Notes for Boot Camp II
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1 Notes for Boot Camp II Mengyuan Zhang Last updated on September 7,
2 The following are notes for Boot Camp II of the Commutative Algebra Student Seminar. No originality is claimed anywhere. The main references are 1. [Eis]: D Eisenbud, 1995, Commutative Algebra: with a view towards Algebraic Geometry. 2. [Eis1]: D Eisenbud, 2005, The Geometry of Syzygies. 3. [Wei]: C Weibel, 1994, An Introduction to Homological Algebra. Throughout, let R denote a commutative unital Noetherian ring and let M denote a finitely generated R- module. We often make stronger statements in the case where (R, m) is a local ring, and these statements remain true in the following situation: if R = k R + is a non-negatively graded Noetherian ring over k with unique maximal homogenous ideal m = R + and M is a bounded below graded R-module and locally finite dimensional over k, where we further require all elements in the statements to be homogenous. 2
3 1 Depth Definition (Regular sequence). A sequence x 1,, x n R is a regular sequence on M (or a M-sequence) if 1. (x 1,, x n )M M; 2. x 1 is a nonzero-divisor on M and x i+1 is a nonzero-divisor on M/(x 1,, x i )M for all i > 1. If in addition x 1,, x n I for some ideal I, then we say x 1,, x n is an M-sequence in I. An M-sequence x 1,, x n in I is maximal if for any y I, y is a zero-divisor on M/(x 1,, x n )M. Note that if (R, m) is a local ring, then any M-sequence is contained in m. Example 1 (Example of maximal regular sequences). Let R = k[x, y, z] and M = R/(xy, y 2 ). We compute the primary decomposition of I = (xy, y 2 ) = (x, y 2 ) (y), and deduce that Ass M = {(x, y), (y)}. Since the set of zero-divisors of M is the union of associated primes of M, we see that z is a nonzero-divisor on M. Now M/zM = R/(xy, y 2, z), and I + (z) = (z, xy, y 2 ) = (x, y 2, z) (y, z) is a primary decomposition, so that Ass M = {(x, y, z), (y, z)}. We cannot find any non-zerodivisor in R + on M/zM, consequently z is a maximal M-sequence. Note that x + z is also a non-zerodivisor on M, and M/(x + z)m = R/(x + z, xy, y 2 ). Since (x + z, xy, y 2 ) = (x, y 2, z) (x + z, y) is a primary decomposition. We see that x + z is also a maximal M-sequence. Example 2 (Order of a regular sequence). Let R = k[x, y, z]/(x 1)z. Then x, (x 1)y is an R-sequence, but (x 1)y, x is not. Proposition 1.1 (Noetherian local case). Suppose (R, m) is local, then x 1,, x n is a M-sequence iff any permutation of it is a M-sequence. Proof. See [Eis] Corollary Theorem 1.2 (Length of maximal regular sequences). Let I be an ideal of R, then any maximal M- sequence in I has the same length. We define this number to be depth(i, M). If IM = M, then we write depth(i, M) =. If (R, m) is local, we abbreviate depth(m, M) simply to depth(m). We also write depth(i) = depth(i, R). Proof. See [Eis] Theorem
4 Proposition 1.3 (Geometric nature of depth). If x 1,, x n is a M-sequence, then for any positive integers l 1,, l n, x l1 1,, xln n is also an M-sequence. It follows that depth(i, M) = depth( I, M) for any ideal I. Proof. See [Eis] Corollary Proposition 1.4 (Depth codim inequality). For any ideal I of R, we have depth(i, M) codim(i + ann M)/ ann M dim R/ ann M = dim M. In particular, depth(r) dim R for local rings R. Proof. See [Eis] Proposition Example 3. If R is Artinian, then depth R = 0. Thus every element in m is a zero-divisor. Theorem 1.5 (Cohomology characterization of depth). The following gives equivalent definitions of depth(i, M): 1. depth(i, M) = inf{i H i (x 1,, x n ; M) 0, I = (x 1,, x n )}; 2. depth(i, M) = inf{i Ext i R(R/I, M) 0}; 3. depth(i, M) = inf{i HI i (M) 0}. Proof. See [Eis] Theorem 17.4; See [Eis] Proposition 18.4; See [Eis] Theorem A4.3. Corollary 1.6 (Depth and short exact sequences). If 0 M M M 0 is an exact sequence of finitely generated R-modules, then Proof. Use long exact sequence of Ext. depth(m ) min(depth(m), depth(n ) 1) depth(m ) min(depth(m), depth(m ) + 1). Theorem 1.7 (Auslander-Buchsbaum). Let (R, m) be a local ring. If pdim(m) <, then Proof. See [Eis] Theorem pdim(m) = depth(r) depth(m). Corollary 1.8. If (R, m) is local and pdim(m) <. If depth(m) = depth(r) then M is free. Example 4. Let (R, m) be local with depth n, and M be a module with pdim(m) < and depth(m) = d. Resolve M n d steps to obtain the n d-th syzygy module Ω n d (M). By depth lemma on exact sequences and induction, we see that depth R depth(ω n d (M)) depth R and thus equality holds. Since pdim(m) <, we also have pdim Ω n d (M) <. Consequently pdim Ω n d (M) = 0 by Auslander-Buchsbaum and Ω n d (M) is free. It follows that pdim(m) = n d indeed. 4
