## Cohen-Macaulay Dominant Dimension

Ozgur Esentepe - University of Leeds
Workshop on Finitistic Dimensions - Bielefeld - June 15, 2023

## We live in a world of homological conjectures.

Let's see some of them in the setting of finite dimensional algebras. Throughout, let $A$ be a finite dimensional algebra.

### Finitistic dimension conjecture

The reason why we are here: it states that $\mathrm{findim}(A) := \{\mathrm{pd}(M) \colon M \in \mathrm{mod } A, \mathrm{pd}(M) < \infty \}$ is always finite.

### The vanishing conjecture

Consider $K^b(\mathrm{inj}(I))$ as a full subcategory of $D^b(A)$. For a complex $X \in D^b(A)$, if we have $\mathrm{Hom}_{D^b(A)}(I,X) =0$ for every $I \in K^b(\mathrm{inj}(I))$, then $X = 0$.

### Nunke's condition

If $X$ is a nonzero $A$-module, then there is an $i \ge 0$ such that $\mathrm{Ext}_A^i(DA,X) \neq 0$.

### Generalized Nakayama conjecture

If $S$ is a simple $A$-module, then there is an $i \ge 0$ such that $\mathrm{Ext}_A^i(DA, S) \neq 0$.

### Auslander-Reiten conjecture

If $M$ is a generator in $\mathrm{mod} A$ and $\mathrm{Ext}_A^i(M,M) =0$ for any $i > 0$, then $M$ is projective.

### Nakayama conjecture

Let $0 \to A \to I^0 \to I^1 \to \dots$ be the minimal injective resolution of $A$. If the dominant dimension $\mathrm{domdim} A := \inf\{n \colon I^n \text{ is not projective.}\}$ is not finite, then $A$ is self-injective.

We have implications \begin{align} \mathrm{FDC} \implies \mathrm{VC} \implies \mathrm{NuC} \implies \mathrm{GNC} \implies \mathrm{NC} \end{align}

• If we can prove the finitistic dimension conjecture, then we have a proof for all the other conjectures.
• If we can find a counterexample to the Nakayama conjecture, then we have a counterexample for all the other conjectures.

## Today's Goal

The main goal is to tell you the story where I thought I found a counterexample to the Nakayama conjecture.

In order to do this, I am going to introduce the framework I work in and define Cohen-Macaulay dominant dimension.

## I am a commutative algebraist.

### Depth

Given a local ring $(R, \mathfrak{m})$ and a finitely generated $R$-module $M$. The depth of $M$ is $\mathrm{depth}(M) = \inf\{i \colon \mathrm{Ext}_R^i(R/\mathfrak{m},M) \neq 0 \}.$

The dimension of $M$, $\mathrm{dim}(M)$, is the Krull dimension of the quotient ring $R/\mathrm{ann}_R(M)$.

We have inequalities $\mathrm{depth}(M) \leq \mathrm{dim}(M) \leq \mathrm{dim} R.$

### Cohen-Macaulayness

We have inequalities $\mathrm{depth}(M) \leq \mathrm{dim}(M) \leq \mathrm{dim} R.$

We say that $R$ is a Cohen-Macaulay ring if $\mathrm{depth} R = \dim R$ and $M$ is a maximal Cohen-Macaulay module if $\mathrm{depth} M = \dim R$. We denote the category of MCM modules by $\mathrm{CM}(R)$.

From now on, we will assume that $R$ is a CM local ring of dimension $d < \infty$ with maximal ideal $\mathfrak{m}$ unless otherwise stated.

### Canonical module

We say that a finitely generated $R$-module $\omega_R$ is a (the) canonical module if

• it is maximal Cohen-Macaulay,
• it has finite injective dimension,
• it satisfies $\mathrm{Ext}^d_R(R/\mathfrak{m}, \omega_R) \cong R/\mathfrak{m}$.

We have $\mathrm{Ext}_R^1(-,\omega_R) = 0$ on $\mathrm{CM}(R)$.

We have an exact duality $D:=\mathrm{Hom}_R(-,\omega_R): \mathrm{CM}(R) \leftrightarrow \mathrm{CM}(R)$.

We say that $R$ is a Gorenstein ring if $\omega_R \cong R$.

## Time to get noncommutative

### Depth

Let $R$ be a commutative Noetherian local ring, $\Lambda$ be a module-finite $R$-algebra with Jacobson radical $J$ and $M$ a finitely generated right $\Lambda$-module. Then,

\begin{align} \mathrm{depth} (M) & = \inf\{ i \colon \mathrm{Ext}_R^i(R/\mathfrak{m},M) \neq 0 \} \\ &= \inf\{ i \colon \mathrm{Ext}_\Lambda^i(\Lambda/J,M) \neq 0 \} \end{align}
The second line does not depend on $R$.
It turns out that the dimension does not depend on $R$ either.

