More commutative algebra

Let B be a Gorenstein local ring, and let M be a finitely generated B-module. It is a standard fact that M is Cohen-Macaulay iff \mathrm{Ext}^{i}_B (M,B) is zero for all i \neq \mathrm{codim}(M)\overset{def}{=}\dim B - \dim M. Moreover, the functor M \mapsto \breve{M}=\mathrm{Ext}^{d}_{B}(M,B) induces an involutive anti-equivalence on the category of Cohen-Macaulay B-modules of codimension d.

Suppose now that we have a flat local map of local Noetherian rings f: A \to B with B Gorenstein, and that M is a CM B-module which is flat over A. It is natural to ask whether \breve{M} is also flat over A. The goal of this post is to prove the following partial result towards this question.

Theorem. Notation and setup as above, \breve{M} is flat over A if A is regular or if \mathrm{projdim}_B (M) < \infty.

Before continuing, note that A and B/\mathfrak{m}_A B are automatically Gorenstein. This will be useful later.

The first key lemma is the following.

Lemma 1. Let B \leftarrow A \rightarrow A' be a diagram of Noetherian commutative rings, and let M be a finitely generated B-module flat over A. Suppose that the maps B \leftarrow A \rightarrow A' are tor-independent (e.g. one of them is flat), and that at least one of the following conditions holds:
1. \mathrm{projdim}_{B}(M) < \infty;
2. A \to A' is of finite tor-dimension.
Then there is a convergent spectral sequence
\mathrm{Tor}_{-i}^{A}(\mathrm{Ext}^j_{B}(M,B),A') \Rightarrow \mathrm{Ext}^{i+j}_{B'}(M',B'), where B' = B\otimes_A A' and M'=M \otimes_B B'=M \otimes_A A'.

This is not so hard to prove. The only real input is Stacks Tag 0A6A – the auxiliary conditions 1. and 2. above corresponds to cases (1) and (4) of that lemma.

Granted this lemma, the auxiliary hypotheses in the theorem let us apply this spectral sequence with A,B,M chosen as in the theorem and with A' = k=A/\mathfrak{m}_A. Since M is Cohen-Macaulay over B of some codimension d, only j = d contributes, and the spec. seq. degenerates to isomorphisms

\mathrm{Tor}_{n}^{A}(\breve{M},k) \cong \mathrm{Ext}^{d-n}_{B_0}(M_0,B_0).

Here and in what follows, I write (-)_0 = (-) \otimes_A k for pullback to the closed fiber of f. By the local criterion of flatness, the theorem follows if we can show that \mathrm{Tor}_1^{A}(\breve{M},k)=0, i.e. that \mathrm{Ext}^{d-1}_{B_0}(M_0,B_0) = 0. Since B_0 is Gorenstein, it clearly suffices to prove that M_0 is a Cohen-Macaulay B_0-module of codimension d. This is the content of the next lemma, which must be well-known, but which I couldn’t find in ten minutes of googling. Do you know a reference for this?

Lemma 2. If A \to B is a flat local map of Gorenstein local rings, and M is a finitely generated B-module which is flat over A, then M Cohen-Macaulay over B implies M_0 Cohen-Macaulay over B_0 of the same codimension. Here (-)_0 = (-) \otimes_A A/\mathfrak{m}_A as above.

To prove this, first note that d:=\mathrm{codim}(M)=\mathrm{depth}(B) - \mathrm{depth}_B(M) since B and M are CM. Then \mathrm{depth}_{B_0}(M_0) = \mathrm{depth}_B(M)-\mathrm{depth}(A) = \mathrm{depth}(B)-d-\mathrm{depth}(A)=\mathrm{depth}(B_0)-d=\mathrm{dim}(B_0)-d, where the first equality holds e.g. by Theorem 23.3 in Matsumura’s Commutative Ring Theory. (One can also read off the equality \mathrm{depth}_{B_0}(M_0) = \mathrm{dim}(B_0)-d from the spectral sequence argument above.)

So now we just need to show that d=\mathrm{dim}(B_0)-\mathrm{dim}(M_0). But we can check this last equality with (-)_0 replaced by -\otimes_A A' where A\twoheadrightarrow A' is any Artinian quotient. In particular we can assume that A'=A/I where I \subset A is an \mathfrak{m}_A-primary ideal generated by a regular sequence. Then

d = \mathrm{dim}(B)-\mathrm{dim}(M) = {least\; n\; with\; \mathrm{Ext}^n_B(M,B) \neq 0}
= {least\; n\; with\; \mathrm{Ext}^n_{B/I}(M/I,B/I) \neq 0} = \mathrm{dim}(B/I)-\mathrm{dim}(M/I)

where the first equality holds by definition, the second and fourth equalities are a well-known property of f.g. modules over CM local rings, and the third equality follows from the facts that B is Gorenstein and that (thanks to our flatness assumptions) I is generated by an M-regular sequence which is also a B-regular sequence. (To actually prove the third equality, use induction on the number of generators of I.)  Thus d = \mathrm{dim}(B/I)-\mathrm{dim}(M/I) = \mathrm{dim}(B_0)-\mathrm{dim}(M_0) as desired.

Q1. Does the theorem hold more generally?

Q2. Is there a reference for Lemma 2 in the literature? This can’t be a new observation.

Q3. Can the Gorenstein condition in Lemma 2 be weakened? This seems unlikely to me.


4 thoughts on “More commutative algebra”

  1. On Lemma 2. Notice dim M= dim M_0+ dim A, because M is flat over A. To spell out the details, since the LHS is always leq to the RHS, it suffices to show the reverse inequality. It suffices to show generalizations lift along the canonical map from Supp M to Spec A (or, in more classical terms, going-down holds for it). But this is clear since if q is a prime ideal in the support of M mapped to p in Spec A then M_q is faithfully flat over A_p and thus Supp M_q maps surjectively to Spec A_p.

    In fact, the same identity holds if you replace “dim” by “depth.” This is also classical, see for example Stacks Project, Lemma 0338. Combining these two identities we get the following.

    Let (A, m) to B be a local homomorphism of Noetherian local rings. Let M be a finite B-module CM of dimension d. If M is flat over A and A is CM, then M/mM is a B-module CM of dimension d-dim A.

    In our situation, apply this to the modules M and B we get Lemma 2 with “Gorenstein” replaced by “CM.”


  2. On the main result. In fact, no additional assumptions are needed to be made. This is because Stacks Project, Lemma 0ATK (which is used to prove 0A6A) holds in the case the complex M is a single (finite) module, the complex L is of finite injective dimension, and K\in D^-_coh(R). This is obviously our case. To see why this is sufficient, compute both sides with a bounded complex J of injective modules that represents L, a (bounded above) complex F of finite free modules that represents K and leave M unchanged as a single R-module. Then we get total complexs Hom(M, J)\otimes F and Hom(M, J\otimes F). Since J is bounded and each term of F is finite free we see that these two complexes are canoncially identified and we win.


    1. Great! Thanks a lot for your comments. Aside from working in more generality, your approach to Lemma 2 is much more natural than the convoluted thing I was doing.

      Let me know if you’re ever in Bonn and I’ll buy you lunch.


      1. I do have the plan to visit various places and spots when the current pandemic cleans out. I will tell you if Bonn will be an item in the list.

        Btw, I am a graduate student at Princeton, and, if you come, let me know as well!


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