Commutative Algebra and Algebraic Geometry. Robert Friedman

Transcription

1 Commutative Algebra and Algebraic Geometry Robert Friedman August 1, 2006

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3 Disclaimer: These are rough notes for a course on commutative algebra and algebraic geometry. I would appreciate all suggestions concerning typos, errors, unclear exposition or obscure exercises, and any other helpful feedback. i

4 Contents 1 Introduction to Commutative Rings Introduction Prime ideals Local rings Factorization in integral domains Motivating questions The prime and maximal spectrum Graded rings and projective spaces Exercises Modules over Commutative Rings Basic definitions Direct and inverse limits Exact sequences Chain conditions Exactness properties of Hom Modules over a PID Nakayama s lemma The tensor product Products of affine algebraic sets Flatness Exercises Localization Basic definitions Ideals in a localization Localization of modules Local properties of rings and modules ii

5 CONTENTS iii 3.5 Regular functions Introduction to sheaves Schemes and ringed spaces Proj as a scheme Zariski local properties Exercises Integral Ring Homomorphisms Definition of an integral homomorphism The going up theorem and dimension The going down theorem More on dimension Noether normalization, version Noether normalization, version Exercises Ideals in Noetherian Rings Irreducible sets and radical ideals Associated primes Primary decomposition Artinian rings Fractional ideals and invertible modules Dedekind domains Higher dimensions Extensions of Dedekind domains Exercises Quasiprojective varieties Projective and quasiprojective varieties Local rings and tangent spaces Derivations, differentials, and tangent spaces Elementary projective geometry Products Grassmannians Exercises

6 iv CONTENTS 7 Graded Rings Filtrations and the Artin-Rees lemma Hilbert functions The degree of a projective variety The dimension theorem Applications of the dimension theorem Regular local rings Completions Hensel s lemma and the implicit function theorem Exercises

7 Chapter 1 Introduction to Commutative Rings 1.1 Introduction Commutative algebra is primarily the study of those rings which most naturally arise in algebraic geometry and number theory.for example, let k be a field (typically algebraically closed, and often the field C of complex numbers). Then (affine) algebraic geometry is to a large extent the study of the ring R = k[x 1,..., x n ] and associated objects, for example ideals I in R or the corresponding quotients R/I. In number theory, one studies the ring Z as well as related rings, for example Z[i], where i = 1, or Z[ 1+ 3 ]. There 2 are other kinds of associated rings, for example the field of quotients Q of Z or of k[x 1,...,x n ] (which is denoted by k(x 1,..., x n ), and called the field of rational functions in n variables). More complicated rings derived from the standard ones are for example the ring of p-adic integers Z p (defined for every prime number p), as well as its polynomial analogue, the ring of formal power series in n variables, denoted by k[[x 1,..., x n ]]. We will discuss the construction of these and other rings later. Although the rings above are very disparate, they have a great deal of structure in common, and one of the goals of commutative algebra is to elucidate this structure. One obvious property that all of the above rings share is that they are commutative, and as its name suggest, commutative algebra is almost exclusively concerned with such rings. In fact, from now on we shall always make the assumptions: 1

8 2 CHAPTER 1. INTRODUCTION TO COMMUTATIVE RINGS Assumption: Every ring R is commutative, with a unity 1. (Note that the zero ring 0 is allowed.) Every ring homomorphism ϕ: R S is required to satisfy ϕ(1) = 1. In other words, ϕ takes the unique unity in R to the corresponding unity in S. In particular, if R is a subring of R, then we require that 1 R. Thus, for example, if R 1 and R 2 are rings, then the Cartesian product R 1 R 2 is also a ring, with unity (1, 1), and the projections π i : R 1 R 2 R i are ring homomorphisms in the above sense. However, in case R 1 and R 2 are nonzero, the subsets R 1 {0} and {0} R 2 are not subrings in the sense described above. In more technical terms, R 1 R 2 is a product in the category of commutative rings (with unity), but not a coproduct. We shall describe the coproduct later. Quite often, rings come in pairs. Thus, implicit in the definition of k[x 1,...,x n ] is the ring homomorphism k k[x 1,..., x n ]; likewise, there are the obvious homomorphisms Z Z[i] or Z Q or Z Z/nZ. Given a ring homomorphism ϕ: R S, not necessarily injective, we call S an R-algebra, and refer (if there is any ambiguity) to the given homomorphism ϕ as the structural homomorphism. Morphisms, subalgebras, and quotients are required to be compatible with the given structural homomorphism. For example, if I is an ideal in R, then R/I is an R-algebra via the natural projection R R/I. Another standard example of an R-algebra is the ring R[x 1,...,x n ] of polynomials in several variables with coefficients in R, with the obvious inclusion R R[x 1,..., x n ]. Here, R[x 1,...,x n ] is the set of all expressions a 1,...,a n 0 r a1,...,a n x a 1 1 xan n, where r a1,...,a n R and r a1,...,a n 0 for only finitely many indices a 1,...,a n. More formally, we can think of R[x 1,..., x n ] as the set of all multi-sequences in R, indexed by n-tuples of nonnegative integers, and such that only finitely many terms are nonzero. Addition of polynomials is defined componentwise: ra1,...,a n x a 1 1 x an n + s a1,...,a n x a 1 1 x an n = (r a1,...,a n +s a1,...,a n )x a 1 1 x an n. Multiplication is defined so as to insure that (x a 1 1 xan n )(xb 1 1 x bn n ) = xa 1+b 1 1 x an+bn n.