### Cohen-Macaulayness

We say that a right $\Lambda$-module $M$ is a Cohen-Macaulay $\Lambda$-module if $M \in \mathrm{CM}(R)$.
We denote the category of CM $\Lambda$-modules by $\mathrm{CM}(\Lambda)$.
The category $\mathrm{CM}(\Lambda)$ does not depend on $R$.
We say that $\Lambda$ is an $R$-order if $\Lambda \in \mathrm{CM}(\Lambda)$.

### Canonical module

When $R$ is a CM local ring, the functor $D=\mathrm{Hom}_R(-,\omega_R)$ again gives an exact duality $\mathrm{CM}(\Lambda) \leftrightarrow \mathrm{CM}(\Lambda^{\mathrm{op}})$.
We denote by $\omega=\omega_\Lambda$ the bimodule $D\Lambda$ and call it the canonical module of $\Lambda$.
We have $\mathrm{Hom}_R(-,\omega_R) \cong \mathrm{Hom}_\Lambda(-,\omega)$ and $\omega$ is an additive generator for injective objects in $\mathrm{CM}(\Lambda)$.

## Cohen-Macaulay injectives

Let's let $\Lambda$ be an $R$-order.

### CM-injective resolutions

Let $M$ be Cohen-Macaulay $\Lambda$-module. By dualizing a (minimal) projective resolution of the $\Lambda^{\mathrm{op}}$-module $DM$, we can obtain a CM-injective resolution of $M$ - an injective resolution in $\mathrm{CM}(\Lambda)$.

This way, we can make sense of CM-injective dimension of modules in $\mathrm{CM}(\Lambda)$.

Example: If $Q$ is an acyclic quiver and $R$ is a Cohen-Macaulay local ring, then the CM-injective dimension of $\Lambda = RQ$ is 1.

### Auslander-Buchsbaum

If the CM-injective dimension of $\Lambda$ is $n$, then for any finitely generated $\Lambda$-module $M$ with finite projective dimension, we have $d \leq \mathrm{pd}_\Lambda (M) + \mathrm{depth} (M) \leq n + d.$

See Joshua Stangle's PhD thesis (2017).
Hence, if $\Lambda$ is an order with finite CM-injective dimension, the "CM"-finitistic dimension conjecture holds.

### Gorenstein orders

We say that $\Lambda$ is a Gorenstein order if its CM-injective dimension is 0. This is equivalent to saying that $\omega$ is a projective module (both as a $\Lambda$ and $\Lambda^{\mathrm{op}}$-module).

A Gorenstein order is non-singular if it has finite global dimension.

### A geometric situation

Suppose that $f: Y \to \mathrm{spec }R$ is a projective birational map where $Y$ and $R$ are both normal and Gorenstein of dimension $d$. Suppose that there is a (noncommutative) ring $\Lambda$ such that $D^b(Y) \cong D^b(\Lambda)$. Then, we have $f$ is crepant if and only if $\Lambda$ is an $R$-order. If $\Lambda$ has finite global dimension, then it is a non-singular order (a noncommutative crepant resolution (vdB)).

## Cohen-Macaulay dominant dimension

Everything (almost always) works the same.

### Definition

Let $0 \to \Lambda \to I^0 \to I^1 \to \dots$ be a minimal CM-injective resolution of $\Lambda$. Then, the CM-dominant dimension of $\Lambda$ is defined as $\mathrm{CMdomdim}(\Lambda) := \inf\{n \colon I^n \text{ is not projective.}.\}$

It measures how far away from being Gorenstein the order $\Lambda$ is.

### A noncommutative phenomenon

This invariant, when applied to commutative rings, is not very interesting.

If $R$ is a commutative Gorenstein local ring, then its CM-dominant dimension is infinite.
If $R$ is a commutative CM local ring which is not Gorenstein, then its CM-dominant dimension is 0.

### Iyama's Auslander-type conditions

In his seminal work on Auslander correspondence", Iyama considers $(m,n)$-conditions. He covers the case where the base ring is a regular local ring. In his language, the CM dominant dimension of an order is at least $\ell$ if and only if the order satisfies the $(d+1, d+\ell)$-condition.

### Regular sequences

If $\mathbf{x}$ is a regular sequence on $\Lambda$, then we have $\mathrm{CMdomdim}(\Lambda) = \mathrm{CMdomdim} (\Lambda/\mathbf{x}\Lambda).$
Therefore, if the length of $\mathbf{x}$ is $d$, we reduce to the case of ordinary dominant dimension.
This means that if you can find a counterexample to the "CM-Nakayama conjecture", then you get a counterexample to the original Nakayama conjecture.
Hence, you can look for counterexamples to the finitistic dimension conjecture by looking for counterexamples to the CM-Nakayama conjecture.

