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 $n \ge 1$?



Which rings are subrings of matrix rings?
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.