9 1.1. INTRODUCTION 3 This can be written efficiently with vector notation: if a = (a 1,...,a n ) is an n-tuple of nonnegative integers, and we denote x a 1 1 x an n by x a, then ( )( ) r a x a s a x a = r b s c x b+ c. a a b+ c= a a If, in the definition of R[x 1,..., x n ], we omit the requirement that only finitely many coefficients r a1,...,a n are nonzero, we obtain a larger ring containing R and R[x 1,...,x n ], the ring of formal power series R[[x 1,...,x n ]]. The R-algebra R[x 1,...,x n ] has the following universal property with respect to R-algebras: if S is an R-algebra, to give an R-algebra homomorphism ϕ: R[x 1,...,x n ] S is equivalent to specifying n elements α 1,...,α n S. In one direction, the α i are just given by ϕ(x i ), so that ϕ determines n elements α 1,...,α n S. Conversely, given n elements α 1,...,α n S, the evaluation homomorphism ev α1,...,α n : R[x 1,...,x n ] S defined by ev α1,...,α n (f(x 1,...,x n )) = f(α 1,...,α n ) defines a homomorphism ϕ: R[x 1,...,x n ] S, and these constructions are clearly inverse. In general, if an R-algebra S is the quotient of R[x 1,...,x n ] (in the sense of R-algebras), then we say that S is a finitely generated R- algebra. If in addition there exists a surjection R[x 1,...,x n ] S whose kernel is a finitely generated ideal of R[x 1,...,x n ], then we say that S is a finitely presented R-algebra. Both algebraic geometry and algebraic number theory are very much relative theories, often concerned not just with a single ring R but rather with a ring R and an R-algebra S, and we shall try to emphasize this aspect a much as possible. The set of ideals I in a ring R, and in particular the sets of prime and maximal ideals, are among the most important objects of study in algebra. They can be generalized to modules: roughly speaking, a module over a ring R is the formal analogy of a vector space over a field k: Definition An R-module M consists of an abelian group M (with group operation denoted by addition), together with a function R M M, whose value at (r, m) is denoted by r m or rm, and is usually referred to as multiplication, satisfying: (i) For all r, s R and m M, r(sm) = (rs)m;

10 4 CHAPTER 1. INTRODUCTION TO COMMUTATIVE RINGS (ii) For all r 1, r 2 R and m M, (r 1 + r 2 )m = r 1 m + r 2 m; (iii) For all r R and m 1, m 2 M, r(m 1 + m 2 ) = rm 1 + rm 2 ; (iv) For all m M, 1 m = m. A submodule N of an R-module module M is defined in the obvious way. If N is a submodule of an R-module module M, then there is a natural structure of an R-module on the abelian group M/N, and we will refer to this structure as the quotient R-module. If M and N are R-modules, then a homomorphism f : M N is a homomorphism of abelian groups satisfying f(rm) = rf(m) for all r R and m M. Isomorphisms of R-modules are defined similarly. We write M = N if M and N are isomorphic. Example The following are standard examples of modules: (i) If k is a field, then a k-module V is the same thing as a k-vector space. (ii) A Z-module M is the same thing as an abelian group. Indeed, given a Z-module M, we associate to M its underlying abelian group (also denoted M). Conversely, if A is an abelian group, for all a A and n Z, defining n a in the usual way gives A the structure of a Z- module. (iii) For R an arbitrary ring, an R-submodule of R is the same thing as an ideal I in R, and a quotient module of R is the same as a quotient R/I. (iv) Again with R arbitrary, the n-fold Cartesian product R n of all ordered n-tuples of elements of R is an R-module in the usual way, via componentwise addition and scalar multiplication: (r 1,...,r n ) + (s 1,...,s n ) = (r 1 + s 1,..., r n + s n ); r(r 1,...,r n ) = (rr 1,...,rr n ). The module R n is called the free R-module of rank n. (v) If M 1 and M 2 are two R-modules, we define the direct sum M 1 M 2 to be the set {(m 1, m 2 ) : m i M i } with addition and scalar multiplication defined componentwise. Thus as a set M 1 M 2 is just the Cartesian

11 1.2. PRIME IDEALS 5 product of M 1 and M 2. The direct sum M 1 M n of finitely many R-modules is defined in a similar way. In particular, R n = R R is the direct sum of n copies of R. We shall consider more general kinds of sums and products of modules in the next chapter. However, at the end of this chapter, we shall need the case of an infinite direct sum indexed by N (or by the set Z 0 of nonnegative integers). Given a collection M n, n N of R-modules, we define n N M n to be the set of all sequences (m 1, m 2,...), with m i M i and m n = 0 for all sufficiently large n, or equivalently for all but finitely many n. Addition and scalar multiplication are, as usual, defined pointwise. Just as it is important to understand all ideals in a ring, we will also try to describe all R-modules up to isomorphism. While this is not in general possible, it is feasible for special classes of rings R and R-modules M, and we will frequently return to this question. 1.2 Prime ideals Fix a ring R. The intersection of an arbitrary collection {I α } α A of ideals in R is an ideal α A I α. Given ideals I and J of R, we can also define the ideal product I J = { i r i s i : r i I, s i J}, and it is an ideal in R contained in I J. For example, RI = I. Products of a finite or infinite number of ideals, and powers I N of an ideal I, are defined similarly. Ideal product is associative, i.e. (IJ)K = I(JK). We can also define the ideal sum I + J = {r + s : r I, s J}. It is the smallest ideal of R containing both I and J. Moreover, the distributive laws hold: I(J + K) = IJ + IK. Infinite sums α A I α are defined similarly.

12 6 CHAPTER 1. INTRODUCTION TO COMMUTATIVE RINGS Given r 1,...,r n R, let (r 1,...,r n )R = (r 1,...,r n ), the ideal generated by r 1,..., r n, be the smallest ideal containing r 1,...,r n. Clearly (r 1,...,r n )R = { i t i r i : t i R}. In particular rr = (r) = {tr : t R} is called the principal ideal generated by r. There is a similar notation for ideals generated by infinitely many elements. Definition A zero divisor in R is an element r R, r 0 such that there exists an s R with s 0 and rs = 0. The ring R is an integral domain if R 0 and R has no zero divisors, in other words if given r, s R with rs = 0, then either r or s is zero. For example, Z is an integral domain, and Z/nZ is an integral domain if and only if n is prime. A field is an integral domain; in particular, a field always has at least two distinct elements 0 and 1. An ideal p in a commutative ring R is a prime ideal if p R and, for all r, s R, if rs p then either r p or s p. Clearly, p is a prime ideal if and only if R/p is an integral domain. A subring of an integral domain is again an integral domain. A field is an integral domain, and therefore every subring of a field is also an integral domain. Conversely, every integral domain R is contained in a field. In fact, one constructs the fraction field or field of quotients of an integral domain R by analogy with the construction of the rational numbers from the integers, and this field is in a sense the smallest field containing R. We shall describe this construction later. Note that if R is an integral domain, then so is R[x] (by considering terms of top degree) and therefore also R[x 1,..., x n ] by induction. In particular, if k is a field, then k[x 1,...,x n ] is an integral domain whose quotient field k(x 1,..., x n ) is called the field of rational functions in x 1,...,x n. Lemma Given a ring homomorphism f : R S and a prime ideal q in S, the ideal f 1 (q) is a prime ideal in R. Proof. The ring R/f 1 (q) is a subring of S/q and hence is an integral domain. Definition A nilpotent element of R is an element r such that r n = 0 for some n > 0. The ring R is reduced if the only nilpotent element is 0.

