminant and SL 2 (Z-equivalence classes ( of binary integral quadratic forms of (Sym the same discriminant D, denoted Cl 2 Z 2 ; D. Binary quadratic forms are thoroughly understood, and number theorists have used this corresondence for centuries to comute the class numbers of quadratic rings. This method, though owerful, was lamentably limited to the secial case of quadratic rings. However, a few years ago, by finding a corresondence between sets of binary quadratic forms and the sace of cubes of integers, denoted Z 2 Z 2 Z 2, Manjul Bhargava of Princeton found that Gauss s corresondence was just a secial case of a more general trend, and found 13 analogous laws for other saces, such as the sace of ideal classes of cubic rings, that is rings in degree 3 extensions of the rationals. These corresondences ease calculations of determinants, simlify the criteria for finding which orders in an algebraic number field are maximal, imrove our understanding of the slitting of rimes in rings of integers, and aid in determining the invertibility of ideal classes of a domain that isn t Dedekind. The corresondences can be remarkably simle. A teaser is laid out below. For a list of the corresondences and seculation as to their imlications, see [2]. For the aer that fleshes out and roves the most imortant and accessible ones, see [3]. Let A = (a, b, c, d, e, f, g, h Z 2 Z 2 Z 2. It s best to think of this as a cube, where ( ( a b e f M1 A = and N c d 1 A = are the front and back, g h This is also known as Minkowski s Theorem. There are robably more. See [2] and [3].

20 20 GRANT LARSEN ( ( a c b d M2 A = and N e g 2 A = are the left and right, and ( ( f h a e c g M3 A = and N b f 3 A = are the to and bottom halves, resectively. d h A little exansion will show that Q A i = det ( Mi Ax N i Ay will give a binary quadratic form for which Disc ( ( ( Q A 1 = Disc Q A 2 = Disc Q A 3, so we can define Disc (A = Disc ( Q A 1. This discriminant can be shown to be the only invariant under the natural action of SL 2 (Z SL 2 (Z SL 2 (Z, which is shown to be equivalent to a more geometrically intuitive action by way of this cube corresondence in [3]. After a lot of work, one can show that, given three binary quadratic forms of the same discriminant D, after modding out by an equivalence relation, this uniquely determines the cube that would ma to them, u to another equivalence relation, i.e. there is a bijection between Cl ( Z 2 Z 2 Z 2 ; D ( ((Sym and Cl 2 Z 2 3 ; D. Moreover, in [3], it is shown how grou laws can be formed on Cl ( Z 2 Z 2 Z 2 ; D ( (Sym and Cl 2 Z 2 ; D such that the above ma comosed with any rojection is a grou homomorhism. References [1] J. A. Beachy. Introductory lectures on rings and modules sulement. htt:// May Unublished sulement: Chater 5: Commutative rings. [2] M. Bhargava. Gauss comosition and generalizations. Lecture Notes in Comuter Science, 2369:1 8, [3] M. Bhargava. Higher comosition laws ı: A new view on gauss comosition, and quadratic generalizations. Annals of Mathematics, 159: , [4] D. S. Dummit and R. M. Foote. Abstract Algebra. John Wile & Sons, Inc., New York, second edition, [5] S. Lang. Algebraic Number Theory. Addison-Wesley Series in Mathematics. Addson-Wesley Publishing Comany, Inc., Redding, Massachusetts, June [6] G. B. Mathews. Theory of Numbers. Chelsea Publication Comany, New York, second edition, 19. Originally ublished as Theory of numbers, art I. [7] J. Milne. Algebraic number theory. htt:// Course notes. [8] M. B. Nathanson. Elementary Methods in Number Theory, volume 195 of Graduate Texts in Mathematics. Sringer-Verlag, New York, [9] J. Neukirch. Algebraic Number Theory, volume 322 of A Series of Comrehensive Studies in Mathematics. Sringer-Verlag, Berlin, March [10] D. G. Northcott. Ideal Theory, volume 42 of Cambridge Texts in Mathematics and Mathematical Physics. Cambridge at the University Press, Cambridge, 1953.