5 2 The Koszul Complex Definition (Tensor product of complexes). Let (F, d), (G, d ) be two chain complexes, then (F G, ) is the chain complex with components (F G) n = F i G j i+j=n and differential (x, y) = (dx, y) + (1) deg(x) (x, d y) for homogeneous x F i and y G j. It is an easy but cumbersome exercise to check that F G = G F and (F G) H = F (G H) as complexes. Definition (Koszul complex and homology). For x R, we define the Koszul complex K(x) to be the chain complex 0 R x R 0. Let x 1,, x n R, and let M be an R-module. We define the Koszul complex to be K(x 1,, x n ; M) = K(x 1 ) K(x n ) M, and the Koszul homology to be H i (x 1,, x n ; M) = H i (K(x 1,, x n ; M)). We define the Koszul cochain complex to be K(x 1,, x n ; M) = Hom(K(x 1 ) K(x n ), M), and the Koszul cohomology to be H i (x 1,, x n ; M) = H i (K(x 1,, x n ; M) ). Example 5 (Koszul complex of length 3). K(x, y, z) : K(x, y) : z y x 0 R R R R K(x) : 0 R x R 0. y [ ] x x y 0 R R R R 0. y z 0 x 0 z [ ] 0 x y x y z R R R R 0. Theorem 2.1 (Duality of Koszul complex). If we regard K(x 1,, x n ; M) as a chain complex and shift degrees to align with K(x 1,, x n ; M), then they are isomorphic. Consequently H i (x 1,, x n ; M) = H n i (x 1,, x n ; M). 5
6 Proof. See [Eis] Theorem Proposition 2.2. If x 1,, x n is an M-sequence, then H i (x 1,, x n ; M) = 0, i < n and H n (x 1,, x n ; M) = M/(x 1,, x n )M 0. In particular, if x 1,, x n is an R-sequence, then the Koszul complex K(x 1,, x n ; R) is a free resolution of R/(x 1,, x n ). Proof. This follows from the fact that depth is equal to the least vanishing Koszul cohomology and duality of Koszul homology/cohomology. Example 6. A word of caution that the above is not sufficient to conclude that x 1,, x n is an M- sequence. We have seen that permutations of regular sequences need not be regular, but all permutations of the sequence have isomorphic Koszul complexes thus the same Koszul cohomology. For example, take R = k[x, y, z]/(x 1)z and (x 1)y, x is not a regular sequence but H 0 ((x 1)y, x; R) = H 1 ((x 1)y, x; R) = 0 and H 2 ((x 1)y, x; R) 0. However, if we assume R is local, the converse is true as well. Proposition 2.3. Suppose (R, m) is local and x 1,, x n m. Then H i (x 1,, x n ; M) = 0 = H i 1 (x 1,, x n ; M) = 0. If H n 1 (x 1,, x n ; M) = 0, then x 1,, x n is an M-sequence. Proof. See [Eis] Theorem Corollary 2.4. Suppose (R, m) is local. x 1,, x n is an M-sequence. If I = (x 1,, x n ) contains an M-sequence of length n, then Proof. Nakayama s Lemma shows that H n (x 1,, x n ; M) = M/IM 0. Since I contains an M-sequence of length n, we must have H i (x 1,, x n ; M) = 0 for i < n. 6
7 3 Projective Dimension Definition. We define pdim(m) = inf{length F F is a projective resolution of M}, and similarly idim(m) = inf{length G G is an injective resolution of M}. Theorem 3.1 (Global dimension theorem). The following numbers are the same 1. sup{pdim(m) M is an R-module}; 2. sup{pdim(m) M is a finitely generated R-module}; 3. sup{pdim(m) M is a cyclic R-module}; 4. sup{idim(m) M is an R-module}; 5. sup{idim(m) M is a finitely generated R-module}; 6. sup{d Ext d R(M, N) 0 for some R-modules M,N}. This number, possibly, is called the global dimension of a ring. Proof. See [Wei] Theorem Example 7. Over R = Z, the module M = Z/n has a free resolution 0 Z n Z of length 1. Since M is not projective, we see that pdim(m) = 1. By 3 of above theorem, we see that gdim(z) = 1. Proposition 3.2 (Projective dimension lemma). The following are equivalent 1. pdim(m) n; 2. Ext i R(M, N) = 0 for i > n and for all R-modules N; 3. Ext n+1 R (M, N) = 0 for all R-modules N; 4. If 0 M n P n 1 P 1 P 0 M 0 is any resolution with P i projective, then M n is projective. Proof. It is clear that 4 = 1 = 2 = 3. Suppose 3 is true, then Ext 1 R(M n, ) = Ext n+1 (M, ) = 0 by dimension shifting. Thus Hom R (M n, ) is exact and M n is projective. 7
