A starting proviso: you didn't require that the map T ---> End_A(A^n)$T \rightarrow End_A(A^n)$ send elements of A to their obvious diagonal representatives. I am going to assume you intended this.
A few partial results:
If A=k[x,y]/x^3-y^2$A=k[x,y]/(x^3-y^2)$, and T$T$ is the integral closure of A, then this can not be done. Let t$t$ be the element y/x$y/x$ of T$T$ and M$M$ the matrix that is supposed to represent it. Then we must have xM=y Id_n$xM=y Id_n$, which has no solutions. More generally, whenever A is a nonnormalnon-normal ring and T$T$ its integral closure, there are no solutions.
If A$A$ is a Dedekind domain the answer is yes. Let V$V$ be the vector space T \otimes Frac(A)$T \otimes Frac(A)$, and V*$V^{\ast}$ the dual vector space. Let T* \subset V*$T^{\ast} \subset V^{\ast}$ be the vectors whose pairing with T$T$ lands in A$A$. Using the obvious action of T$T$ on itself, we get an action of T^{op}$T^{op}$ on T*$T^{\ast}$. Since T$T$ is commutative, this is an action of T$T$ on T*$T^{\ast}$. Now, T \oplus T*$T \oplus T^{\ast}$ 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$A$ is a polynomial ring, T*$T^{\ast}$ is always free. Can anyone confirm or refute this?
- A case which I think is impossible, but can't quite prove at this hour: Let T = k[x,y]$T = k[x,y]$ and let A$A$ be the subring k[x^2, xy, y^2]$k[x^2, xy, y^2]$. I am convinced that we cannot realize T$T$ inside the ring of matrices with entries in A$A$, but the proof fell apart when I tried to write it down.
