# Homological Methods in Commutative Algebra

Size: px
Start display at page:

Transcription

1 Homological Methods in Commutative Algebra Olivier Haution Ludwig-Maximilians-Universität München Sommersemester 2017

2 1 Contents Chapter 1. Associated primes 3 1. Support of a module 3 2. Associated primes 5 3. Support and associated primes 7 Chapter 2. Krull dimension 9 1. Dimension of a module 9 2. Length of a module 9 3. Principal ideal Theorem Flat base change 12 Chapter 3. Systems of parameters Alternative definition of the dimension Regular local rings 16 Chapter 4. Tor and Ext Chain complexes Projective Resolutions The Tor functor Cochain complexes The Ext functor 25 Chapter 5. Depth M-regular sequences Depth Depth and base change 30 Chapter 6. Cohen-Macaulay modules Cohen-Macaulay modules Cohen-Macaulay rings Catenary rings 36 Chapter 7. Normal rings Reduced rings Locally integral rings Normal rings 39 Chapter 8. Projective dimension Projective dimension over a local ring The Auslander-Buchsbaum formula 45

3 Contents 2 Chapter 9. Regular rings Homological dimension Regular rings 47 Chapter 10. Factorial rings Locally free modules The exterior algebra Factorial rings 53 Bibliography 57

4 3 CHAPTER 1 Associated primes Basic references are [Bou98, Bou06, Bou07], [Ser00], and [Mat89]. All rings are commutative, with unit, and noetherian. A local ring is always nonzero. We will use the convention that R will denote a (noetherian, commutative, unital) ring, A a local ring, m its maximal ideal, and k its residue field. The letter M will either denote a R-module, or an A-module. A prime will mean a prime ideal of R, or of A. When p is a prime, we denote by κ(p) the field R p /(pr p ), or A p /(pa p ). 1. Support of a module Definition Let M be an R-module, and m M. The annihilator Ann(m) is the set of elements x R such that xm = 0. This is an ideal of R. We write Ann(M), or Ann R (M), for the intersection of the ideals Ann(m), where m M. Definition The set of prime ideals of R is denoted Spec(R). The support of an R-module M, denoted Supp(M), or Supp R (M), is the subset of of Spec(R) consisting of those primes p such that M p 0. Observe that if p Supp(M) and q Spec(R) with p q, then q Supp(M). Lemma The support of M is the set of primes containing the annihilator of some element of M. Proof. Let p Spec(R). Then M p 0 if and only if there exists m M such that tm 0 for all t p, or equivalently Ann(m) p. Lemma Let M be a finitely generated R-module. Then Supp(M) is the set of primes containing Ann(M). Proof. Since for any m M, we have Ann(M) Ann(m), it follows from Lemma that any element of Supp(M) contains Ann(M) (we did not use the assumption that M is finitely generated). Conversely assume that M is finitely generated, and let p be a prime containing Ann(M). We claim that there is m M such that Ann(m) p; by Lemma this will show that p Supp(M). Assuming the contrary, let m 1,, m n be a finite generating family for M. We can find s i Ann(m i ) such that s i p, for i = 1,, n. Then the product s 1 s n belongs to Ann(M), hence to p. Since p is prime, it follows that s j p for some j, a contradiction. Lemma Consider an exact sequence of R-modules: 0 M M M 0.

5 1. Associated primes 4 Then Supp(M) = Supp(M ) Supp(M ). Proof. For every prime p, we have an exact sequence 0 M p M p M p 0, and therefore M p = 0 if and only if M p = 0 and M p = 0. Lemma (Nakayama s Lemma). Let (A, m) be a local ring, and M a finitely generated A-module. If mm = M then M = 0. Proof. Assume that M 0. Let M be a maximal proper (i.e. M) submodule of M, and M = M/M (if no proper submodule were maximal, then we could build an infinite ascending chain of submodules in M, a contradiction since A is noetherian and M finitely generated). Then by maximality of M, the module M is simple, i.e. has exactly two submodules (0 and M ). But a simple module is isomorphic to A/m (it is generated by a single element, hence is of the type A/I for an ideal I; but A/I is simple if and only if I = m). Therefore mm = 0, hence mm M. This is a contradiction with mm = M. Definition If (A, m) and (B, n) are two local rings, a ring morphism φ: A B is called a local morphism if φ(m) n. Lemma Let A B be a local morphism of local rings, and M a finitely generated A-module. If M A B = 0, then M = 0. Proof. Assume that M 0 and let k be the residue field of A. By Nakayama s Lemma 1.1.6, the k-vector space M A k is nonzero hence admits a one-dimensional quotient. This gives a surjective morphism of A-modules M k. Then k A B vanishes, being a quotient of M A B. But since A B is local, the residue field of B is a quotient of k A B, a contradiction. Proposition Let ϕ: R S be a ring morphism, and M a finitely generated R-module. Then Supp S (M R S) = {q Spec(S) ϕ 1 q Supp R (M)}. Proof. Let q Spec(S) and p = ϕ 1 q. Then the morphism R p S q is local. We have an isomorphism of S q -modules (M R S) q M p Rp S q, and the result follows from Lemma Corollary Let M be a finitely generated R-module, and I an ideal of R. Then Supp R (M/IM) = {p Supp(M) I p}. Proof. Let ϕ: R R/I be the quotient morphism. Any prime p containing I may be written as ϕ 1 q for some q Spec(R/I). If in addition p Supp(M), then by Proposition we have q Supp R/I (M/IM). By Lemma there is m M/IM such that Ann R/I (m) q, hence Ann R (m) = ϕ 1 Ann R/I (m) ϕ 1 q = p, proving that p Supp R (M/IM). This proves one inclusion. The other inclusion is clear.

6 5 1. Associated primes 2. Associated primes Definition A prime p of R is an associated prime of M if there is m M such that p = Ann(m). The set of associated primes is written Ass(M), or Ass R (M). In other words we have p Ass(M) if and only if there is an injective R-module morphism R/p M. Proposition Any maximal element of the set {Ann(m) m M, m 0}, ordered by inclusion, is prime. Proof. Let I = Ann(m) be such a maximal element. Let x, y R, and assume that xy I. If y I, then ym 0. Then I = Ann(m) Ann(ym). By maximality I = Ann(ym). Since xym = 0, we have x Ann(ym), hence x I. Corollary We have M 0 if and only if Ass(M). Proof. Since R is noetherian, the set of Proposition admits a maximal element as soon as it is not empty. Lemma Let p be a prime in R. Then Ass R (R/p) = {p}. Proof. Let m R/p be a nonzero element. Then p Ann R (m). Conversely, let x Ann R (m). If r R p is the preimage of m R/p, we have xr p, and since p is prime, it follows that x p. Thus p = Ann R (m). Proposition Consider an exact sequence of R-modules: 0 M M M 0. Then Ass(M ) Ass(M) Ass(M ) Ass(M ). Proof. If p Ass(M ), then M contains a module isomorphic to R/p. Since M M, it follows that M also contains such a module, hence p Ass(M). Now let p Ass(M). Then M contains a submodule E isomorphic to R/p. By Lemma we have Ass(E) = {p}. Let F = M E. The inclusion proved above implies that Ass(F ) Ass(E) = {p} and Ass(F ) Ass(M ). If F 0, we have Ass(F ) by Corollary 1.2.3, so that Ass(F ) = {p}, and therefore p Ass(M ). If F = 0, then the morphism E M is injective, so that {p} = Ass(E) Ass(M ). Lemma Let M α be a family of submodules of M such that M = α M α. Then Ass(M) = α Ass(M α ). Proof. Since M α M, we have Ass(M α ) Ass(M). Conversely if p = Ann(m) Ass(M), then there is α such that m M α. Then p Ass(M α ). Proposition Let Φ Ass(M). Then there is a submodule N of M such that Ass(N) = Φ and Ass(M/N) = Ass(M) Φ.

7 1. Associated primes 6 Proof. Consider the set Σ of submodules P of M such that Ass(P ) Φ. This set is non-empty since 0 Σ, and ordered by inclusion. Moreover Σ is stable under taking reunions of totally ordered subsets by Lemma By Zorn s lemma, we can find a maximal element N Σ (when M is finitely generated over the noetherian ring R, we do not need Zorn s lemma). Let p Ass(M/N). Then M/N contains a submodule isomorphic to R/p, of the form N /N with N N M. By Proposition and Lemma 1.2.4, we have Ass(N ) Ass(N) Ass(N /N) Φ {p}. By maximality of N, we have Ass(N ) Φ. It follows that p Φ and p Ass(N ). Since N is a submodule of M, we have Ass(N ) Ass(M), and therefore p Ass(M) Φ. Thus we have inclusions Ass(M/N) Ass(M) Φ and Ass(N) Φ. Since Ass(M) Ass(N) Ass(M/N) by Proposition 1.2.5, the above inclusions are in fact equalities. Definition An element of R is called a zerodivisor in M if it annihilates a nonzero element of M, a nonzerodivisor otherwise. Any element of an associated prime of M is a zerodivisor in M. The converse is true: Lemma The set of zerodivisors in M is the union of the associated primes of M. Proof. Assume that r Ann(x) with x M 0. Then Ann(x) is contained in a maximal element of the set {Ann(m) m M, m 0} (otherwise we could construct an ascending chain of ideals in the noetherian ring R). Proposition says that this maximal element is an associated prime of M. Recall that when S is a multiplicatively closed subset of R, the map p S 1 p induces a bijection {p Spec(R) p S = } Spec(S 1 R). Proposition Let S be a multiplicatively closed subset of R. Then Ass S 1 R(S 1 M) = {S 1 p p Ass R (M) and p S = }. Proof. If M contains an R-submodule isomorphic to R/p, then (by exactness of the localisation) S 1 M contains an (S 1 R)-submodule isomorphic to S 1 (R/p). The latter is isomorphic to (S 1 R)/(S 1 p). Conversely, as recalled above any element of Ass S 1 R(S 1 M) is of the form S 1 p for a unique p Spec(R) satisfying S p =. We need to prove that p Ass R (M). Let m M and s S be such that S 1 p = Ann S 1 R(m/s). Let p 1,, p n be a set of generators of the R-module p. For every i = 1,, n, we have p i m/s = 0 in S 1 M, which means that we can find t i S such that t i p i m = 0 in M. Let m = t 1 t n m M. Since each p i belongs to Ann R (m ), it follows that p Ann R (m ). Conversely if x Ann R (m ), then xt 1 t n /1 Ann S 1 R(m/s) = S 1 p. Thus uxt 1 t n p for some u S. Since ut 1 t n S and S p =, it follows from the primality of p that x p. Therefore Ann R (m ) = p, and p Ass R (M).

