In this question, all rings are commutative with a $1$, unless we explicitly say so, and all morphisms of rings send $1$ to $1$.

Let $A$ be a Noetherian local integral domain. Let $T$ be a non-zero $A$-algebra which, as an A-module, is finitely-generated and torsion-free.

Can one realise $T$ as a subring of the (not necessarily commutative) ring $End_A(A^n)$ for some $n \ge 1$?

  • $\begingroup$ The obvious guess is "always". I realize that the obvious proof (consider T acting on itself) doesn't work. However, do you have any reason to believe this is not the correct answer? $\endgroup$ Oct 30, 2009 at 2:18
  • $\begingroup$ I should add two things: 1) as I forgot to say, but as David guessed, yes I want the map from T to the endomorphism ring to be an A-algebra homomorphism. 2) David's first counterexample made me realise that really I am only interested in the case where A is regular, and in fact probably only in the case where A is a power series ring in finitely many variables over a field of characteristic zero. $\endgroup$ Nov 9, 2009 at 21:07

3 Answers 3


A starting proviso: you didn't require that the map $T \rightarrow End_A(A^n)$ send elements of A to their obvious diagonal representatives. I am going to assume you intended this.

A few partial results:

1) If $A=k[x,y]/(x^3-y^2)$, and $T$ is the integral closure of A, then this can not be done. Let $t$ be the element $y/x$ of $T$ and $M$ the matrix that is supposed to represent it. Then we must have $xM=y Id_n$, which has no solutions. More generally, whenever A is a non-normal ring and $T$ its integral closure, there are no solutions.

2) If $A$ is a Dedekind domain the answer is yes. Let $V$ be the vector space $T \otimes Frac(A)$, and $V^{\ast}$ the dual vector space. Let $T^{\ast} \subset V^{\ast}$ be the vectors whose pairing with $T$ lands in $A$. Using the obvious action of $T$ on itself, we get an action of $T^{op}$ on $T^{\ast}$. Since $T$ is commutative, this is an action of $T$ on $T^{\ast}$. Now, $T \oplus T^{\ast}$ is free as an A-module, so this gives us the desired representation.

2') A conjectural variant of the above: I have a vague recollection that, if $A$ is a polynomial ring, $T^{\ast}$ is always free. Can anyone confirm or refute this?

3) A case which I think is impossible, but can't quite prove at this hour: Let $T = k[x,y]$ and let $A$ be the subring $k[x^2, xy, y^2]$. I am convinced that we cannot realize $T$ inside the ring of matrices with entries in $A$, but the proof fell apart when I tried to write it down.

  • $\begingroup$ I asked this question for a reason (to do with parameter spaces of p-adic automorphic forms) and now I realise that I'm suddenly very interested in the case where A is regular. So now I have a new question: if A is a power series ring in finitely many variables over a field, is it true? In particular I am now quite interested in the comments about 2' above. Should I ask another question or somehow bump this one? $\endgroup$ Nov 6, 2009 at 22:04
  • $\begingroup$ Just wanted to let you know I saw this, and am thinking about it. I don't understand how the front page is sorted, so maybe the best way to attract other people would be to ask another question. $\endgroup$ Nov 7, 2009 at 20:32
  • $\begingroup$ As far as I can tell, the front page is sorted by "activity," which is defined as new answers or edits to old answers or the question itself. $\endgroup$ Nov 9, 2009 at 23:16

Hello, I just want to add a few minor comments here, since this is a topic very close to my heart:

1) Finite MCM modules are not known to exist in dimension 3. What Hochster proved for equicharacteristic case and also in general dimension 3 (based on Ray Heitmann result) is that non-finitely generated MCM modules exist.

2) If R is a N-graded domain over a perfect field of char p > 0 and R is locally Cohen-Macaulay on the punctured spectrum, then R admits a finite MCM. A proof can be found in:


3) A possible candidate for a counter-example is the local ring at the origin of the cone over some abelian surfaces.

4) Many module-theoretic consequence of existence of finite MCM can be deduced from existence of non-finitely generated MCM and other approaches to the homological conjectures, so it may be helpful to what you want to do.


