Which rings are subrings of matrix rings? 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 $n \ge 1$?
 A: 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:
http://www.math.utah.edu/vigre/minicourses/algebra/hochster.pdf 
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.
Cheers,
A: 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?
A: 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.