8 7 1. Associated primes 3. Support and associated primes Proposition The set Supp(M) is the set of primes of R containing an element of Ass(M). Proof. If p contains an associated prime Ann(m) for some m M, then p Supp(M) by Lemma Let now p Supp(M). Then M p 0, hence by Corollary we can find a prime in Ass Rp (M p ), which corresponds by Proposition to a prime q Ass R (M) such that q p. Corollary We have Ass(M) Supp(M), and these sets have the same minimal elements. Corollary Minimal elements of Supp(M) consist of zerodivisors in M. Proof. Combine Proposition with Lemma Definition The non-minimal elements of Ass(M) are called embedded primes of M. Proposition Assume that M is finitely generated. Then there is a chain of submodules 0 = M 0 M 1 M n = M such that M i /M i 1 R/p i with p i Spec(R) for i = 1,, n. We have Ass(M) {p 1,, p n } Supp(M), and these sets have the same minimal elements. Proof. Assume that we have constructed a chain 0 = M 0 M 1 M j M such that M i /M i 1 R/p i with p i prime, for i = 1,, j. If M j = M, then the first part of the statement is proved. Otherwise, by Corollary we can find p j+1 Ass(M/M j ). Thus M/M j contains a submodule isomorphic to R/p j+1, which is necessarily of the form M j+1 /M j with M j M j+1 M. This process must stop, since R is noetherian and M finitely generated. This proves the first part. By Proposition 1.2.5, we have Ass(M i ) Ass(M i 1 ) Ass(R/p i ). We obtain that Ass(M) {p 1,, p n } using Lemma and induction on i. By Lemma 1.1.5, we have Supp(R/p i ) Supp(M i 1 ) Supp(M i ). In particular p i Supp(R/p i ) Supp(M i ). Since M i M, we have Supp(M i ) Supp(M). This proves that {p 1,, p n } Supp(M). The last statement follows from Proposition Corollary Assume that M is finitely generated. Then: (i) The set Ass(M) is finite. (ii) The set of minimal elements of Supp(M) is finite. Corollary Assume that M is finitely generated and nonzero. Then Supp(M) possesses at least one minimal element. Remark Corollary may also be proved directly using Zorn s Lemma.

9

10 9 CHAPTER 2 Krull dimension 1. Dimension of a module Definition The length of a chain of primes p 0 p n in R is the integer n. The dimension of a finitely generated R-module M is the supremum of the lengths of the chains of primes in Supp(M). It is denoted dim M, or dim R M. The height of a prime p of R is the supremum of the lengths n of chains p 0 p n = p of primes in R. In other words: height p = dim R p. The dimension of the zero module is. By Lemma 1.1.4, we have dim M = dim R/ Ann(M). Remark Note that dim R/p+dim R p is the supremum of the lengths of chains of primes of R with p appearing in the chain, so that dim R/p + dim R p dim R. Later we will provide conditions on R ensuring that it is an equality. Proposition Let R S be a ring homomorphism. Let M be an S-module, finitely generated as an R-module. Then dim R M = dim S M. Proof. Let m 1,, m n be generators of the S-module M. The morphism of S- modules S M n sending s to (sm 1,, sm n ) has kernel Ann S (M). This makes S/ Ann S (M) an S-submodule of M n, which is therefore finitely generated as an R-module (R is noetherian). The ring morphism R/ Ann R (M) S/ Ann S (M) is injective, and, as we have just seen, integral. In this situation chains of primes are in bijective correspondence (see e.g. [AM69, Corollary 5.9 and Theorem 5.10]). Proposition Let M be a finitely generated R-module. Then dim M = max dim R/p = max dim R/p. p Ass(M) p Supp(M) Proof. This follows from Lemma and Proposition Length of a module Definition The length of a chain of submodules 0 = M 0 M n = M is the integer n. The chain is called maximal if for each i there is no submodule N satisfying M i N M i+1. The length of an R-module M is the supremum of the lengths of the chains of submodules of M. It is denoted length M. The zero module is the only module of length zero.

11 2. Krull dimension 10 Lemma Consider an exact sequence of R-modules 0 M M M 0. Then we have length M = length M + length M. Proof. If length M = or length M =, then length M =. Assume that length M = e < and length M =. Let n be an integer. We may find a chain M 0 M n+e in M. Let M i = M i M and M i = M i /M i. There are at least n indices i such that M i = M i+1, and for such i we have M i M i+1. Thus from the family M i we may extract a chain of length n of submodules of M. This proves that length M n. Since n was arbitrary, we deduce that length M =. So we may assume that all modules have finite length. The statement is true if length M = length M or if length M = length M, for then M = M or M = M. Thus we may assume that length M < length M and length M < length M, and proceed by induction on length M. Let 0 = M 0 M r = M be a chain of maximal length, so that r = length M. Let N = M r 1. Then length N = length M 1, and length M/N = 1. Form the commutative diagram with exact rows and columns N N 0 M M 0 P P N M P Then length P = 1, hence either P = 0 or P = P. In any case, we have Then, using induction length M = length N + 1 length P + length P = 1. = length N + length N + length P + length P = length M + length M. Proposition The length of any maximal chain of submodules of M is equal to the length of M. Proof. If M contains an infinite chain, then length M =. Let 0 = M 0 M r = M be a maximal chain. We prove that r = length M by induction on r. If r = 0, then M = 0, hence length M = 0. Assume that r > 0, and let N = M r 1. We have length M/N = 1 by maximality of the chain. In addition, the chain 0 = M 0

12 11 2. Krull dimension M r 1 = N is maximal in N, so that length N = r 1 by induction. Therefore, by Lemma length M = length N + length M/N = r = r. Lemma Let R be an integral domain. Then R has finite length as an R-module if and only if it is a field. Proof. If R is a field, it has exactly two ideals (0 and R), and thus has length 1. Now assume that R has finite length, and let x R {0}. The sequence of ideals x i+1 R x i R R must stabilise, hence x n = ax n+1 for some n N and some a R. Thus x n (1 ax) = 0. If R is an integral domain then ax = 1, showing that x is invertible in R. Lemma Assume that M is finitely generated. Then dim M = 0 if and only if M is nonzero and has finite length. Proof. We may assume that M 0. Let us choose M i, p i as in Proposition Then by induction M has finite length if and only if each R/p i has finite length. This is so if and only if each p i is a maximal ideal of R by Lemma Since {p 1,, p n } and Supp(M) have the same minimal elements, each p i is maximal if and only if Supp(M) consists of maximal ideals of R, or equivalently dim M = Principal ideal Theorem Definition When S is a subset of R, and p Spec(R), we say that p is minimal over S if it is a minimal element of the set of primes containing S. Theorem (Krull). Assume that R is an integral domain. Let x R {0}, and p be a prime minimal over {x}. Then height p = 1. Proof. The ring R p is an integral domain, and the image of x in R p is nonzero. Thus we may replace R with R p, and assume that R is local with maximal ideal p. Let q be a prime such that q p. It will suffice to prove that q = 0. We view R as a subring of R q. For each integer n 0, we consider the ideal of R defined as q n = (q n R q ) R = {u R su q n for some s R q}, (and called the n-th symbolic power of the ideal q). The ring R/xR has dimension zero by minimality of p, hence finite length by Lemma It follows that the chain of ideals q n+1 /(q n+1 xr) q n /(q n xr) of R/xR must stabilise. Therefore we can find an integer n such that q n q n+1 + xr. Thus for any y q n, we may find a R such that y ax q n+1. Note that x q by minimality of p, hence x becomes invertible in R q. But ax q n q n R q, and therefore a = axx 1 q n R q. Since a R, it follows that a q n. We have proved that q n = q n+1 + xq n. Consider the finitely generated R-module N = q n /q n+1. We have xn = N with x in the maximal ideal p of R. Applying Nakayama s Lemma we obtain that N = 0, or equivalently q n = q n+1. Observe that q m R q = q m R q = (qr q ) m for any m. Thus (qr q ) n = (qr q ) n+1. We now apply Nakayama s Lemma to the finitely generated R q -module (qr q ) n and conclude that (qr q ) n = 0. This shows that any element of the maximal ideal qr q of R q is nilpotent; but R q is a domain, so that qr q = 0, and finally q = 0.

13 2. Krull dimension 12 Lemma Let p 0 p n be a chain of primes, and let x p n. Then we can find a chain of primes p 0 p n with p 0 = p 0, p n = p n, and x p 1. Proof. We proceed by induction on n, and we may assume that n 2. It will suffice to find a prime p n 1 containing x and such that p n 2 p n 1 p n (then we find by induction a chain of primes p 0 p n 2 such that p 0 = p 0, p n 2 p n 1, and x p 1). If x p n 1, we may take p n+1 = p n+1. Thus we assume that x p n 1. Then we can find a prime p n 1 containing {x} p n 2, contained in p n, and minimal for these properties (it corresponds to a minimal element of the support of the R pn - module R pn /(p n 2 R pn + xr pn ), which exists by Corollary since p n 2 R pn + xr pn p n R pn R pn ). Then the prime ideal p n 1/p n 2 of R/p n 2 is minimal over the image of x in R/p n 2, and therefore has height 1 by Theorem Since the prime ideal p n /p n 2 of R/p n 2 has height 2, it cannot be equal to p n 1/p n 2. Thus we have p n 2 p n 1 p n, with x p n 1, as required. Proposition Let (A, m) be a local ring, x m, and M a finitely generated A-module. Then dim M/xM dim M 1, with equality if and only x belongs to no prime p Supp(M) such that dim A/p = dim M. Proof. Let p 0 p n be a chain of primes in Supp(M). Replacing p n with m, we may assume that p n = m. By Lemma we can assume that x p 1. This gives a chain of primes p 1 p n of length n 1 in Supp(M) Supp(A/xA) = Supp(M/xM) (the last equality follows from Corollary ), which proves that dim M/xM n 1. Now a prime p Supp(M) contains x if and only if p Supp(M/xM) by Corollary Thus the second statement follows from Proposition applied to the module M/xM. Corollary Let (A, m) be a local ring and M a finitely generated A-module. Let x m be a nonzerodivisor in M. Then dim M/xM = dim M 1. Proof. This follows from Corollary and Proposition Flat base change Definition An R-module M is called flat if for every exact sequence of R- modules N 1 N 2 N 3 the induced sequence M R N 1 M R N 2 M R N 3 is exact. We say that a ring morphism R S is flat if S is flat as an R-module. Lemma Let ϕ: (A, m) (B, n) be a flat local morphism. Then (i) For any A-module M, the morphism M B A M is injective. (ii) The morphism Spec B Spec A is surjective. Proof. (i): Let m M {0}. The ideal I = Ann(m) is contained in m. The exact sequence I A m M induces by flatness an exact sequence B A I B 1 m B A M. The image of B A I B is the ideal J generated by ϕ(i) in B. Since ϕ is local and I m, we have J n. If 1 m = 0 B A M, then B = J, a contradiction. (ii): Let p Spec(A). Then κ(p) B A κ(p) is injective by (i), hence B A κ(p) 0. Thus Spec(B A κ(p)), which means that there is q Spec B such that ϕ 1 q = p (by the description of the set of primes in a quotient or a localisation).

