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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>=1?

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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? – David Speyer Oct 30 '09 at 2:18
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. – Kevin Buzzard Nov 9 '09 at 21:07
up vote 7 down vote accepted

A starting proviso: you didn't require that the map T ---> 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 nonnormal 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* the dual vector space. Let T* \subset V* 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*. Since T is commutative, this is an action of T on T*. Now, T \oplus T* 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* 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.

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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? – Kevin Buzzard Nov 6 '09 at 22:04
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. – David Speyer Nov 7 '09 at 20:32
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. – Qiaochu Yuan Nov 9 '09 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.


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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. – Kevin Buzzard Nov 28 '09 at 20:36

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?

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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. " – Kevin Buzzard Nov 9 '09 at 23:17
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. – Graham Leuschke Nov 10 '09 at 3:06
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. – David Speyer Nov 10 '09 at 13:03
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. – Graham Leuschke Nov 10 '09 at 19:46

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