13 1.2. PRIME IDEALS 7 The radical of R (written 0) is the set of r R such that there exists an n > 0 such that r n = 0, i.e. such that r is nilpotent. Thus R is reduced if and only if 0 = 0. More generally, for an ideal I, we define the radical of I to be the set I = {r R : r n I for some n > 0 }. Thus I is the preimage in R of the set of nilpotent elements of R/I. An ideal I is a radical ideal if I = I. The ideal I is a radical ideal if and only if R/I is reduced. Lemma For every ideal I, I is an ideal of R. Proof. First consider the case I = (0). If r n = 0 and s m = 0, then (r + s) n+m = 0, by applying the binomial theorem to (r + s) n+m. Thus 0 is closed under addition, and if r n = 0, then clearly (sr) n = 0 for all s R. So 0 is an ideal. For an arbitrary ideal I, I is the inverse image of the radical of R/I under the natural map R R/I, and so is an ideal in R. Definition A maximal ideal m of R is an ideal m such that m R, and if I is an ideal such that m I, then I = m or I = R. Lemma The ideal m is maximal if and only if R/m is a field. Hence a maximal ideal is prime. Proof. The ideal m is maximal if and only if R/m has no proper ideals. Thus we must show that a nonzero commutative ring k is a field if and only if it has no proper ideals. Clearly if k is a field, then it has no proper ideals. Conversely, if k is a commutative ring without proper ideals, then given a k, a 0, the ideal (a) is a nonzero ideal and so (a) = R. Thus 1 (a), so that there exists x R with ax = 1. Thus k is a field. The following will be used repeatedly: Lemma If R is a commutative ring and I is an ideal of R such that I R, then I is contained in a maximal ideal m. Proof. Consider the set of all ideals containing I and not equal to R, partially ordered by inclusion. We claim that we can apply Zorn s lemma to this set to produce a maximal element, in other words a maximal ideal. It suffices to note that if J α, α A is a totally ordered set of proper ideals containing I, then α J α is an ideal (because every two elements are contained in a J β

14 8 CHAPTER 1. INTRODUCTION TO COMMUTATIVE RINGS for some β, so we can add), containing I, and R (since if 1 α J α, then 1 J α for some α). Thus the hypotheses of Zorn s lemma are satisfied, and a maximal element of this set is a maximal ideal containing I. Corollary If R 0, then there exists a maximal ideal of R. Corollary An element u of R is a unit if and only if u does not lie in any maximal ideal. Proof. If u is a unit, then an ideal containing u contains 1 and so is equal to R, so that u is not contained in a maximal ideal. Conversely, if u is not a unit, then (u) is a proper ideal and thus it is contained in a maximal ideal m. So u m. Another application of Zorn s lemma gives another description of the radical of an ideal: Lemma Let I R be an ideal. Then I is the intersection of the prime ideals containing I. Proof. First consider the case where I = (0) and 0 is the ideal of all nilpotent elements of R. Clearly, if r is nilpotent, then r p for every prime ideal p of R. Conversely, suppose that r is not nilpotent. Consider the set I of all ideals J of R such that, for all n > 0, r n / J (note that such an ideal is automatically not equal to R). Clearly I, since (0) I. A standard application of Zorn s lemma to I, partially ordered by inclusion, shows that I contains a maximal element p. To see that p is prime, we must show that for all a / p, b / p, the product ab / p. Since a / p, the ideal (a) + p strictly contains p and hence does not lie in I. Thus there exist s 1 R, p 1 p, and n 1 > 0 such that r n 1 = as 1 + p 1. Likewise there exist s 2 R, p 2 p, and n 2 > 0 such that r n 2 = bs 2 + p 2. But then r n 1+n 2 = abs 1 s 2 + p for some p p, and so (ab) + p strictly contains p as well. Thus ab / p. The case of a general I now follows by applying the above to the ring R/I and noting that the prime ideals of R containing I are in one-to-one correspondence with the prime ideals of R/I. Corollary An ideal I is an intersection of prime ideals if and only if I is a radical ideal, i.e. I = I.

15 1.3. LOCAL RINGS 9 Proof. Clearly, if I = I, then I = I is the intersection of all of the prime ideals containing I. Conversely, if I is an intersection of prime ideals, then an easy exercise shows that I = I. In fact, if R has stronger finiteness hypotheses, for example if R = k[x 1,...,x n ], then every radical ideal is an intersection of finitely many prime ideals, and the prime ideals which appear are essentially unique if we impose an appropriate minimality condition. This is one analogue of unique factorization into primes. For ideals which are not radical ideals, there is a more subtle kind of factorization, called primary decomposition, which we shall discuss in Chapter Local rings Definition A ring R is local if there is a unique maximal ideal m of R. For example, a field k is a local ring. If R is a local ring with maximal ideal m and S is a local ring with maximal ideal n, then a ring homomorphism ϕ: R S is local if ϕ(m) n. Note that ϕ(m) n if and only if m ϕ 1 (n), if and only if m = ϕ 1 (n) since n and hence ϕ 1 (n) are proper ideals. The following is immediate from Corollary Lemma R is local if and only if the set of all r in R such that R is not a unit is an ideal of R. For example, the ring k[[x 1,...,x n ]] of formal power series with coefficients in a field k is a local ring with maximal ideal m = (x 1,...,x n ). It suffices to see that, if f(x 1,...,x n ) k[[x 1,...,x n ]] has constant term equal to 1, then f is invertible. This follows by formally taking the inverse: g = 1 g + g2, and this infinite series makes sense if g m. Likewise the ring Z p of p-adic integers is a local ring. We shall discuss such rings in more detail later, and give more elementary methods for constructing local rings. Local rings are of fundamental importance in commutative algebra and many questions about rings and modules can be reduced to questions about local rings. More generally, there is the following definition:

16 10 CHAPTER 1. INTRODUCTION TO COMMUTATIVE RINGS Definition The Jacobson radical of R is the intersection of all the maximal ideals of R. For example, if R is local with maximal ideal m, then its Jacobson radical is just m. Lemma The Jacobson radical of R is the set of all r R such that, for every s R, 1 + rs is a unit. Proof. Suppose that r lies in m for every maximal ideal m. Then so does rs for every s R. Thus 1 + rs / m, since otherwise 1 m. It follows that 1 + rs is contained in no maximal ideal, and is thus a unit. Conversely, suppose that r is not contained in some maximal ideal m. Then the ideal (r, m) generated by r and m is an ideal properly containing m and so is R. Hence we can write 1 = t + ry where t m and y R. Thus 1 + r( y) m is not a unit. 1.4 Factorization in integral domains Throughout this section, R denotes an integral domain. Our goal is to collect some elementary results about factorization. Definition The element r R divides r (which we write as r r ) if there exists x R such that r = xr. Two elements r, r are associates if r = ur, where u is a unit. Equivalently, r and r are associates if (r) = (r ). An element r R {0} is irreducible if r is not a unit and if whenever r = st for s, t R, then s or t is a unit. Thus r 0 is irreducible if and only if the only elements of R which divide r are either units or associates of r. Lemma If (r) is a prime ideal and r 0, then r is irreducible. Proof. If r = st, then s or t lies in (r). We may suppose that s (r), say s = ru. Then r = rut, so that ut = 1 and u is a unit. However, the converse to the above need not hold: if r is irreducible, then (r) need not be a prime ideal. For example, in Z[ 5], it is easy to check that 2 is irreducible, but (2) is not prime since 6 = 2 3 = (1+ 5)(1 5), and neither nor 1 5 is in (2). One important case where this does hold is the case where R has unique factorization:

17 1.4. FACTORIZATION IN INTEGRAL DOMAINS 11 Definition An integral domain R is a unique factorization domain (UFD) if, given r not a unit in R, r = p 1 p n, where the p i are irreducible elements, and moreover if r = q 1 q m, where the q i are also irreducible, then n = m and possibly after relabeling the q i, p i and q i are associates for all i. In a UFD, two nonzero elements r and s are relatively prime if, whenever t r and t s, then t is a unit. We can define a greatest common divisor (gcd) of two nonzero elements r and s to be an element d such that d r, d s, and whenever e r and e s, then e d. Two gcds of r and s are associates. Greatest common divisors for a finite collection of elements of R are defined similarly. An easy argument shows that gcds exist in a UFD. The gcd of tr 1,...,tr n is t times the gcd of r 1,..., r n. If r is an irreducible element and s, t 0, if r st, then by unique factorization r s or r t. Lemma Let R be a UFD. Then r is irreducible if and only if (r) is a nonzero prime ideal. Moreover, in this case (r) is a minimal nonzero prime ideal, and every minimal nonzero prime ideal p is of the form (r) where r is irreducible. Proof. We have seen that (r) nonzero and prime = r irreducible. Conversely, suppose that r is irreducible, and suppose that st (r). Then st = rx for some x R, and we claim that either s or t lies in R. We may clearly assume that s and t are both nonzero. Then as r st, either r s or r t and so either s or t lies in (r). Now suppose that p is any nonzero prime ideal. Then p contains a nonzero element x which is not a unit. At least one irreducible factor r of x must lie in p. Then (r) p, so that every nonzero prime ideal p contains a nonzero prime ideal of the form (r). Thus if p is a minimal nonzero prime ideal, then p = (r) for some irreducible element r. If now r is irreducible and p is a nonzero prime ideal contained in (r), then there exists an irreducible r such that (r ) p (r). Thus r = ru for some u R. Since r is irreducible and r is not a unit, u is a unit and thus (r) = (r ) = p. Thus (r) is minimal. Basic examples of UFD s come from the following definition: Definition An integral domain R is a principal ideal domain (PID) if every ideal in R is principal.

18 12 CHAPTER 1. INTRODUCTION TO COMMUTATIVE RINGS For example, the ring Z is a PID. Likewise, if k is a field, then k[x] is a PID. In fact, for a field k we always have the following (long division with remainder for polynomials): Proposition Let k be a field and let f(x), g(x) k[x] with g 0. Then there exist unique q, r k[x] such that f = gq + r with r = 0 or deg r < deg g. Proof. Existence: induct on deg f, the cases f = 0 or deg f < deg g = d are clear. Write g = d i=0 a ix i, with a d 0, and suppose that f = n i=0 b ix i with n d and b n 0. Then h = f a 1 d b nx n d g has degree strictly less than n (or is zero). Thus h = gq + r with r = 0 or deg r < deg g, and so f = g(q + a 1 d b nx n d ) + r = gq + r as desired. To see uniqueness, if gq+r = gq +r with r = 0 or deg r < deg g as well, then g (r r ) and deg(r r ) < deg g, which is only possible if r r = 0, so that r = r. In this case gq = gq, and so q = q as well. Remark More generally, if R is any ring (not necessarily an integral domain), and g(x) is monic (i.e. the leading coefficient of g is 1), then the statement and proof of the proposition continue to hold. Corollary If k is a field, then k[x] is a PID. Proof. If I is an ideal of k[x] and I 0, choose a nonzero element g I of minimal degree. If f I, write f = gq+r with deg r < deg g or r = 0. Since r = f gq I, we must have r = 0 by the choice of g. Thus I = (g). As usual, we define a root of f k[x] in a field K k to be an element α K such that f(α) = 0. Corollary If α k, then α is a root of f if and only if (x α) divides f. If f k[x] has degree n, then there are at most n elements α k such that f(α) = 0. Proof. Given α k, we can write f(x) = (x α)q + r, where r k is a constant polynomial. Thus f(α) = r, and so f(α) = 0 if and only if (x α) f. Clearly, if α 1 α 2, the polynomials (x α 1 ) and (x α 2 ) are irreducibles which are not associates. Thus if α 1,...,α k are roots of f, then (x α 1 ) (x α k ) f. It follows that k deg f.