14 13 2. Krull dimension Proposition (Going down). Let ρ: R S be a flat ring morphism. Let q Spec(S) and p Spec(R) be such that p ρ 1 q. Then we may find q Spec(S) such that q q and ρ 1 q = p. Proof. The morphism R p S q is flat and local. Therefore by Lemma (ii) the prime p R p has a preimage in Spec(S q ), necessarily of the form q S q with q q. The primes ρ 1 q and p coincide because they are contained in p and localise to the same prime of R p. Corollary Let R S be a flat ring morphism and M a finitely generated R- module. Then the morphism Spec S Spec R sends minimal elements of Supp S (S R M) to minimal elements of Supp R (M). Proof. Let q be a minimal element Supp S (S R M). Then its image p Spec(R) belongs to Supp R (M) by Proposition If p Supp R (M) is such that p p, then by Proposition we may find a preimage q of p such that q q. Then q Supp S (S R M) by Proposition 1.1.9, hence q = q by minimality of q. Thus p = p, proving that p is a minimal element of Supp R (M). Proposition (Prime avoidance). Let I, p 1,, p n be ideals of R. Assume that p i is prime for i 3. If I p 1 p n then I p i for some i {1,, n}. Proof. We assume that I is contained in no p i and find x I belonging to no p i. This is clear for n = 0, 1. If n = 2, we x i I p i for i = 1, 2. We may assume that x 1 p 2 and x 2 p 1 (otherwise the statement is proved). Then x = x 1 + x 2 works. Now assume that n > 2, and proceed by induction on n. For each j = 1,, n, we can find by induction x j I which is in none of the p i for i j, and we may assume as above that x j p j. Then x = x n + x 1 x 2 x n 1 works, since p n is prime (n 3). Proposition Let ϕ: A B be a local morphism of local rings and M a finitely generated A-module. Let m be the maximal ideal of A, and k its residue field. Then with equality if B is flat as an A-module. dim B B A M dim A M + dim B B A k, Proof. We may assume that M 0, and proceed by induction on dim A M. First assume that dim A M = 0. Then {m} = Supp A (M) = Supp A (k), hence Supp B (B A M) = Supp B (B A k) by Proposition and thus dim B B A M = dim B A k, proving the statement in this case. Assume that dim A M > 0. Then m is not a minimal element of Supp A (M). By prime avoidance (Proposition 2.4.5) and finiteness of the set of minimal primes (Corollary 1.3.6), we may find x m belonging to no minimal primes of Supp A (M). By Proposition we have dim A M/xM = dim A M 1, so that we may use the induction hypothesis for the module M/xM and obtain (2.4.a) dim B B A (M/xM) dim A M 1 + dim B B A k, with equality if ϕ is flat. Applying Proposition to the B-module B A M and the element ϕ(x) B, we obtain (2.4.b) dim B B A M dim B B A (M/xM) + 1,

15 2. Krull dimension 14 with equality if ϕ(x) belongs to no minimal primes of Supp B (B A M). The latter condition is fulfilled if ϕ is flat by Corollary The statement follows by combining (2.4.a) and (2.4.b).

16 15 CHAPTER 3 Systems of parameters 1. Alternative definition of the dimension In this section (A, m, k) is a local ring, and M a finitely generated A-module. Lemma The following conditions are equivalent: (i) dim M = 0. (ii) Supp(M) = {m}. (iii) Ass(M) = {m}. (iv) The A-module M has finite length and is nonzero. (v) M 0 and there is an integer n such that m n M = 0. Proof. (i) (ii): Indeed, dim M is the supremum of the lengths of chains of primes in Supp(M), and m Supp(M) as soon as Supp(M). (ii) (iii): This follows from Proposition (iv) (i): This was proved in Lemma (iv) (v): The sequence of submodules m i+1 M m i M must stabilise, hence there is n such that m n+1 M = m n M. By Nakayama s Lemma (applied to m n M) we obtain m n M = 0. (v) (ii): If p Supp(M), then m n Ann(M) p. Thus for any x m we have x n p. Since p is prime, this implies x p, proving that m = p. Proposition Assume that M 0. Then dim M is finite, and coincides with the smallest integer n for which there exists elements x 1,, x n m such that the module M/{x 1,, x n }M satisfies the conditions of Lemma Proof. If x 1,, x m m are such that dim M/{x 1,, x m }M = 0, then dim M m by Proposition If x 1,, x m is a finite set of generators of the ideal m (which exists since A is noetherian), then the module M/{x 1,, x m }M = M/mM satisfies the condition (v) of Lemma 3.1.1, hence dim M m <. We prove by induction on n = dim M that we may find x 1,, x n m such that dim M/{x 1,, x n }M = 0. The case n = 0 being clear, let us assume that n > 0. By prime avoidance (Proposition 2.4.5), we may find an element x n m belonging to no p Supp(M) such that dim A/p = n (by Corollary there are only finitely many such p, since they are among the minimal elements of Supp(M)). Then dim M/x n M = n 1 by Proposition Applying the induction hypothesis to the module N = M/x n M, we find x 1,, x n 1 m such that N/{x 1,, x n 1 }N = M/{x 1,, x n }M satisfies the conditions of Lemma Definition A set {x 1,, x n } as in Proposition (with n = dim M) is called a system of parameters for M.

17 3. Systems of parameters 16 If V is a k-vector space, we denote by dim k vect V its dimension in the sense of linear algebra (that is, the cardinality of a k-basis). Proposition The minimal number of generators of the ideal m is equal to dim k vect (m/m 2 ). Proof. Let n = dim k vect (m/m 2 ), and x 1,, x n m a family which reduces modulo m 2 to a k-basis of m/m 2. Let I m be the ideal generated by x 1,, x n. Then m = I + m 2. Thus the finitely generated A-module M = m/i satisfies mm = M, hence vanishes by Nakayama s Lemma This prove that m = I can be generated by n elements. Conversely if the A-module m is generated by m elements, then the k-vector space m/m 2 is generated by their images modulo m 2, so that dim k vect (m/m 2 ) m. Corollary We have dim k vect (m/m 2 ) dim A. Proof. Since the A-module k = A/mA satisfies the conditions of Lemma 3.1.1, this follows from Proposition applied with M = A, and Proposition Regular local rings Definition We will say that a local (noetherian) ring A is regular if dim A = dim k vect (m/m 2 ), or equivalently (Proposition 3.1.4) if m can be generated by dim A elements. A system of parameters for A generating the maximal ideal is called a regular system of parameters. Example A local ring of dimension zero is a regular local ring if and only if it is a field. Indeed let m be its maximal ideal. Then dim k vect (m/m 2 ) = 0 if and only if m = m 2. By Nakayama s Lemma 1.1.6, this condition is equivalent to m = 0. Example (Exercise) A local ring of dimension one is a regular local ring if and only if it is a discrete valuation ring. Lemma Let (A, m) be a regular local ring, and x m m 2. Then A/xA is a regular local ring of dimension dim A 1. Proof. Consider the local ring B = A/xA, and let n = m/xa be its maximal ideal. Note that k = A/m = B/n. There is a surjective morphism m/m 2 n/n 2 of k-vector spaces whose kernel contains the 1-dimensional k-vector space generated by x mod m 2. It follows that dim k vect (n/n 2 ) dim k vect (m/m 2 ) 1 = dim A 1 dim B, where we use Proposition for the last inequality. Since dim k vect (n/n 2 ) dim B by Corollary 3.1.5, we conclude that dim k vect (n/n 2 ) = dim B = dim A 1. A partial converse is given by the following. Lemma Let (A, m) be a local ring, and x m a nonzerodivisor in A. If A/xA is a regular local ring then so is A. Proof. Let n = dim A. By Corollary we have dim A/xA = n 1. Let x 1,, x n 1 be elements of m reducing modulo xa to a regular system of parameters for the local ring A/xA. Then the n elements x, x 1,, x n 1 generate the ideal m, and thus form a regular system of parameters for A.

18 17 3. Systems of parameters Proposition A regular local ring is an integral domain. Proof. Let (A, m) be a regular local ring. We prove that A is an integral domain by induction on dim A. If dim A = 0, then A is a field by Example 3.2.2, and in particular an integral domain. If dim A > 0, then m 0, hence m m 2 by Nakayama s Lemma Thus by prime avoidance (Proposition 2.4.5) we may find an element x m not belonging to m 2 nor to any of the finitely many minimal primes of A (Corollary 1.3.6). The local ring A/xA is regular and has dimension dim A 1 by Lemma By the induction hypothesis it is an integral domain, which means that xa is a prime ideal of A. So xa contains a minimal prime q; by the choice of x we have x q. For any y q, we can write y = xa for some a A. Since q is prime and x q we have a q. Thus q = xq, hence q = mq and by Nakayama s Lemma we have q = 0, proving that A is an integral domain.

