Showing that the stable module category of a ring $R$ restricted to maximal Cohen-Macaulay objects is trivial if $\text{gldim } R < \infty$

(In the following, a (not necessarily commutative) ring $R$ is Gorenstein if it has finite injective dimension as a module over itself on either side, and a finitely generated (right) $R$-module is maximal Cohen-Macaulay (MCM) if $\text{Ext}_R^i(M,R)=0$ for all $i \geqslant1$.)

In the paper Maximal Cohen-Macaulay Modules and Tate-Cohomology Over Gorenstein Rings, Buchweitz demonstrates that for a noetherian, Gorenstein ring the stabilized derived category of $R$ is equivalent to the stable module category of $R$, restricted to MCM objects, written $\underline{\text{MCM}}(R)$.

It is clear that the stabilized derived category of a noetherian, Gorenstein ring $R$ is trivial if $\text{gldim } R < \infty$, but (without making use of the above equivalence) I can't see why $\underline{\text{MCM}}(R)$ should be trivial in this case. It seems that an equivalent problem would be to show that for MCM modules $M$ and $N$, we have $\text{Hom}_R(M,N) = NM^*$, where $M^* = \text{Hom}_R(M,R)$, but I haven't made much progress with this either.

• So that I understand: are you assuming that every $R$-module has finite injective dimension? If so, at least in the commutative, Noetherian case, that is equivalent to assuming that $R$ is a regular ring, cf. Theorem 19.2 of Matsumura and Lemma 2 immediately preceding). – Jason Starr Aug 12 '15 at 14:55
• I'm assuming that the global dimension of $R$ is finite, so that every $R$-module has finite injective dimension. However, I'm not assuming commutativity. – lokodiz Aug 12 '15 at 14:57

If $M$ is a maximal Cohen-Macaulay module for a Noetherian Gorenstein ring $R$, then it follows easily by induction on the projective dimension of $N$ that $\operatorname{Ext}^i_R(M,N)=0$ for all $i>0$ if $N$ is finitely generated of finite projective dimension.

So if $R$ also has finite global dimension, then $\operatorname{Ext}^i(M,N)=0$ for $i>0$ for all finitely generated modules $N$, and so $M$ is projective.

So when $R$ has finite global dimension, all $MCM$ modules are projective, so the stable module category of $MCM$ modules is trivial.

• Thanks, I'll work through the details of this tomorrow and then accept this. – lokodiz Aug 12 '15 at 19:22

The OP confirms that in the commutative case he is considering only regular rings, so then the problem is trivial. I thought the OP was only assuming that $R$ has finite injective dimension (the usual definition of Gorenstein). The OP is interested in the non-commutative case, not the commutative case. At any rate, I am including below the argument in the commutative case when $R$ is not assumed regular, but $N$ is assumed to have finite injective dimension.

The following addresses your second question in the case that $R$ is commutative, about $\text{Hom}_R(M,N)$. First of all, the result is false if there is not some additional hypothesis. Let $R$ be the zero-dimensional, local, Noetherian, Gorenstein ring $k[x]/\langle x^2 \rangle$. Let $M$ and $N$ both be $R/xR = k$. Then $\text{Id}_M\in \text{Hom}_R(M,M)$ is not in the image of the morphism $\text{Hom}_R(M,R)\otimes_R M \to \text{Hom}_R(M,M)$. The only reasonable additional hypothesis I can see is that $N$ has finite injective dimension. However, in this case it seems to me that $N$ is a finitely generated projective $R$-module, so that the problem becomes trivial. I am including the following argument for any case.

Original argument with additional hypothesis -- probably overkill. For every commutative Noetherian ring $R$, for all maximal Cohen-Macaulay $R$-modules $M$ and $N$, assuming that $N$ has finite injective dimension, then the natural $R$-module homomorphism, $$\beta_{R,M,N}:\text{Hom}_R(M,R)\otimes_R N \to \text{Hom}_R(M,N),$$ is an isomorphism.

First, since $\beta_{R,M,N}$ is a natural transformation of additive functors, this is obviously an isomorphism when $N$ is a finite free $R$-module. (Edit. This is probably enough to conclude, since a maximal Cohen-Macaulay module of finite injective dimension over a Gorenstein ring appears to be projective; this probably follows also from Jeremy Rickard's argument.) Second, since this is a local problem, it suffices to prove the case when $R$ is a local, Noetherian, Gorenstein ring of some dimension $d$ (local Noetherian rings have finite Krull dimension). The result is proved by induction on $d$. I will use some results from Chapter 21 of the following. (Aside: This is the second time I have mentioned Chapter 21 on MathOverflow in two days.)