19 1.4. FACTORIZATION IN INTEGRAL DOMAINS 13 Corollary If k is an infinite field and f k[x] induces the zero function from k to itself, then f = 0. We leave the corresponding statement about f k[x 1,...,x n ] as an exercise. We now show that every principal ideal domain is in fact a unique factorization domain. Theorem A PID is a UFD. Proof. The proof will be in several steps. Lemma Let R be an integral domain with the property that, if (a 1 ) (a 2 ) (a n ) (a n+1 ) is an increasing sequence of principal ideals, then the sequence is eventually constant. Then every nonzero r R which is not a unit factors into a product of irreducibles. We can paraphrase the hypothesis of the lemma by saying that R satisfies the ascending chain condition (a.c.c) on principal ideals. Proof of Lemma If not, let r R be an element, not zero or a unit, which does not factor into a product of irreducibles. In particular, r itself is not irreducible, so that r = r 1 s 1 where neither r 1 nor s 1 is a unit. Thus (r) is properly contained in (r 1 ) and in (s 1 ). Clearly, we can assume that at least one of r 1, s 1, say r 1, does not factor into irreducibles (if both so factor, so does the product). By applying the above to r 1, we see that (r 1 ) is strictly contained in a principal ideal (r 2 ), where r 2 does not factor into a product of irreducibles. Continuing in this way, we can produce an infinite chain of principal ideals (r 1 ) (r 2 ), each properly containing the previous one, contradicting the hypothesis on R. Lemma Suppose that R is an integral domain in which every nonzero element which is not a unit factors into a product of irreducibles. Moreover, suppose that r R is irreducible if and only if (r) is prime. Then R is a UFD. Proof. It suffices to show, in the usual way, that if an irreducible element r divides a product s 1 s n, then r divides s i for some i. This is an immediate consequence of the fact that (r) is prime.

20 14 CHAPTER 1. INTRODUCTION TO COMMUTATIVE RINGS To complete the proof of Theorem , it suffices to show that a PID R satisfies the hypotheses of Lemma and Lemma First suppose that (r 1 ) (r 2 ) is an increasing sequence of ideals of R. It is easy to check that I = i (r i) is again an ideal. Since R is a PID, I = (r) for some r R. Necessarily r (r N ) for some N. But then (r) (r N ) (r N+1 ) i (r i) = (r). Thus all inclusions are equalities, and (r n ) = (r N ) for all n N. Thus the sequence is eventually constant. Finally we must show that, if r R is irreducible, then (r) is a prime ideal. We begin by showing that gcds exist in R and are of a very special form: Lemma Let R be a PID, and let a, b R. Let d be a generator of the ideal (a, b), so that d = ar + bs for some r, s R. Then d r, d s, and if e R is such that e r and e s, then e d. Proof. Since a (a, b) = (d), d a, and by symmetry d b. Moreover, if e r and e s, then e ar + bs for all r, s R, and hence e d. For a, b R, let gcd(a, b) = d where d is any generator of (a, b). Clearly, d is uniquely specified up to a unit. Moreover, d is uniquely specified by the property that d r, d s, and if e R is such that e r and e s, then e d. Corollary Let R be a PID, and let r R be an irreducible. Suppose that r st for some s, t R. Then either r s or r t. Proof. Since gcd(r, s) r, either gcd(r, s) is an associate of r or gcd(r, s) is a unit. In the first case, r s. In the second case, write 1 = rx + sy for some x, y R. Then t = rxt + sty. Since r st, r t. Now suppose that r is irreducible, and that st (r), i.e. that r st. Since r is irreducible, it must divide one of s, t, and thus either t or s lies in (r). Thus (r) is a prime ideal. This then finishes the proof of Theorem In a PID, the gcd d of two elements r and s exists and is of the form ar + bs. This result fails for most UFD s. For example, in k[x, y], the gcd of x and y is 1, but 1 cannot be written as ax + by. Related to the above is the following special property of a PID: Lemma If R is a PID, every nonzero prime ideal of R is maximal.

21 1.4. FACTORIZATION IN INTEGRAL DOMAINS 15 Proof. A nonzero prime ideal p in R is of the form (r) where r is irreducible. If p I, where I is an ideal and I R, then I = (r ) for some r. Since r r and r is irreducible, r and r must be associates. Thus I = p and so p is maximal. The ascending chain condition and the arguments of Theorem are so fundamental that we generalize them as follows: Proposition For a ring R, the following two conditions are equivalent: (i) Every ideal I of R is finitely generated: if I is an ideal of R, then I = (r 1,...,r n ) for some r i R. (ii) Every increasing sequence of ideals is eventually constant, in other words if I 1 I 2 I n I n+1, where the I n are ideals of R, then there exists an N N such that for all k N, I k = I N. If the ring R satisfies either of the equivalent conditions above, then R is called a Noetherian ring. Proof. (i) = (ii): given an increasing sequence of ideals I 1 I 2, let I = n I n. Then as in the proof of Theorem , I is an ideal, and hence I = (r 1,..., r n ) for some r i R. Thus r i I ni for some n i. If N = max i n i, then r i I N for every i. Hence, for all k N, I = (r 1,...,r n ) I N I k I. It follows that I k = I N = I for all k N. (ii) = (i): Let I be an ideal of R and choose r 1 I. Set I 1 = (r 1 ). If I = I 1, stop. Otherwise there exists an r 2 I I 1. Set I 2 = (r 1, r 2 ). If I = I 2, stop, otherwise there exists an r 3 I I 2. Inductively suppose that we have found I k = (r 1,...,r k ) with I k I. If I = I k we are done, otherwise there exists r k+1 I I k and we set I k+1 = (r 1,..., r k+1 ). So if I is not finitely generated, we have constructed a strictly increasing sequence I 1 I 2, contradicting the assumption on R. Thus I is finitely generated. Clearly, the proof of Lemmas and together with Lemma imply the following:

22 16 CHAPTER 1. INTRODUCTION TO COMMUTATIVE RINGS Theorem Suppose that R is a Noetherian integral domain. Then R is a UFD if and only if, for every nonzero r R, the element r is irreducible if and only if (r) is prime. Next we turn to the study of polynomials over a UFD. The main result here is: Theorem If R is a UFD, then so is R[x]. By induction, we have the obvious corollary: Corollary The rings Z[x 1,...,x n ] and k[x 1,...,x n ] for k a field are UFD s. We note that Z[x 1,..., x n ] is not a PID if n 1 and that, if k is a field, then k[x 1,...,x n ] is not a PID if n 2. Proof of Theorem We begin with a preliminary definition. Given a nonzero polynomial f(x) = i a ix i R[x], define the content c(f) to be the gcd of the coefficients a i. Here c(f) is well-defined up to a unit, or as an element of the multiplicative set R/R. In general f(x) = c(f)f 0 (x), where the gcd of the coefficients of f 0 is 1. Define f to be primitive if c(f) = 1, in other words if the gcd of the coefficients of f is 1. If f = rf 0 where f 0 is primitive, then c(f) = r. More generally, if r R, r 0, then c(rf) = rc(f). We will show: Claim Let R be a UFD with quotient field K. (i) If r R is irreducible, then r is an irreducible element of R[x]. (ii) If f 0 (x) R[x] is a primitive polynomial which is irreducible in K[x], then f 0 (x) is an irreducible element of R[x]. (iii) Every element of R[x] factors into a product of elements which are either irreducible elements in R or primitive polynomials which are irreducible in K[x]. Moreover such a factorization is unique up to order and associates. Clearly the claim implies Theorem , indeed somewhat more as it describes all of the irreducibles in R[x]. Proof of Claim It is easy to check that all of the elements described in (i) and (ii) of the claim are indeed irreducible. Next we prove:

23 1.4. FACTORIZATION IN INTEGRAL DOMAINS 17 Lemma (Gauss lemma). If f and g are primitive, then fg is primitive. Proof. Let r be an irreducible element of R such that r divides all coefficients of fg. Given a polynomial in R[x], we use a bar to denote its reduction in R/rR, which is an integral domain since r is irreducible and R is a UFD. By hypothesis fg = 0, and so fḡ = 0. Since (R/rR)[x] is an integral domain, either f = 0 or ḡ = 0. In particular either f or g is not primitive, a contradiction. Corollary For all f, g R[x], c(f g) = c(f)c(g). Proof. If f = c(f)f 0 and g = c(g)g 0 where f 0 and g 0 are primitive, then fg = c(f)c(g)f 0 g 0, where, by the lemma, f 0 g 0 is primitive. Hence c(fg) = c(f)c(g). Returning to the proof of Claim , given f R[x], let g 1,...,g n be the irreducible factors of f in K[x]. After clearing denominators, we can write g i = λ i h i with h i R[x], and by factoring out c(h i ) we can assume that h i is primitive. So f = (λ 1 λ k )h 1 h k, where the λ i K and the h i R[x] are primitive. Thus h 1 h k is primitive as well, by the Gauss lemma and induction. Now there is r R such that r(λ 1 λ k ) R. Thus c(rf) = rc(f) = c(r(λ 1 λ k )h 1 h k ) = r(λ 1 λ k ). It follows that c(f) = λ 1 λ k and in particular that λ 1 λ k R. We have now written f as ch 1 h k where the h i are primitive polynomials in R[x] which are irreducible in K[x]. If we further factor c into irreducibles in R, we have a factorization as claimed. We must finally show that this factorization is unique. Suppose that f = r 1 r n h 1 h k = r 1 r mh 1 h l, where the r i are irreducible in R and the h i are primitive polynomials in R[x] which are irreducible in K[x]. Up to a unit r 1 r n = r 1 r m since they are both equal to c(f), and so we can reorder until n = m and r i = r i up to associates by unique factorization in R. We also know by unique factorization

24 18 CHAPTER 1. INTRODUCTION TO COMMUTATIVE RINGS in K[x] that k = l and that after reordering we can assume that h i and h i are associates in K[x]. Thus h i = th i for some t K, t 0. Choosing an s such that st R, we have sh i = sth i, and taking contents we find that s and st are associates. Thus t is a unit in R, and so h i and h i are associates in R[x]. As an easy corollary of Theorem , we can prove the famous Eisenstein criterion for a polynomial to be irreducible: Theorem (Eisenstein criterion). Let R be a UFD with quotient field K, let r R be an irreducible and let f = n i=0 a ix i R[x] satisfy: (i) For all i < n, r a i ; (ii) r a n, so that a n 0 and hence n = deg f; (iii) r 2 a 0. Then f is irreducible in K[x]. Proof. Since r a n, the gcd of r and c(f) is a unit. Thus, if we write f = c(f)f 0, then f 0 is a primitive polynomial satisfying the same hypotheses as f. So we may as well assume that f is primitive in R[x]. In this case, f is irreducible in K[x] if and only if it is irreducible in R[x]. As usual, we shall let a bar stand for the image of a polynomial in (R/rR)[x]. If f = gh where g, h R[x] and both g and h have positive degree, then f = gh = ḡ h. On the other hand, f = cx n, where c 0. Thus ḡ = c 1 x d and h = c 2 x n d, and both have constant term equal to zero. It follows that r divides the constant terms of both g and h, so that r 2 a 0. This contradicts (iii). Thus f is irreducible. 1.5 Motivating questions in algebraic geometry and number theory Now that we have some preliminary results concerning commutative rings, we describe some of the questions which arise naturally in the two main applications of commutative algebra. We begin with the case of algebraic geometry. Fix an algebraically closed field k (the classical case is k = C). In its crudest form, algebraic geometry

25 1.5. MOTIVATING QUESTIONS 19 seeks to study the common zeroes of polynomials in n variables. Traditionally, the vector space k n is denoted by A n k in algebraic geometry, and is called affine n-space (over k). The allowable functions on this space are the functions which can be defined algebraically via the field operations, namely the polynomial functions, and these form the ring k[x 1,...,x n ]. (It is easy to see, using the fact that k is algebraically closed, that a rational function f/g is well-defined on all of A n k if and only if it is a polynomial.) Given a polynomial f(x 1,..., x n ) k[x 1,...,x n ], its set of zeroes is V (f) = {(x 1,...,x n ) A n k : f(x 1,...,x n ) = 0}. More generally, given a finite collection f 1,...f k k[x 1,..., x n ], their set of common zeroes is defined in a similar way: V (f 1,...,f k ) = {(x 1,...,x n ) A n k : f i (x 1,...,x n ) = 0 for all i }. Thus V (f 1,...,f k ) = k i=1 V (f i). Clearly, V (f 1,...,f k ) only depends on the ideal I = (f 1,...,f k ) generated by the f i, and we shall also use the notation V (I) for V (f 1,...,f k ). A priori, it might seem that we could also consider infinite collections of polynomials. However, a famous theorem of Hilbert says that it is always enough to look at the common zeroes of finitely many. In the language of ideals, the theorem is as follows: Theorem (Hilbert s Basis Theorem). Every ideal in k[x 1,...,x n ] is finitely generated. In fact, we shall prove the following: Theorem (Hilbert s Basis Theorem II). If R is a Noetherian ring, then R[x] is Noetherian. Corollary If R is a Noetherian ring, then R[x 1,...,x n ] is Noetherian. In particular, the rings k[x 1,...,x n ] (k a field) and Z[x 1,...,x n ] are Noetherian. Proof of the basis theorem. Let I be an ideal in R[x]. Set I k = {r R : there exists f(x) I with f(x) = k a i x i and a k = r}. i=0