19

20 19 CHAPTER 4 Tor and Ext In this section R is a commutative unital ring. 1. Chain complexes Definition A chain complex (of R-modules) C is a collection of R-modules C i and morphisms of R-modules d C i : C i C i 1 for i Z satisfying d C i 1 dc i = 0. The R-module H i (C) = ker d C i / im d C i+1 is called the i-th homology of the chain complex C. The chain complex C is called exact if H i (C) = 0 for all i. A morphism of chain complexes f : C C is a collection of morphisms f i : C i C i such that f i 1 d i = d i f i. Such a morphism induces a morphism of the homology modules H i (C) H i (C ). We say that the morphism C C is a quasi-isomorphism if the induced morphism H i (C) H i (C ) is an isomorphism for all i. Definition We say that two morphisms of chain complexes f, g : C C are homotopic if there is a collection of morphisms s i : C i C i+1 such that f i g i = d C i+1 s i + s i 1 d C i. A morphism of chain complexes f : M N is a homotopy equivalence if there is a morphism of chain complexes g : N M such that f g is homotopic to id N and g f is homotopic to id M. We say that two chain complexes are homotopy equivalent if there is a homotopy equivalence between them. Proposition Homotopic morphisms induce the same morphism in homology. Proof. In the notations of Definition 4.1.2, the morphism d C i s i has image contained in im d C i+1 and the morphism s i 1 d C i has kernel contained in ker d C i. They induce the zero morphism in homology by construction. Corollary Homotopy equivalent chain complexes are quasi-isomorphic. Definition A sequence of chain complexes 0 C C C 0 is called exact if the sequence is exact for each i. 0 C i C i C i 0

21 4. Tor and Ext 20 Proposition An exact sequence of chain complexes induces an exact sequence of modules 0 C C C 0 H i+1 (C ) H i (C ) H i (C) H i (C ) H i 1 (C) Proof. We only describe the morphism : H i+1 (C ) H i (C ). Any element x i+1 ker d C i+1 lifts to x i+1 C i+1. Let x i = d C i+1 (x i+1) C i. The image of x i in C i is d C i+1 (x i+1 ) = 0, hence x i is the image of some x i C i. In addition the image of dc i (x i ) C i 1 in C i 1 is d C i d C i+1 (x i) = 0. Since C i 1 C i 1 is injective, it follows that x i ker dc i. We define (x) as the class of x i H i(c ) = ker d C i / im d C i+1. We leave it as an exercise to check that is well-defined and that the sequence is exact. 2. Projective Resolutions Lemma Let M be a module. Then there is a surjective morphism F M with F free. If M finitely generated, then F may be chosen to be finitely generated. Proof. First assume that G M is a generating set for the R-module M, and let F be the free module on the basis {e g g G}. Then there is a surjective morphism F M given by e g g. We may always take G = M. If M is finitely generated, we may find a finite generating set G; in this case F is finitely generated. Definition An R-module P is projective if for every surjective R-module morphism M M, the natural morphism Hom R (P, M) Hom R (P, M ) is surjective. Lemma A module is projective if and only if it is a direct summand of a free module. Proof. If P is a projective R-module, we may find a surjective R-module morphism p: F P with F free by Lemma Since P is projective, there is an R-module morphism s: P F such that p s = id P. This gives a decomposition F = P ker p. Let L be a free module with basis l α, and M M be a surjective morphism. Let g : L M be a morphism. For each α, choose an element of m α M mapping to g(l α ). Then the unique morphism L M mapping l α to m α is a lifting of g. This proves that L is projective. Let now A be a direct summand of a free module L, which means that there are morphisms A L and L A such that the composite A L A is the identity. Let A M be a morphism. As we have just seen, the morphism L A M lifts to a morphism L M. The composite A L M is then a lifting of the morphism A M. This proves that the module A is projective. Lemma A projective module is flat. Proof. Using the fact that tensor products commutes with (possibly infinite) direct sums, we see that a direct summand of a flat module is flat, and that a free module is flat. The lemma then follows from Lemma

22 21 4. Tor and Ext Definition Let M be an R-module. A resolution C M is a chain complex C such that C i = 0 for i < 0, together with a morphism C 0 M such that the augmented chain complex C 1 C 0 M 0 is exact. This may be reformulated as follows. We denote by C(M) the chain complex such that C(M) i = 0 for i 0 and C(M) 0 = M (and thus d C(M) i = 0 for all i). A resolution of M is a chain complex C such that C i = 0 for i < 0, together with a quasi-isomorphism C C(M). A resolution C M is said to be projective, resp. free, resp. finitely generated, if each C i is so. Proposition Every module admits a free resolution. If R is noetherian, any finitely generated R-module admits a finitely generated free resolution. Proof. Let M be a module. We construct a chain complex D as follows. We let D i = 0 for i < 0 and D 1 = M. Assuming that D i 1 D i 2 is constructed for some i 0, by Lemma we may find a surjection D i ker(d i 1 D i 2 ) with D i free (resp. free and finitely generated). Then the sequence of modules D i D i 1 D i 2 is exact. The resolution C M is obtained by letting C i = D i for i 0 and C 0 = 0. Proposition Let E and P be two chain complexes. Assume that P i = E i = 0 for i < 1. P i is projective for i 0. E is exact. Let g : P 1 E 1 be a morphism of modules. Then there is a morphism of chain complexes f : P E such that f 1 = g. This morphism is unique up to homotopy. Proof. We construct f i inductively, starting with f 1 = g. Assume that i 0 and that f i 1 is constructed. The composite f i 1 d P i : P i E i 1 lands into ker d E i 1, because d E i 1 f i 1 d P i = d E i d E i 1 = 0. By exactness of the complex E, the morphism E i ker d E i 1 induced by de i is surjective, hence by projectivity of P i, we may find a morphism f i : P i E i such that d E i f i = f i 1 d P i. Now let f, f : P E be two morphisms of chain complexes extending g. We construct for each i a morphism s i : P i E i+1 such that f i f i = d E i+1 s i + s i 1 d P i by induction on i. We let s i = 0 for i < 1. Assume that s i 1 is constructed. Then d E i (f i f i) = (f i 1 f i 1) d P i = d E i s i 1 d P i + s i 2 d P i 1 d P i = d E i s i 1 d P i, so that (f i f i ) s i 1 d P i : P i E i has image in ker d E i. By exactness of the complex E, the morphism E i+1 ker d E i is surjective. By projectivity of P i, we obtain a morphism s i : P i E i+1 such that d E i+1 s i = (f i f i ) s i 1 d P i, as required. Corollary Two projective resolutions of the same module are canonically homotopy equivalent.

23 4. Tor and Ext 22 Proof. Let C M and C M be two projective resolutions. Then the identity of M extends to morphisms of chain complexes C C and C C by the existence part of Proposition The composite C C C and the identity of C are two extensions of the identity of M. They must be homotopic by the unicity part of Proposition For the same reason, the composite C C C is homotopic to the identity of C. Lemma Let C and C be two chain complexes. Assume that C i = C i = 0 for i < 1 C i is projective for i 0. C is exact. Then any exact sequence of modules 0 C 1 M C 1 0 is the degree 1 part of an exact sequence of chain complexes 0 C C C 0. In addition: (i) If the chain complex C is exact, then so is C. (ii) For each i 0, the exact sequence of modules 0 C i C i C i 0 splits (i.e. induces a decomposition C i = C i C i ). (iii) If C i is projective, then so is C i. Proof. Let us first prove (i) (ii) (iii) assume the first part of lemma. (i): This follows from the homology long exact sequence Proposition (ii): This follows from the fact that C i is projective. (iii): This follows from (ii), since a direct sum of projective modules is projective (e.g. by Lemma 4.2.3). Let us now prove the first part of the lemma. We let C i = C i C i with the natural morphisms C i C i C i. We construct by induction a morphism dc i : C i C i 1 such that d C i 1 dc i = 0 making the following diagram commute C i C i C i C i 1 and moreover such that the sequence C i 1 C i 1 0 Z i Z i Z i 0 is exact, where Z i = ker d C i, Z i = ker dc i, Z i = ker d C i. We let d C 1 = 0. Assume d C i 1 constructed for some i 0. The morphism dc i : C i Z i 1 C i 1 is the sum of the morphism C i Z i 1 Z i 1 and a morphism C i Z i 1 lifting the morphism C i Z i 1, which exists since C i is projective and Z i 1 Z i 1 is surjective. It only remains to prove that Z i Z i is surjective. Any x i Z i C i lifts to an element x i C i. Let x i 1 = d C i (x i) C i 1. Then the image of x i 1 in C i 1 is d C i (x i ) = 0, hence x i 1 is the image of some element of x i 1 C i 1. In addition, the

24 23 4. Tor and Ext image of d C i 1 (x i 1 ) in C i 2 is dc i 1 (x i 1) = d C i 1 dc i (x i) = 0, hence d C i 1 (x i 1 ) = 0 by injectivity of C i 2 C i 2. Since the complex C is exact, we may find x i C i such that d C i (x i ) = x i 1. Let y i C i be the image of x i. Then x i y i C i maps to x i in C i, and satisfies d i (x i y i ) = 0, i.e. belongs to Z i. 3. The Tor functor When C is a chain complex, and N a module, we denote by C R N the chain complex such that that (C R N) i = C i R N and d C RN i = d C i id N. A morphism of chain complexes f : C C induces a morphism of chain complexes f R N : C R N C R N. If f is homotopic to g, then f R N is homotopic to g R N. Thus a homotopy equivalence C C induces a homotopy equivalence C R N C R N, and in particular a quasiisomorphism by Corollary Definition Let M, N be two modules and n an integer. Let C M be a projective resolution. Then the module H n (C R N) is independent of the choice of C, up to a canonical isomorphism by Corollary and the discussion above. We denote this module by Tor n (M, N), or Tor R n (M, N). A morphism g : N N induces a morphism Tor n (M, g): Tor n (M, N) Tor n (M, N ). Let now M be another module, and C M be a projective resolution. By Proposition any morphism of modules f : M M extends to a morphism of complexes C C. The latter induces a morphism Tor n (f, N): Tor n (M, N) Tor n (M, N) which does not depend on any choice by the unicity part of Proposition and Proposition Proposition (i) Tor 0 (M, N) M R N. (ii) Tor n (M, N) = 0 for n < 0. (iii) If N is flat, then Tor n (M, N) = 0 for n > 0. (iv) If M is projective, then Tor n (M, N) = 0 for n > 0. (v) If f, g : M M are two morphisms and λ R, then Tor n (f + λg, N) = Tor n (f, N) + λ Tor n (g, N). (vi) If a, b: N N are two morphisms and µ R, then Tor n (M, a + µb) = Tor n (M, a) + µ Tor n (M, b). Proof. If C M is a projective resolution of M, then M = coker(c 1 C 0 ), hence by right-exactness of the tensor product, we have M R N = coker(c 1 R N C 0 R N) = H 0 (C R N). This proves (i). Since C n = 0 for n < 0, we have C n R N = 0, and thus H n (C R N) = 0, proving (ii). If N is flat, then C R N M R N is a resolution, hence H n (C R N) = 0 for n > 0. This proves (iii). Now if M is projective, we may use the trivial projective resolution C(M) M (see Definition 4.2.5) to compute Tor n (M, N), so that Tor n (M, N) = 0 for n > 0. This proves (iv). The two remaining statements follow easily from the construction of the Tor functor. Proposition Consider an exact sequence of modules 0 M M M 0