8 Example 8. Let R = Z/p 2 and M = Z/p. Then R p R p R is an infinite projective resolution of M. Further Ext n R(M, M) = M for all n, thus pdim(m) = gdim(r) =. Definition (Minimal free resolution). Let (R, m) be local. A free resolution F of M is called minimal if F R k has 0 differential. Equivalently, all the entries of the matrices of the differentials are in m. Theorem 3.3 (Existence and uniqueness of a minimal resolution). Let (R, m) be local. For any M there exists a minimal free resolution F of M. Any free resolution G of M has F as a direct summand. Thus there is a unique up to isomorphism minimal free resolution of M. Proof. Existence is by Nakayama s Lemma. We pick a k-basis of M/mM, and lift to surjection R n M 0. Now take the kernel of this map and do the same. Uniqueness is by [Eis] Theorem Proposition 3.4 (Projective dimension in the local case). Let (R, m) be local, then pdim(m) n iff Tor R n+1(m, k) = 0. In particular, pdim(m) is the length of the minimal free resolution of M. Furthermore, gdim(r) = pdim(k). Proof. If pdim(m) n, then Tor R n+1(m, k) = 0 trivially. Conversely, let F be a minimal free resolution of M. Then 0 = Tor R n+1(m, k) = H n+1 (F R k) = F n+1, and thus F has length n and pdim(m) n. If pdim M > n, then Tor R n+1(m, k) 0, thus pdim k > n. It follows that pdim k pdim M for any M and thus gdim(r) = pdim(k). 8
9 4 Regular Local Rings Definition (Regular local rings). Let (R, m) be local, and let x 1,, x n be a minimal set of generators of m, i.e. the images x 1,, x n in m/m 2 forms a basis over R/m = k. Krull s PIT implies that dim R n. We say (R, m) is regular is equality holds, i.e. m can be generated by dim R elements. Example 9 (Low dimensional cases). The zero dimensional regular local rings are exactly fields. The one dimensional regular local rings are exactly the DVRs. For example, R = k[x] (x), or R = Z (p). Theorem 4.1 (Auslander-Buchsbaum). Any regular local ring is a UFD. Proof. See [Eis] Theorem Corollary 4.2. If (R, m) is a regular local ring of dimension n, then any n elements that generate m forms a regular sequence. Proof. Let x 1,, x n be a minimal set of generators of m. Since R is a domain, x 1 is a nonzero-divisor. Now R = R/(x 1 ) is a regular local ring because dim R = n 1 and m = mr can be minimally generated by x 2,, x n. Thus R is a domain and x 2 is a nonzero-divisor. Continuing this way, we see that x 1,, x n is a regular sequence. Example 10. The converse of the above is not true. Consider R = k[x] (x), then x 2 is a non-zerodivisor and forms a maximal regular sequence, but it doesn t generate the maximal ideal. Corollary 4.3 (Hilbert Syzygy Theorem). If (R, m) is a regular local ring of dimension n, and x 1,, x n is any set of minimal generators of m, then the Koszul complex K(x 1,, x n ; R) is a minimal free resolution of k. It follows that gdim(r) = pdim(k) = n = dim(r). Thus if M is any finitely generated R-module, then pdim(m) n. Theorem 4.4 (Serre). A Noetherian local ring (R, m) is regular iff gdim(r) <. Proof. If (R, m) is regular of dimension n, then gdim(r) = n <. Conversely, assume gdim(r) = pdim(k) <. Let F be the minimal free resolution of k, and let x 1,, x n be a minimal set of generators of m. Then K(x 1,, x n ; R) is a subcomplex of F by [Eis] Lemma It follows from Auslander-Buchsbaum that we have dim R depth R = depth k n and equality holds. Definition (Regular rings). A Noetherian ring R is regular if gdim R <. 9
10 Proposition 4.5. If R is regular local, then R[U 1 ] is regular for any multiplicative set U of R. Proof. Let F be a minimal free resolution of k, then F[U 1 ] is a minimal free resolution of R[U 1 ]/mr[u 1 ] = k, which is of finite length. Example 11. Let R = k[x, y, z], and M = R/(xy, y 2 ). We have seen depth M = 1. By regularity of R, we have pdim M pdim k = dim R = 3. Thus Auslander-Buchsbaum says that pdim M = 2. Let us actually compute a minimal free resolution of M. y [ x xy, y 2 ] 0 R( 3) R( 2) 2 R M 0. 10
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