26 20 CHAPTER 1. INTRODUCTION TO COMMUTATIVE RINGS Clearly I k is an ideal in R, and I k I k+1 since if f(x) = k i=0 a ix i with a k = r, then xf(x) = k+1 i=1 a i 1x i, so that the coefficient of x k+1 is r as well. Thus, since R is Noetherian, there exists an n 0 such that I k = I n0 for all k n 0, and I n0 = (r 1,...,r A ) for r i R. By construction, for each i, 1 i A, there exists an f i (x) I of degree n 0 and leading coefficient r i. For k n 0, the ideal I k is finitely generated. Thus, after enlarging the collection {f i (x)} we can further assume that the ideals I k, k n 0, are generated by leading coefficients of suitable polynomials f i (x) I of degree k. We claim that I = (f 1 (x),...,f A (x)). Clearly (f 1 (x),...,f A (x)) I. Conversely, let f(x) I, and suppose that deg f(x) = k. By induction, starting with k = 0 n 0, we can assume that all polynomials in I of degree less than k lie in (f 1 (x),...,f A (x)). If k n 0, then the leading coefficient of f is in the ideal generated by the leading coefficients of the f i, and since each f i has degree n 0, it follows that f i xa i s i f i has degree < k for suitable integers a i and elements s i R. If k n 0, then by the construction of the f i there exists a subcollection of the f i of degree k such that f i s if i has degree < k for appropriate s i R. In all cases, there exists a g (f 1 (x),..., f A (x)) such that f g has degree < k and lies in I, and so by induction lies in (f 1 (x),...,f A (x)). Thus f (f 1 (x),...,f A (x)) as well. While the proof we have given of Hilbert s theorem is nonconstructive, a great deal of interest has centered on finding explicit generators for any given ideal, which might be the best in some sense. For example, what is the minimal number of generators of a given ideal? What are the minimal degrees needed? A subset X of A n k of the form V (I) will be called a (closed) algebraic set in A n k. We have the following: Lemma (i) If I 1 I 2, then V (I 2 ) V (I 1 ). (ii) V (0) = A n k and V (1) =. (iii) V (I 1 +I 2 ) = V (I 1 ) V (I 2 ). More generally, V ( α A I α) = α A V (I α). (iv) V (I 1 I 2 ) = V (I 1 I 2 ) = V (I 1 ) V (I 2 ). Proof. The proofs of (i) (iii) are easy and left to the reader. To see (iv), note that I 1 I 2 I 1 I 2 I 1. Thus V (I 1 ) V (I 1 I 2 ) V (I 1 I 2 ). Since the same holds for V (I 2 ), we have V (I 1 ) V (I 2 ) V (I 1 I 2 ) V (I 1 I 2 ).

27 1.5. MOTIVATING QUESTIONS 21 On the other hand, if p / V (I 1 ) V (I 2 ), then p / V (I 2 ) and p / V (I 2 ). So there exist g 1 I 1 such that g 1 (p) 0, and likewise g 2 I 2 such that g 2 (p) 0. Thus g 1 g 2 (p) 0, so that p / V (I 1 I 2 ). It follows that V (I 1 I 2 ) V (I 1 ) V (I 2 ), so that equality holds throughout. This proves (iv). Given a subset X of A n k, we can consider the ideal I(X) of all functions f k[x 1,..., x n ] such that f(p) = 0 for all p X. Clearly, we have: Lemma (i) If X 1 X 2, then I(X 2 ) I(X 1 ). (ii) I(X 1 X 2 ) = I(X 1 ) I(X 2 ). (iii) X V (I(X)). (iv) I I(V (I)). The ideal I(X) is interesting for the following reason. Just as with A n k, we want to describe those functions on X which can be defined algebraically. It is natural to look at restrictions of polynomials in k[x 1,...,x n ] to the subset X. (We will see later that it makes no difference if we try to enlarge this to the set of rational functions which are everywhere defined on X.) Two polynomials have the same restriction to functions on X if and only if their difference lies in I(X). Thus the natural ring of algebraically defined functions on X is the quotient ring k[x 1,...,x n ]/I(X). Definition Let X be a closed algebraic subset of A n k. We call X an affine algebraic set. The affine coordinate ring A(X) is the ring k[x 1,..., x n ]/I(X). Given a closed algebraic set X = V (I), what is the relation between I and I(X)? On the one hand, we have the following easy result: Proposition V (I(X)) is the smallest algebraic subset of A n k containing X. In particular, if X is an algebraic subset of A n k, then V (I(X)) = X. Proof. Clearly, V (I(X)) is an algebraic subset of A n k containing X. If Y = V (J) contains X, then J I(Y ) I(X), and hence V (I(X)) V (J) = Y. Thus V (I(X)) is the smallest algebraic subset of A n k containing X. The last statement is then clear. In this context, Hilbert proved another fundamental result, which we shall prove later:

28 22 CHAPTER 1. INTRODUCTION TO COMMUTATIVE RINGS Theorem (Hilbert s Nullstellensatz). In the above notation, I(V (I)) = I, where I is the ideal of k[x 1,...,x n ] defined by I = {f k[x1,...,x n ] : f N I for some positive integer N }. Since I I(V (I)), I I(V (I)), because if a power of a function f is zero at all points of a set, then the same must be true for the function f. So the point of the theorem is the opposite inclusion I(V (I)) I. Corollary The function I V (I) is an order reversing bijection from the set of radical ideals of k[x 1,...,x n ] to the set of closed algebraic subsets of A n k, with inverse X I(X). The proof we shall give for the Nullstellensatz will also be nonconstructive. However, a great deal of interest has centered on finding effective versions of the Nullstellensatz. For example, if f vanishes on V (g 1,...,g k ), then one wants to find an effective bound for an integer m such that f m = i h ig i as well as bound the degrees of the h i. One very important special case of the Nullstellensatz, from which the full theorem can be derived by a formal trick, is the following: Theorem Let m be a maximal ideal of k[x 1,...,x n ]. Then there exists a point p = (c 1,...,c n ) A n k such that m = (x 1 c 1,...,x n c n ) = {f k[x 1,...,x n ] : f(p) = 0} = I({p}). In particular, maximal ideals in k[x 1,...,x n ] are in one-to-one correspondence with points of A n k. Note that the correspondence is as follows: the maximal ideal m corresponds to the point p = V (m), and the point p corresponds to the maximal ideal I({p}). Corollary If X is an algebraic subset of A n k, then the maximal ideals of A(X) are in one-to-one correspondence with the points of X. Proof. A maximal ideal of A(X) is the same thing as a maximal ideal m of k[x 1,...,x n ] containing I(X). But I(X) m if and only if V (m) V (I(X)) = X.

29 1.5. MOTIVATING QUESTIONS 23 The above corollary explains the significance of the maximal ideals in k[x 1,...,x n ]. It is natural to characterize those algebraic sets corresponding more generally to prime ideals. This is the goal of the following definition: Definition A nonempty closed algebraic subset X of A n k is irreducible if, whenever X = X 1 X 2, where the X i are closed algebraic subsets of A n k, then either X = X 1 or X = X 2. For example, a point is an irreducible closed subset of A n k. Proposition An algebraic subset X of A n k if I(X) is a prime ideal. is irreducible if and only Proof. First suppose that X = X 1 X 2 with neither X 1 nor X 2 equal to X. In particular X i X and so I(X) I(X i ). We cannot have I(X) = I(X i ), for then X = V (I(X)) = V (I(X i )) = X i. So there must exist f i I(X i ) I(X), i = 1, 2. But then f 1 f 2 I(X 1 X 2 ) = I(X), and so I(X) is not prime. Conversely, suppose that I(X) is not prime, and choose f i / I(X), i = 1, 2, such that f 1 f 2 I(X). (Note that I(X) is a proper ideal since X. Set X i = V (f i ) X. Note that X i X, for otherwise X V (f i ) and so f i I(X). Moreover, since f 1 f 2 I(X), X V (f 1 f 2 ) = V (f 1 ) V (f 2 ) and thus X = (V (f 1 ) X) (V (f 2 ) X) = X 1 X 2. Thus X is not irreducible. Definition An irreducible closed algebraic subset X of A n k is called an (affine) algebraic variety. Equivalently, X is an affine algebraic variety if and only if A(X) is an integral domain. In this case, the quotient field K(X) of A(X) is called the function field or field of rational functions on X. We shall see that every affine algebraic set X is a union of finitely many irreducible algebraic sets X i, and the X i are unique up to order if there are no containment relations among them, i.e. X i is not contained in X j if i j. The ideals which are of the form I(X) are not necessarily prime, but have the weaker property that they are radical ideals, in other words ideals I such that I = I. The Nullstellensatz sets up a correspondence between algebraic sets in A n k and radical ideals I in k[x 1,...,x n ]. Of course, we might wish to study all ideals in k[x 1,...,x n ]. For example, in case n = 1, every ideal is principal, and the study of ideals in k[x] is the study of monic polynomials in k[x]. But for n = 2, the study of ideals in k[x, y] is already considerably more complicated. We can say a little bit about prime ideals as follows:

30 24 CHAPTER 1. INTRODUCTION TO COMMUTATIVE RINGS Proposition In k[x 1,...,x n ], every nonzero minimal prime ideal is principal. Proof. This is a consequence of Lemma and Corollary Thus, in k[x 1, x 2 ], there are three kinds of prime ideals: maximal ideals (x 1 a 1, x 2 a 2 ), principal ideals (f(x 1, x 2 )) which are prime (equivalently, f(x 1, x 2 ) is irreducible), and the zero ideal (0). In fact, it is not hard to show (Exercise 1.15) that these are the only prime ideals. Note that, in this case, the maximal ideals are generated by two elements (and no fewer), and the minimal nonzero primes are generated by one element. We shall show later that every proper closed subset of A 2 k is a union of finitely many points and closed subsets of the form V (f), where f is irreducble. However, for n > 2, the picture is much more complicated. While every maximal ideal in k[x 1, x 2, x 3 ] is generated by three elements, and every minimal nonzero prime ideal is generated by one element, there exist intermediate prime ideals p. Given such a prime ideal p, i.e. a prime ideal which is neither maximal nor minimal, we shall show in Chapter 4 that every prime ideal which properly contains it is maximal, and every nonzero prime ideal properly contained in it is a minimal nonzero prime ideal. We say that the ring k[x 1, x 2, x 3 ] has dimension three. However, while some of intermediate prime ideals can be generated by two elements, some can only be generated by three elements, and it is not easy to characterize those which can be generated by only two. Let us make some (much briefer) comments about some motivating questions in number theory. A great deal of elementary number theory is concerned with factorization in the PID Z: in particular, every natural number factors into a product of primes, in an essentially unique way. A similar statement holds in the ring Z[i] of Gaussian integers; this result is at the heart of the description of which natural numbers are sums of two squares, i.e. for which natural numbers n the Diophantine equation x 2 + y 2 = n has a solution. On the other hand, the ring Z[ 5] is not a PID, and so the Diophantine equation x 2 + 5y 2 = n is harder to study. Of course, the most famous ring of this sort is the ring Z[ζ k ], where ζ k = e 2πi/k is a primitive k th root of unity, connected to the Fermat equation x k + y k = n. However, the ring Z[ 5], as well as many of the rings of algebraic numbers encountered in number theory, does have a very remarkable property which is a weaker substitute for unique factorization: Every ideal in Z[ 5] (and in much more general rings of algebraic numbers) has a unique factorization (in the