25 4. Tor and Ext 24 Let N be a module. Then we have an exact sequence Tor n+1 (M, N) Tor n (M, N) Tor n (M, N) Tor n (M, N) Proof. Let C M and C M be projective resolutions. By Lemma 4.2.9, we find a projective resolution C M and an exact sequence of chain complexes 0 C C C 0 extending the exact sequence of modules 0 M M M 0. Since each exact sequence 0 C i C i C i 0 is split, the sequence of chain complexes 0 C R N C R N C R N 0 is exact. The corresponding long exact sequence (Proposition 4.1.6) is the required sequence. Proposition Consider an exact sequence of modules 0 N N N 0 Let M be a module. Then we have an exact sequence Tor n+1 (M, N ) Tor n (M, N ) Tor n (M, N) Tor n (M, N ) Proof. Let C M be a projective resolution. Since each C i is projective, hence flat by Lemma 4.2.4, we have an exact sequence of complexes 0 C R N C R N C R N 0. The corresponding long exact sequence (Proposition 4.1.6) is the required sequence. Proposition The modules Tor n (N, M) and Tor n (M, N) are isomorphic. Proof. We proceed by induction on n, the case n = 0 being the symmetry of the tensor product. Let 0 K P N 0 be an exact sequence with P projective (this is possible by Lemma 4.2.1). Since P is both projective and flat (Lemma 4.2.4), so that Tor n (P, M) = Tor(P, M) = 0 for n > 0 by Proposition Applying Proposition and Proposition 4.3.4, we obtain a commutative diagram with exact rows (recall that Tor 1 (P, M) = Tor 1 (M, P ) = 0) 0 Tor 1 (M, N) M R K M R P 0 Tor 1 (N, M) K R M P R M Since horizontal arrows are isomorphisms, we conclude that Tor 1 (M, N) Tor 1 (N, M). Let now n > 1. Using Proposition and the vanishing of Tor n (M, P ) and Tor n 1 (M, P ) we deduce that Tor n (M, N) Tor n 1 (M, K). Using Proposition and the vanishing of Tor n (P, M) and Tor n 1 (P, M) we deduce that Tor n (N, M) Tor n 1 (K, M). By induction Tor n 1 (M, K) Tor n 1 (K, M), and the result follows. 4. Cochain complexes Definition A cochain complex (of R-modules) C is a collection of R-modules C i and morphisms of R-modules d i C : Ci C i+1 for i Z satisfying d i+1 C di C = 0. The R-module H i (C) = ker d i C/ im d i 1 C is called the i-th cohomology of the cochain complex C. A morphism of cochain complexes f : C C is a collection of morphisms f i : C i C i such that f i+1 d i = d i f i. Such a morphism induces a morphism of the cohomology modules H i (C) H i (C ). We say that

26 25 4. Tor and Ext the morphism C C is a quasi-isomorphism if the induced morphism H i (C) H i (C ) is an isomorphism for all i. Definition We say that two morphisms of cochain complexes f, g : C C are homotopic if there is a collection of morphisms s i : C i C i 1 such that f i g i = d i 1 C si + s i+1 d i C. A morphism of cochain complexes f : M N is a homotopy equivalence if there is a morphism of cochain complexes g : N M such that f g is homotopic to id N and g f is homotopic to id M. We say that two cochain complexes are homotopy equivalent if there is a homotopy equivalence between them. Proposition Homotopic morphisms induce the same morphism in cohomology. Proof. In the notations of the definition, the morphism d i 1 C si has image contained in im d C i 1 and the morphism si+1 d i C has kernel contained in ker di C. They induce the zero morphism in cohomology by construction. Corollary Homotopy equivalent cochain complexes are quasi-isomorphic. Definition A sequence of cochain complexes is called exact if the sequence is exact for each i. 0 C C C 0 0 C i C i C i 0 Proposition An exact sequence of cochain complexes induces an exact sequence of modules 0 C C C 0 H i 1 (C ) H i (C ) H i (C) H i (C ) H i+1 (C) 5. The Ext functor When M, N are two R-modules, we denote by Hom R (M, N) the R-module of R- module morphisms M N. When C is a chain complex and N a module, we denote by Hom R (C, N) the cochain complex such that (Hom R (C, N)) i = Hom R (C i, N) and d i Hom R (C,N) : Hom R(C i, N) Hom R (C i+1, N) is the morphism induced by left-composition with d C i+1. Definition Let M, N be two modules and n an integer. Let C M be a projective resolution. Then the module H n (Hom R (C, N)) is independent of the choice of C, up to a canonical isomorphism. We denote this module by Ext n (M, N), or Ext n R(M, N). A morphism g : N N induces a morphism Ext n (M, g): Ext n (M, N) Ext n (M, N ). A morphism f : M M induces a morphism Ext n (f, N): Ext n (M, N) Ext n (M, N). Proposition (i) Ext 0 (M, N) Hom R (M, N). (ii) Ext n (M, N) = 0 for n < 0. (iii) If M is projective, then Ext n (M, N) = 0 for n > 0.

27 4. Tor and Ext 26 (iv) If f, g : M M are two morphisms and λ R, then Ext n (f + λg, N) = Ext n (f, N) + λ Ext n (g, N). (v) If a, b: N N are two morphisms and µ R, then Ext n (M, a + µb) = Ext n (M, a) + µ Ext n (M, b). Proof. If C M is a (projective) resolution of M, then M = coker(c 1 C 0 ), hence by left-exactness of the contravariant functor Hom R (, N), we have Hom R (M, N) = ker(hom R (C 0, N) Hom R (C 1, N)) = H 0 (Hom R (C, N)). This proves the first statement. Since C n = 0 for n < 0, we have Hom R (C n, N) = 0, and thus H n (Hom R (C, N)) = 0, proving the second statement. Now if M is projective, we may use the trivial projective resolution C(M) M (see Definition 4.2.5) to compute Ext n (M, N), so that Ext n (M, N) = 0 for n > 0. This proves the third statement. The two remaining statements follow easily from the construction of the Tor functor. Proposition Consider an exact sequence of modules 0 M M M 0 Let N be a module. Then we have an exact sequence Ext n 1 (M, N) Ext n (M, N) Ext n (M, N) Ext n (M, N) Proof. Let C M and C M be projective resolutions. By Lemma 4.2.9, we find a projective resolution C M and an exact sequence of chain complexes 0 C C C 0 extending the exact sequence of modules 0 M M M 0. Since each exact sequence 0 C i C i C i 0 is split, the sequence of cochain complexes 0 Hom R (C, N) Hom R (C, N) Hom R (C, N) 0 is exact. The corresponding long exact sequence (Proposition 4.4.6) is the required sequence. Proposition Consider an exact sequence of modules 0 N N N 0 Let M be a module. Then we have an exact sequence Ext n 1 (M, N ) Ext n (M, N ) Ext n (M, N) Ext n (M, N ) Proof. Let C M be a projective resolution. Since each C i is projective, we have an exact sequence of cochain complexes 0 Hom R (C, N ) Hom R (C, N) Hom R (C, N ) 0. The corresponding long exact sequence (Proposition 4.4.6) is the required sequence.

28 27 CHAPTER 5 Depth In this chapter (A, m) is a noetherian local ring, and M a finitely generated A-module. 1. M-regular sequences Definition A finite tuple (x 1,, x n ) of elements of m is called an M-regular sequence if for all i the element x i is a nonzerodivisor in M/{x 1,, x i }M. The integer n is the length of the M-regular sequence. The M-regular sequence is called maximal if there is no x n+1 m such that (x 1,, x n+1 ) is an M-regular sequence. Lemma If M 0, then a maximal M-regular sequence exists. Proof. If not, we may find x i m for i N such that (x 1,, x n ) is an M-regular sequence for all n. By Nakayama s Lemma 1.1.6, the A-module M/{x 1,, x n 1 }M is nonzero, hence we may find an element m M such that m {x 1,, x n 1 }M. Assume that x n {x 1,, x n 1 }A. Then x n m {x 1,, x n 1 }M, hence x n is a zerodivisor in M/{x 1,, x n 1 }M, a contradiction. It follows that the sequence of ideals {x 1,, x n }A {x 1,, x n+1 }A of A is strictly increasing, which is impossible since A is noetherian. Definition A finite subset S of m is called secant for M if dim M/SM = dim M s, where s is the cardinal of S. We will say that a sequence (s 1,, s n ) is secant for M if the set {s 1,, s n } is secant for M. Proposition Any M-regular sequence is secant. Proof. By induction it is enough to consider the case of a sequence of length 1, in which case the statement is Corollary Depth Definition The depth of M is defined as depth M = depth A M = inf{i N Ext i (k, M) 0}. This is an element of N { }. When M = 0, we have depth M =. Proposition Let x m be a nonzerodivisor in M. Then depth M/xM = depth M 1.