MR1322960 (97a:13001)
Eisenbud, David(1-BRND)
Commutative algebra. With a view toward algebraic geometry.
Graduate Texts in Mathematics, 150. Springer-Verlag, New York, 1995. xvi+785 pp.
ISBN: 0-387-94268-8; 0-387-94269-6
13-01 (14A05)

In the base case that $d=0$, by Proposition 21.11, $N$ is a finite free $R$-modules. Thus, by the first observation above, $\beta_{R,M,N}$ is an isomorphism.

Next, by way of induction, assume that $d>0$, and assume that the result has been proved for smaller values of $d$. Since $R$ is Gorenstein, it is Cohen-Macaulay, so that $\text{depth}_R(M) = d > 0$. So there exists a system of parameters $(x_1,\dots,x_d)$ for $R$. By Proposition 21.9, this system of parameters is simultaneously an $M$-sequence and an $N$-sequence. Denote $\overline{R} = R/x_d R$, $\overline{M} = M/x_d M$, and $\overline{N} = N/x_d N$. In particular, consider the $\overline{R}$-module homomorphism induced from $\beta_{R,M,N}$, $$\overline{\beta_{R,M,N}} : \text{Hom}_R(M,R)\otimes_R N \otimes_R \overline{R} \to \text{Hom}_R(M,N)\otimes_R \overline{R}.$$ By Proposition 21.13, to prove that $\beta_{R,M,N}$ is an isomorphism, it is equivalent to prove that $\overline{\beta_{R,M,N}}$ is an isomorphism.

Since $R$ is a local Gorenstein ring of dimension $d$ and $x_d\in \mathfrak{m}_R$ is a nonzerodivisor, the quotient ring $\overline{R}$ is a local, Noetherian, Gorenstein ring (there are other references, but this does follow from Exercise 21.20). Also, since $(x_1,\dots,x_d)$ is both an $M$-sequence and an $N$-sequence, by definition, also the induced system of parameters $(\overline{x}_1,\dots,\overline{x}_{d-1})$ is both an $\overline{M}$-sequence and an $\overline{N}$-sequence. Therefore, by Proposition 21.9 again, both $\overline{M}$ and $\overline{N}$ are maximal Cohen-Macaulay $\overline{R}$-modules. By Proposition 21.10, since $N$ has finite injective dimension as an $R$-module, also $\overline{N}$ has finite injective dimension as an $\overline{R}$-module. Thus, by the induction hypothesis, the $\overline{R}$-module homomorphism,$$\beta_{\overline{R},\overline{M},\overline{N}}: \text{Hom}_{\overline{R}}(\overline{M},\overline{R})\otimes_{\overline{R}} \overline{N} \to \text{Hom}_{\overline{R}}(\overline{M}, \overline{N}),$$ is an isomorphism. Finally, since $R$, $M$ and $N$ are maximal Cohen-Macaulay $R$-modules, by Proposition 21.12(b) (applied twice), the $\overline{R}$-module homomorphism $\beta_{\overline{R},\overline{M},\overline{N}}$ is equivalent to the $\overline{R}$-module homomorphism $\overline{\beta_{R,M,N}}$. Therefore $\beta_{R,M,N}$ is an isomorphism. By induction, the result holds for all $d$.

• Isn't $\text{gldim } \Bbbk[x]/\langle x^2 \rangle = \infty$? – lokodiz Aug 12 '15 at 14:37
• @Simon: So that I understand, are you hypothesizing that $R$ has finite injective dimension as an $R$-module (the usual definition of Gorenstein), or are you hypothesizing that every $R$-module has finite injective dimension? Certainly $R=k[x]/\langle x^2\rangle$ has finite injective dimension as an $R$-module: it is an injective $R$-module. – Jason Starr Aug 12 '15 at 14:47
• In the non-commutative case I'm not sure whether every map from an $MCM$ module to a module of finite injective dimension factors through a projective module. I don't think the argument in my answer generalizes straightforwardly to prove this, but I'll think about it. – Jeremy Rickard Aug 12 '15 at 18:07
• @JasonStarr: Sorry, I got confused by the fact that $\text{pdim}_{\Bbbk[x]} \Bbbk [x]/\langle x^2 \rangle = \infty$. – lokodiz Aug 12 '15 at 19:20