  • $\begingroup$ Thanks for your comments. I'm the OP (I posted under a temporary ID). I'm now mainly interested in the case where A is a power series ring in several variables over a characteristic zero field. $\endgroup$ Nov 28, 2009 at 20:36
  • $\begingroup$ About 3), I thought any non-trivial line bundle in $Pic^0$ gave a rank one MCM module over the cone over an abelian surface thanks to Kodaira vanishing. $\endgroup$
    – Mohan
    May 13, 2018 at 13:18

Kevin Buzzard informs me, by e-mail and in the above comment, that what he cares about is the case where $A$ is regular. In particular, it would be good to know whether every $T$ is a matrix ring in this case. I thought about it but didn't solve this, so here is a record of my ideas. Let $n$ be the dimension of $A$.

(1) A subring of a matrix is obviously a matrix ring. So, if we knew that $T$ was a matrix ring whenever $T$ was normal, this would establish that $T$ was always a matrix ring.

(2) As explained in my previous answer, if $T$ is free as an $A$-module, then $T$ is a matrix ring. We have the following implications: if $T$ is Cohen-Macaulay then $T$ torsion-free and finite over $A$ implies $T$ flat over $A$; if $A$ is a polynomial ring then $T$ flat over $A$ implies $T$ free over $A$. So, if $A$ is a polynomial ring and $T$ is Cohen-Macaulay, then $T$ is a matrix ring.

In particular, if $n=1$ or $2$, then $T$ normal implies $T$ Cohen-Macaulay. So, in these cases, and with $A$ a polynomial ring, $T$ is a matrix ring.

(3) As explained above, if $T^\*$ is free over $A$, we also get to conclude that $T$ is a matrix ring. Unfortunately, this can fail when $n \geq 3$.

(4) If there is some $T$-module $M$ on which $T$ acts without kernel, and $T$ is free as an $A$-module, then $T$ is a matrix algebra. Restated geometrically, if there is any coherent sheaf on $\mathrm{Spec}(T)$, with support on the whole of $\mathrm{Spec}(T)$, whose pushforward to $\mathrm{Spec}(A)$ is a trivial vector bundle, then $T$ is a matrix algebra. If we restrict our attention to $A$ a local ring, or a polynomial ring, then the adjective "trivial" comes for free.

So, if there were to be a counter-example, we would want $n \geq 3$ and we would want $T$ to be normal but not Cohen-Macaulay. Moreover, we would want that basically any $T$-module is not free as an $A$-module.

Any ideas?

  • $\begingroup$ I have now independently heard that the case of $A$ regular is "open and difficult". If T is finite as an A-module then (and I don't understand what I'm writing here) "this is precisely the issue of whether T has a finitely generated maximal Cohen-Macaulay module. This is true in dim < 3 and not known in dimension 3 in any characteristic, although it is known to be true in many instances. " $\endgroup$ Nov 9, 2009 at 23:17
  • 2
    $\begingroup$ I haven't yet thought at all about whether this is equivalent to T having a MCM module, but just wanted to point out that (a) any local ring containing a field admits a MCM module by work of Hochster in the 70s; (b) in dimensions 1 and 2 existence of MCM modules is trivial; and (c) MCM modules exist in dimension 3 (mixed characteristic) by work of Heitmann from ~2002. $\endgroup$ Nov 10, 2009 at 3:06
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    $\begingroup$ Let A be local. As I understand it, an MCM module would mean a T-module M for which there is a regular sequence of length n? That's definitely true if T is a subring of a matrix ring: If T is contained in the m x m matrices then A^m is a T-module and, since it is free as an A-module, the generators of the maximal idea of A give a regular sequence. $\endgroup$ Nov 10, 2009 at 13:03
  • 4
    $\begingroup$ Right. Conversely, if M is an MCM T-module, then it is an MCM A-module since T is finite over A; as A is regular, every MCM A-module is free over A by Auslander-Buchsbaum. Since every T-endomorphism of M is also an A-endomorphism, we have T \subseteq \End_T(M) \subseteq \End_A(M), so T is a matrix ring. $\endgroup$ Nov 10, 2009 at 19:46

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