29 5. Depth 28 Proof. From the exact sequence 0 M Proposition an exact sequence x M M/xM 0 we deduce using Ext i 1 (k, M/xM) Ext i (k, M) x Ext i (k, M) In view of Proposition (iv), the A-module Ext i (k, M) is annihilated by Ann(k) = m, and in particular multiplication by x is zero in this module. We obtain for each i an exact sequence 0 Ext i 1 (k, M) Ext i 1 (k, M/xM) Ext i (k, M) 0. Therefore Ext i 1 (k, M/xM) 0 if and only if Ext i 1 (k, M) 0 or Ext i (k, M) 0. The result follows. Corollary Let (x 1,, x n ) be an M-regular sequence. Then and in particular depth M n. depth(m/{x 1,, x n }M) = depth M n, Lemma The following conditions are equivalent: (i) depth M = 0. (ii) Every element of m is a zerodivisor in M. (iii) m Ass(M). Proof. A nonzero A-linear morphism k M is necessarily injective, therefore Ext 0 (k, M) = Hom A (k, M) is nonzero if and only if there is an injective A-modules morphism k M. This proves that (i) (iii). By Lemma 1.2.9, the set of nonzerodivisors in M is the union of the associated primes of M. Since Ass(M) is finite (Corollary 1.3.6), we see using prime avoidance (Proposition 2.4.5) that (ii) (iii). Lemma Let (x 1,, x n ) be an M-regular sequence. The following conditions are equivalent: (i) depth M = n. (ii) The M-regular sequence (x 1,, x n ) is maximal. (iii) m Ass(M/{x 1,, x n }M). Proof. In view of Corollary 5.2.3, we see that (i) is equivalent to the condition depth(m/{x 1,, x n }M) = 0. On the other hand (ii) means that every element of m is a zerodivisor in M/{x 1,, x n }M. So the lemma is just a reformulation of Lemma Proposition Assume that M 0. Then depth M is finite, and coincides with the length of any maximal M-regular sequence. Proof. If (x 1,, x n ) is a maximal M-regular sequence, then depth M = n by Lemma Such a sequence always exists by Lemma Combining Proposition and Proposition 5.1.4, we obtain: Corollary If M 0, then depth M dim M. We can be more precise: Proposition We have depth M dim A/p for every p Ass(M).

### COHEN-MACAULAY RINGS SELECTED EXERCISES. 1. Problem 1.1.9

COHEN-MACAULAY RINGS SELECTED EXERCISES KELLER VANDEBOGERT 1. Problem 1.1.9 Proceed by induction, and suppose x R is a U and N-regular element for the base case. Suppose now that xm = 0 for some m M. We

### On the vanishing of Tor of the absolute integral closure

On the vanishing of Tor of the absolute integral closure Hans Schoutens Department of Mathematics NYC College of Technology City University of New York NY, NY 11201 (USA) Abstract Let R be an excellent

### CRing Project, Chapter 16

Contents 16 Homological theory of local rings 3 1 Depth........................................... 3 1.1 Depth over local rings.............................. 3 1.2 Regular sequences................................

### Injective Modules and Matlis Duality

Appendix A Injective Modules and Matlis Duality Notes on 24 Hours of Local Cohomology William D. Taylor We take R to be a commutative ring, and will discuss the theory of injective R-modules. The following

### INJECTIVE MODULES: PREPARATORY MATERIAL FOR THE SNOWBIRD SUMMER SCHOOL ON COMMUTATIVE ALGEBRA

INJECTIVE MODULES: PREPARATORY MATERIAL FOR THE SNOWBIRD SUMMER SCHOOL ON COMMUTATIVE ALGEBRA These notes are intended to give the reader an idea what injective modules are, where they show up, and, to

### REPRESENTATION THEORY WEEK 9

REPRESENTATION THEORY WEEK 9 1. Jordan-Hölder theorem and indecomposable modules Let M be a module satisfying ascending and descending chain conditions (ACC and DCC). In other words every increasing sequence

### CRing Project, Chapter 5

Contents 5 Noetherian rings and modules 3 1 Basics........................................... 3 1.1 The noetherian condition............................ 3 1.2 Stability properties................................

### Cohomology and Base Change

Cohomology and Base Change Let A and B be abelian categories and T : A B and additive functor. We say T is half-exact if whenever 0 M M M 0 is an exact sequence of A-modules, the sequence T (M ) T (M)

### Lecture 8. Throughout this lecture, A will be a ring and M, N etc will be A-modules.

Lecture 8 1. Associated Primes and Primary decomposition for Modules We would like to describe in more detail the structure of submodules of a finitely generated module M over a Noetherian ring A. We will

### The most important result in this section is undoubtedly the following theorem.

28 COMMUTATIVE ALGEBRA 6.4. Examples of Noetherian rings. So far the only rings we can easily prove are Noetherian are principal ideal domains, like Z and k[x], or finite. Our goal now is to develop theorems

### Lecture 2. (1) Every P L A (M) has a maximal element, (2) Every ascending chain of submodules stabilizes (ACC).

Lecture 2 1. Noetherian and Artinian rings and modules Let A be a commutative ring with identity, A M a module, and φ : M N an A-linear map. Then ker φ = {m M : φ(m) = 0} is a submodule of M and im φ is

### Commutative Algebra. Contents. B Totaro. Michaelmas Basics Rings & homomorphisms Modules Prime & maximal ideals...

Commutative Algebra B Totaro Michaelmas 2011 Contents 1 Basics 4 1.1 Rings & homomorphisms.............................. 4 1.2 Modules........................................ 6 1.3 Prime & maximal ideals...............................

### Homological Dimension

Homological Dimension David E V Rose April 17, 29 1 Introduction In this note, we explore the notion of homological dimension After introducing the basic concepts, our two main goals are to give a proof

### Notes for Boot Camp II

Notes for Boot Camp II Mengyuan Zhang Last updated on September 7, 2016 1 The following are notes for Boot Camp II of the Commutative Algebra Student Seminar. No originality is claimed anywhere. The main

### Honors Algebra 4, MATH 371 Winter 2010 Assignment 4 Due Wednesday, February 17 at 08:35

Honors Algebra 4, MATH 371 Winter 2010 Assignment 4 Due Wednesday, February 17 at 08:35 1. Let R be a commutative ring with 1 0. (a) Prove that the nilradical of R is equal to the intersection of the prime

### Commutative Algebra. B Totaro. Michaelmas Basics Rings & homomorphisms Modules Prime & maximal ideals...

Commutative Algebra B Totaro Michaelmas 2011 Contents 1 Basics 2 1.1 Rings & homomorphisms................... 2 1.2 Modules............................. 4 1.3 Prime & maximal ideals....................

### Rings and groups. Ya. Sysak

Rings and groups. Ya. Sysak 1 Noetherian rings Let R be a ring. A (right) R -module M is called noetherian if it satisfies the maximum condition for its submodules. In other words, if M 1... M i M i+1...

### PRIMARY DECOMPOSITION OF MODULES

PRIMARY DECOMPOSITION OF MODULES SUDHIR R. GHORPADE AND JUGAL K. VERMA Department of Mathematics, Indian Institute of Technology, Mumbai 400 076, India E-Mail: {srg, jkv}@math.iitb.ac.in Abstract. Basics

### LIVIA HUMMEL AND THOMAS MARLEY

THE AUSLANDER-BRIDGER FORMULA AND THE GORENSTEIN PROPERTY FOR COHERENT RINGS LIVIA HUMMEL AND THOMAS MARLEY Abstract. The concept of Gorenstein dimension, defined by Auslander and Bridger for finitely

### MATH 221 NOTES BRENT HO. Date: January 3, 2009.

MATH 22 NOTES BRENT HO Date: January 3, 2009. 0 Table of Contents. Localizations......................................................................... 2 2. Zariski Topology......................................................................

### Cohen-Macaulay Dimension for Coherent Rings

University of Nebraska - Lincoln DigitalCommons@University of Nebraska - Lincoln Dissertations, Theses, and Student Research Papers in Mathematics Mathematics, Department of 5-2016 Cohen-Macaulay Dimension

### Formal power series rings, inverse limits, and I-adic completions of rings

Formal power series rings, inverse limits, and I-adic completions of rings Formal semigroup rings and formal power series rings We next want to explore the notion of a (formal) power series ring in finitely

### ALGEBRA HW 4. M 0 is an exact sequence of R-modules, then M is Noetherian if and only if M and M are.

ALGEBRA HW 4 CLAY SHONKWILER (a): Show that if 0 M f M g M 0 is an exact sequence of R-modules, then M is Noetherian if and only if M and M are. Proof. ( ) Suppose M is Noetherian. Then M injects into

### Assigned homework problems S. L. Kleiman, fall 2008

18.705 Assigned homework problems S. L. Kleiman, fall 2008 Problem Set 1. Due 9/11 Problem R 1.5 Let ϕ: A B be a ring homomorphism. Prove that ϕ 1 takes prime ideals P of B to prime ideals of A. Prove

### THE AUSLANDER BUCHSBAUM FORMULA. 0. Overview. This talk is about the Auslander-Buchsbaum formula:

THE AUSLANDER BUCHSBAUM FORMULA HANNO BECKER Abstract. This is the script for my talk about the Auslander-Buchsbaum formula [AB57, Theorem 3.7] at the Auslander Memorial Workshop, 15 th -18 th of November

### Course 311: Michaelmas Term 2005 Part III: Topics in Commutative Algebra

Course 311: Michaelmas Term 2005 Part III: Topics in Commutative Algebra D. R. Wilkins Contents 3 Topics in Commutative Algebra 2 3.1 Rings and Fields......................... 2 3.2 Ideals...............................