## A failed attempt

I thought I had a counterexample.

### The complete (2,1)-scroll

Let $k$ be an infinite field and let $R = k[[x,y,z,u,v]]/I$ where $I$ is the ideal generated by the $2 \times 2$ minors of the matrix $\begin{bmatrix} x & y & u \\ y & z & v \end{bmatrix}.$
Then, $R$ is a CM local ring of dimension 3.
The endomorphism ring $\Lambda = \mathrm{End}_R(R \oplus \omega_R)$ is an order of global dimension 4 and thus CM-injective dimension 1.
Therefore, $\mathrm{CMdomdim} \Lambda$ is 0, 1 or infinite. Note that $\Lambda$ is not a Gorenstein order.

### I should have known.

Since $\Lambda$ has finite CM-injective dimension, by the Auslander-Buchsbaum type of inequality, the CM-finitistic dimension conjecture holds and therefore the CM-Nakayama conjecture should hold. Therefore, it is clear that $\Lambda$ does not have infinite CM-dominant dimension.

But I did not know. So, I considered the following arguments.

There is one (and only one) mistake in the slides with red background.

Let $R$ be any CM local ring and $M$ be an MCM $R$-module which has $R$ and $\omega_R$ as a direct summand. Consider $DM$ and a projective resolution $\mathbb{P}$ of $DM$ and dualize it to get a CM-injective resolution $D\mathbb{P}$ of $M$. Then, apply the functor $\mathrm{Hom}_R(M,-)$ to it. We have a complex $0 \to \Gamma = \mathrm{End}_R(M) \to \mathrm{Hom}_R(M,\omega_R)^{n_0} \to \mathrm{Hom}_R(M,\omega_R)^{n_1} \to \ldots$ which may or may not have cohomology. In fact, the cohomology of this is isomorphic to $\mathrm{Ext}^*_R(M,M)$.

We have that $\Gamma$ is an $R$-order if and only if $\mathrm{Ext}_R^{i}(M,M)=0$ for $i = 1, \ldots, d-2$.

Now, let us focus on $\mathrm{Hom}_R(M,\omega_R)$.

Since we assume that $M$ contains $R$ as a direct summand, we have that $M_\Gamma$ is a direct summand of $\Gamma_\Gamma$. Indeed, if we have $M = R\oplus M'$, then we have $\Gamma_\Gamma = \mathrm{Hom}_R(_R M, _R M_\Gamma) = \mathrm{Hom}_R(_R R \oplus _R M', _R M_\Gamma) = M_\Gamma \oplus \mathrm{Hom}_R(M',M)$. So, $\mathrm{Hom}_R(M,\omega_R)$ is a CM-injective $\Gamma$-module.

Similarly, since $R$ is assumed to contain $\omega_R$ as a direct summand, we can show that $\mathrm{Hom}_R(M,\omega_R)$ is a projective $\Gamma$-module.
So, we have that $\mathrm{Hom}_R(M,\omega_R)$ is a CM-projective-injective module.

So, we have $\color{black}{0 \to \Gamma \to \mathrm{Hom}_R(M,\omega_R)^{n_0} \to \mathrm{Hom}_R(M,\omega_R)^{n_1}} \to \mathrm{Hom}_R(M,\omega_R)^{n_2} \to \ldots$ where the terms are CM-projective-injective $\Gamma$-modules. The black portion is always exact by the half-exactness of $\mathrm{Hom}_R(M,-)$ and we may get more exactness by Ext vanishing.

Now, say, we have exactness until $\mathrm{Hom}_R(M,\omega_R)^{n_m}$. Then, after that point, we don't have exactness. BUT.. we can just complete that part to a CM-injective resolution!
Therefore, if $M$ is an MCM-module that contains $R$ and $\omega_R$ as a direct summand, then the CM-dominant dimension of $\Gamma=\mathrm{End}_R(M)$ is at least 2 and the vanishing of self extensions of $M$ tells us how high the CM-dominant dimension can get.

### Were you able to find the mistake?

I sent my "finding" to people.

Graham Leushcke said "I too am confused. It is not a fun feeling."
He also said "It seems like this is a question the normal representation theory people should know the answer to."
And finally, on December 1st, 2021 at 14:51, after I told him the response I received from Osamu Iyama, he said:
Aha! Well, thank heaven for Osamu.
The mistake is that unless the cokernel of $\mathrm{Hom}_R(M, \omega_R)^{n_t} \to \mathrm{Hom}_R(M, \omega_R)^{n_{t+1}}$ is MCM, we can not complete our complex into a CM-injective resolution.

# Thank you

Question: Do you need to assume that "minimal" CM-injective resolutions exist?

Answer: Yes, I forgot to add the semi-perfect assumption on $\Lambda$ when I was making the slides. You can guarantee this by assuming that $R$ is complete, for example.