### COURSE SUMMARY FOR MATH 508, WINTER QUARTER 2017: ADVANCED COMMUTATIVE ALGEBRA

COURSE SUMMARY FOR MATH 508, WINTER QUARTER 2017: ADVANCED COMMUTATIVE ALGEBRA JAROD ALPER WEEK 1, JAN 4, 6: DIMENSION Lecture 1: Introduction to dimension. Define Krull dimension of a ring A. Discuss

### FILTERED RINGS AND MODULES. GRADINGS AND COMPLETIONS.

FILTERED RINGS AND MODULES. GRADINGS AND COMPLETIONS. Let A be a ring, for simplicity assumed commutative. A filtering, or filtration, of an A module M means a descending sequence of submodules M = M 0

### NOTES FOR COMMUTATIVE ALGEBRA M5P55

NOTES FOR COMMUTATIVE ALGEBRA M5P55 AMBRUS PÁL 1. Rings and ideals Definition 1.1. A quintuple (A, +,, 0, 1) is a commutative ring with identity, if A is a set, equipped with two binary operations; addition

### 11. Finitely-generated modules

11. Finitely-generated modules 11.1 Free modules 11.2 Finitely-generated modules over domains 11.3 PIDs are UFDs 11.4 Structure theorem, again 11.5 Recovering the earlier structure theorem 11.6 Submodules

### ALGEBRAIC GEOMETRY COURSE NOTES, LECTURE 2: HILBERT S NULLSTELLENSATZ.

ALGEBRAIC GEOMETRY COURSE NOTES, LECTURE 2: HILBERT S NULLSTELLENSATZ. ANDREW SALCH 1. Hilbert s Nullstellensatz. The last lecture left off with the claim that, if J k[x 1,..., x n ] is an ideal, then

### Lecture 6. s S} is a ring.

Lecture 6 1 Localization Definition 1.1. Let A be a ring. A set S A is called multiplicative if x, y S implies xy S. We will assume that 1 S and 0 / S. (If 1 / S, then one can use Ŝ = {1} S instead of

### ADVANCED COMMUTATIVE ALGEBRA: PROBLEM SETS

ADVANCED COMMUTATIVE ALGEBRA: PROBLEM SETS UZI VISHNE The 11 problem sets below were composed by Michael Schein, according to his course. Take into account that we are covering slightly different material.

### A NEW PROOF OF SERRE S HOMOLOGICAL CHARACTERIZATION OF REGULAR LOCAL RINGS

A NEW PROOF OF SERRE S HOMOLOGICAL CHARACTERIZATION OF REGULAR LOCAL RINGS RAVI JAGADEESAN AND AARON LANDESMAN Abstract. We give a new proof of Serre s result that a Noetherian local ring is regular if

### 4.4 Noetherian Rings

4.4 Noetherian Rings Recall that a ring A is Noetherian if it satisfies the following three equivalent conditions: (1) Every nonempty set of ideals of A has a maximal element (the maximal condition); (2)

### Primal, completely irreducible, and primary meet decompositions in modules

Bull. Math. Soc. Sci. Math. Roumanie Tome 54(102) No. 4, 2011, 297 311 Primal, completely irreducible, and primary meet decompositions in modules by Toma Albu and Patrick F. Smith Abstract This paper was

### NOTES ON BASIC HOMOLOGICAL ALGEBRA 0 L M N 0

NOTES ON BASIC HOMOLOGICAL ALGEBRA ANDREW BAKER 1. Chain complexes and their homology Let R be a ring and Mod R the category of right R-modules; a very similar discussion can be had for the category of

### Lie Algebra Cohomology

Lie Algebra Cohomology Carsten Liese 1 Chain Complexes Definition 1.1. A chain complex (C, d) of R-modules is a family {C n } n Z of R-modules, together with R-modul maps d n : C n C n 1 such that d d

### Math 210B. Artin Rees and completions

Math 210B. Artin Rees and completions 1. Definitions and an example Let A be a ring, I an ideal, and M an A-module. In class we defined the I-adic completion of M to be M = lim M/I n M. We will soon show

### Adic Spaces. Torsten Wedhorn. June 19, 2012

Adic Spaces Torsten Wedhorn June 19, 2012 This script is highly preliminary and unfinished. It is online only to give the audience of our lecture easy access to it. Therefore usage is at your own risk.

### Tensor Product of modules. MA499 Project II

Tensor Product of modules A Project Report Submitted for the Course MA499 Project II by Subhash Atal (Roll No. 07012321) to the DEPARTMENT OF MATHEMATICS INDIAN INSTITUTE OF TECHNOLOGY GUWAHATI GUWAHATI

### A Primer on Homological Algebra

A Primer on Homological Algebra Henry Y Chan July 12, 213 1 Modules For people who have taken the algebra sequence, you can pretty much skip the first section Before telling you what a module is, you probably

### A TALE OF TWO FUNCTORS. Marc Culler. 1. Hom and Tensor

A TALE OF TWO FUNCTORS Marc Culler 1. Hom and Tensor It was the best of times, it was the worst of times, it was the age of covariance, it was the age of contravariance, it was the epoch of homology, it

### FLAT RING EPIMORPHISMS AND UNIVERSAL LOCALISATIONS OF COMMUTATIVE RINGS

FLAT RING EPIMORPHISMS AND UNIVERSAL LOCALISATIONS OF COMMUTATIVE RINGS LIDIA ANGELERI HÜGEL, FREDERIK MARKS, JAN ŠŤOVÍČEK, RYO TAKAHASHI, JORGE VITÓRIA Abstract. We study different types of localisations

### EXT, TOR AND THE UCT

EXT, TOR AND THE UCT CHRIS KOTTKE Contents 1. Left/right exact functors 1 2. Projective resolutions 2 3. Two useful lemmas 3 4. Ext 6 5. Ext as a covariant derived functor 8 6. Universal Coefficient Theorem

### TCC Homological Algebra: Assignment #3 (Solutions)

TCC Homological Algebra: Assignment #3 (Solutions) David Loeffler, d.a.loeffler@warwick.ac.uk 30th November 2016 This is the third of 4 problem sheets. Solutions should be submitted to me (via any appropriate

### Math 121 Homework 5: Notes on Selected Problems

Math 121 Homework 5: Notes on Selected Problems 12.1.2. Let M be a module over the integral domain R. (a) Assume that M has rank n and that x 1,..., x n is any maximal set of linearly independent elements

### Introduction Non-uniqueness of factorization in A[x]... 66

Abstract In this work, we study the factorization in A[x], where A is an Artinian local principal ideal ring (briefly SPIR), whose maximal ideal, (t), has nilpotency h: this is not a Unique Factorization

### VANISHING THEOREMS FOR COMPLETE INTERSECTIONS. Craig Huneke, David A. Jorgensen and Roger Wiegand. August 15, 2000

VANISHING THEOREMS FOR COMPLETE INTERSECTIONS Craig Huneke, David A. Jorgensen and Roger Wiegand August 15, 2000 The starting point for this work is Auslander s seminal paper [A]. The main focus of his

### ALGEBRA EXERCISES, PhD EXAMINATION LEVEL

ALGEBRA EXERCISES, PhD EXAMINATION LEVEL 1. Suppose that G is a finite group. (a) Prove that if G is nilpotent, and H is any proper subgroup, then H is a proper subgroup of its normalizer. (b) Use (a)

### Integral Extensions. Chapter Integral Elements Definitions and Comments Lemma

Chapter 2 Integral Extensions 2.1 Integral Elements 2.1.1 Definitions and Comments Let R be a subring of the ring S, and let α S. We say that α is integral over R if α isarootofamonic polynomial with coefficients

### Advanced Algebra II. Mar. 2, 2007 (Fri.) 1. commutative ring theory In this chapter, rings are assume to be commutative with identity.

Advanced Algebra II Mar. 2, 2007 (Fri.) 1. commutative ring theory In this chapter, rings are assume to be commutative with identity. 1.1. basic definitions. We recall some basic definitions in the section.

### DOUGLAS J. DAILEY AND THOMAS MARLEY

A CHANGE OF RINGS RESULT FOR MATLIS REFLEXIVITY DOUGLAS J. DAILEY AND THOMAS MARLEY Abstract. Let R be a commutative Noetherian ring and E the minimal injective cogenerator of the category of R-modules.

### Ring Theory Problems. A σ

Ring Theory Problems 1. Given the commutative diagram α A σ B β A σ B show that α: ker σ ker σ and that β : coker σ coker σ. Here coker σ = B/σ(A). 2. Let K be a field, let V be an infinite dimensional

### Reid 5.2. Describe the irreducible components of V (J) for J = (y 2 x 4, x 2 2x 3 x 2 y + 2xy + y 2 y) in k[x, y, z]. Here k is algebraically closed.

Reid 5.2. Describe the irreducible components of V (J) for J = (y 2 x 4, x 2 2x 3 x 2 y + 2xy + y 2 y) in k[x, y, z]. Here k is algebraically closed. Answer: Note that the first generator factors as (y

### FROBENIUS AND HOMOLOGICAL DIMENSIONS OF COMPLEXES

FROBENIUS AND HOMOLOGICAL DIMENSIONS OF COMPLEXES TARAN FUNK AND THOMAS MARLEY Abstract. It is proved that a module M over a Noetherian local ring R of prime characteristic and positive dimension has finite

### Spring 2016, lecture notes by Maksym Fedorchuk 51

Spring 2016, lecture notes by Maksym Fedorchuk 51 10.2. Problem Set 2 Solution Problem. Prove the following statements. (1) The nilradical of a ring R is the intersection of all prime ideals of R. (2)

### 1. Algebraic vector bundles. Affine Varieties

0. Brief overview Cycles and bundles are intrinsic invariants of algebraic varieties Close connections going back to Grothendieck Work with quasi-projective varieties over a field k Affine Varieties 1.

### Solutions Ark3. 1 It is also an exercise in the aikido toho iai of writing diagrams in L A TEX

Solutions Ark3 From the book: Number 1, 2, 3, 5 and 6 on page 43 and 44. Number 1: Let S be a multiplicatively closed subset of a ring A, and let M be a finitely generated A-module. Then S 1 M = 0 if and

### MATH 101B: ALGEBRA II PART A: HOMOLOGICAL ALGEBRA 23

MATH 101B: ALGEBRA II PART A: HOMOLOGICAL ALGEBRA 23 6.4. Homotopy uniqueness of projective resolutions. Here I proved that the projective resolution of any R-module (or any object of an abelian category

### Semidualizing Modules. Sean Sather-Wagstaff

Semidualizing Modules Sean Sather-Wagstaff Department of Mathematics, North Dakota State University Department # 2750, PO Box 6050, Fargo, ND 58108-6050, USA E-mail address: sean.sather-wagstaff@ndsu.edu

### Commutative Algebra. Instructor: Graham Leuschke & Typist: Caleb McWhorter. Spring 2016

Commutative Algebra Instructor: Graham Leuschke & Typist: Caleb McWhorter Spring 2016 Contents 1 General Remarks 3 2 Localization 3 2.1 Localization......................................... 3 2.6 The Construction

### On Extensions of Modules

On Extensions of Modules Janet Striuli Department of Mathematics, University of Kansas, Lawrence, KS 66045, USA Abstract In this paper we study closely Yoneda s correspondence between short exact sequences

### The Depth Formula for Modules with Reducible Complexity

The Depth Formula for Modules with Reducible Complexity Petter Andreas Bergh David A Jorgensen Technical Report 2010-10 http://wwwutaedu/math/preprint/ THE DEPTH FORMULA FOR MODULES WITH REDUCIBLE COMPLEXITY

### Notes on p-divisible Groups

Notes on p-divisible Groups March 24, 2006 This is a note for the talk in STAGE in MIT. The content is basically following the paper [T]. 1 Preliminaries and Notations Notation 1.1. Let R be a complete

### (1) A frac = b : a, b A, b 0. We can define addition and multiplication of fractions as we normally would. a b + c d

The Algebraic Method 0.1. Integral Domains. Emmy Noether and others quickly realized that the classical algebraic number theory of Dedekind could be abstracted completely. In particular, rings of integers

### COMMUTATIVE ALGEBRA, LECTURE NOTES

COMMUTATIVE ALGEBRA, LECTURE NOTES P. SOSNA Contents 1. Very brief introduction 2 2. Rings and Ideals 2 3. Modules 10 3.1. Tensor product of modules 15 3.2. Flatness 18 3.3. Algebras 21 4. Localisation

### Primary Decomposition and Associated Primes

Chapter 1 Primary Decomposition and Associated Primes 1.1 Primary Submodules and Ideals 1.1.1 Definitions and Comments If N is a submodule of the R-module M, and a R, let λ a : M/N M/N be multiplication

### Noetherian property of infinite EI categories

Noetherian property of infinite EI categories Wee Liang Gan and Liping Li Abstract. It is known that finitely generated FI-modules over a field of characteristic 0 are Noetherian. We generalize this result

### Quasi-primary submodules satisfying the primeful property II

Hacettepe Journal of Mathematics and Statistics Volume 44 (4) (2015), 801 811 Quasi-primary submodules satisfying the primeful property II Hosein Fazaeli Moghimi and Mahdi Samiei Abstract In this paper

### Subrings of Finite Commutative Rings

Subrings of Finite Commutative Rings Francisco Franco Munoz arxiv:1712.02025v1 [math.ac] 6 Dec 2017 Abstract We record some basic results on subrings of finite commutative rings. Among them we establish

### ON THE REPRESENTABILITY OF Hilb n k[x] (x) Roy Mikael Skjelnes

ON THE REPRESENTABILITY OF Hilb n k[x] (x) Roy Mikael Skjelnes Abstract. Let k[x] (x) be the polynomial ring k[x] localized in the maximal ideal (x) k[x]. We study the Hilbert functor parameterizing ideals

### 12. Projective modules The blanket assumptions about the base ring k, the k-algebra A, and A-modules enumerated at the start of 11 continue to hold.

12. Projective modules The blanket assumptions about the base ring k, the k-algebra A, and A-modules enumerated at the start of 11 continue to hold. 12.1. Indecomposability of M and the localness of End

### n P say, then (X A Y ) P

COMMUTATIVE ALGEBRA 35 7.2. The Picard group of a ring. Definition. A line bundle over a ring A is a finitely generated projective A-module such that the rank function Spec A N is constant with value 1.

### Math 711: Lecture of September 7, Symbolic powers

Math 711: Lecture of September 7, 2007 Symbolic powers We want to make a number of comments about the behavior of symbolic powers of prime ideals in Noetherian rings, and to give at least one example of

### SEQUENCES FOR COMPLEXES II

SEQUENCES FOR COMPLEXES II LARS WINTHER CHRISTENSEN 1. Introduction and Notation This short paper elaborates on an example given in [4] to illustrate an application of sequences for complexes: Let R be

### 214A HOMEWORK KIM, SUNGJIN

214A HOMEWORK KIM, SUNGJIN 1.1 Let A = k[[t ]] be the ring of formal power series with coefficients in a field k. Determine SpecA. Proof. We begin with a claim that A = { a i T i A : a i k, and a 0 k }.

### FLAT RING EPIMORPHISMS AND UNIVERSAL LOCALISATIONS OF COMMUTATIVE RINGS

FLAT RING EPIMORPHISMS AND UNIVERSAL LOCALISATIONS OF COMMUTATIVE RINGS LIDIA ANGELERI HÜGEL, FREDERIK MARKS, JAN ŠŤOVÍČEK, RYO TAKAHASHI, JORGE VITÓRIA Abstract. We study different types of localisations

### THE RADIUS OF A SUBCATEGORY OF MODULES

THE RADIUS OF A SUBCATEGORY OF MODULES HAILONG DAO AND RYO TAKAHASHI Dedicated to Professor Craig Huneke on the occasion of his sixtieth birthday Abstract. We introduce a new invariant for subcategories

### MATH 233B, FLATNESS AND SMOOTHNESS.

MATH 233B, FLATNESS AND SMOOTHNESS. The discussion of smooth morphisms is one place were Hartshorne doesn t do a very good job. Here s a summary of this week s material. I ll also insert some (optional)

### Structure of rings. Chapter Algebras

Chapter 5 Structure of rings 5.1 Algebras It is time to introduce the notion of an algebra over a commutative ring. So let R be a commutative ring. An R-algebra is a ring A (unital as always) together

### Primary Decomposition and Secondary Representation of Modules over a Commutative Ring

Georgia State University ScholarWorks @ Georgia State University Mathematics Theses Department of Mathematics and Statistics 4-21-2009 Primary Decomposition and Secondary Representation of Modules over

### REFLEXIVE MODULES OVER GORENSTEIN RINGS

REFLEXIVE MODULES OVER GORENSTEIN RINGS WOLMER V. VASCONCELOS1 Introduction. The aim of this paper is to show the relevance of a class of commutative noetherian rings to the study of reflexive modules.

### MATH 326: RINGS AND MODULES STEFAN GILLE

MATH 326: RINGS AND MODULES STEFAN GILLE 1 2 STEFAN GILLE 1. Rings We recall first the definition of a group. 1.1. Definition. Let G be a non empty set. The set G is called a group if there is a map called

### J A S O N M C C U L L O U G H S T I L L M A N S Q U E S T I O N A N D T H E E I S E N B U D - G O T O C O N J E C T U R E

J A S O N M C C U L L O U G H S T I L L M A N S Q U E S T I O N A N D T H E E I S E N B U D - G O T O C O N J E C T U R E Copyright 2017 Jason McCullough tufte-latex.googlecode.com This work is licensed

### REPRESENTATION THEORY, LECTURE 0. BASICS

REPRESENTATION THEORY, LECTURE 0. BASICS IVAN LOSEV Introduction The aim of this lecture is to recall some standard basic things about the representation theory of finite dimensional algebras and finite

### Commutative Algebra. Timothy J. Ford

Commutative Algebra Timothy J. Ford DEPARTMENT OF MATHEMATICS, FLORIDA ATLANTIC UNIVERSITY, BOCA RA- TON, FL 33431 Email address: ford@fau.edu URL: http://math.fau.edu/ford Last modified January 9, 2018.

### Rees Algebras of Modules

Rees Algebras of Modules ARON SIMIS Departamento de Matemática Universidade Federal de Pernambuco 50740-540 Recife, PE, Brazil e-mail: aron@dmat.ufpe.br BERND ULRICH Department of Mathematics Michigan

### Modules over Principal Ideal Domains

Modules over Principal Ideal Domains Let henceforth R denote a commutative ring with 1. It is called a domain iff it has no zero-divisors, i.e. if ab = 0 then either a or b is zero. Or equivalently, two

### SCHEMES. David Harari. Tsinghua, February-March 2005

SCHEMES David Harari Tsinghua, February-March 2005 Contents 1. Basic notions on schemes 2 1.1. First definitions and examples.................. 2 1.2. Morphisms of schemes : first properties.............

### ON STRONGLY PRIME IDEALS AND STRONGLY ZERO-DIMENSIONAL RINGS. Christian Gottlieb

ON STRONGLY PRIME IDEALS AND STRONGLY ZERO-DIMENSIONAL RINGS Christian Gottlieb Department of Mathematics, University of Stockholm SE-106 91 Stockholm, Sweden gottlieb@math.su.se Abstract A prime ideal

### ABSTRACT NONSINGULAR CURVES

ABSTRACT NONSINGULAR CURVES Affine Varieties Notation. Let k be a field, such as the rational numbers Q or the complex numbers C. We call affine n-space the collection A n k of points P = a 1, a,..., a

### NOTES ON CHAIN COMPLEXES

NOTES ON CHAIN COMPLEXES ANDEW BAKE These notes are intended as a very basic introduction to (co)chain complexes and their algebra, the intention being to point the beginner at some of the main ideas which

### ALGEBRAIC GEOMETRY COURSE NOTES, LECTURE 9: SCHEMES AND THEIR MODULES.

ALGEBRAIC GEOMETRY COURSE NOTES, LECTURE 9: SCHEMES AND THEIR MODULES. ANDREW SALCH 1. Affine schemes. About notation: I am in the habit of writing f (U) instead of f 1 (U) for the preimage of a subset

### COMMUNICATIONS IN ALGEBRA, 15(3), (1987) A NOTE ON PRIME IDEALS WHICH TEST INJECTIVITY. John A. Beachy and William D.

COMMUNICATIONS IN ALGEBRA, 15(3), 471 478 (1987) A NOTE ON PRIME IDEALS WHICH TEST INJECTIVITY John A. Beachy and William D. Weakley Department of Mathematical Sciences Northern Illinois University DeKalb,

### BENJAMIN LEVINE. 2. Principal Ideal Domains We will first investigate the properties of principal ideal domains and unique factorization domains.

FINITELY GENERATED MODULES OVER A PRINCIPAL IDEAL DOMAIN BENJAMIN LEVINE Abstract. We will explore classification theory concerning the structure theorem for finitely generated modules over a principal

### Schemes via Noncommutative Localisation

Schemes via Noncommutative Localisation Daniel Murfet September 18, 2005 In this note we give an exposition of the well-known results of Gabriel, which show how to define affine schemes in terms of the

### HARTSHORNE EXERCISES

HARTSHORNE EXERCISES J. WARNER Hartshorne, Exercise I.5.6. Blowing Up Curve Singularities (a) Let Y be the cusp x 3 = y 2 + x 4 + y 4 or the node xy = x 6 + y 6. Show that the curve Ỹ obtained by blowing