When does the forgetful functor S-Mod -> R-Mod induce injective maps on Ext-groups? Assume we have a complete regular local ring $R$ and an $R$-algebra $S$.
Is there a class of such algebras $S$ with the following property:
Given two $S$-modules $M,N$, then the maps induced by the forgetful functor $S-Mod \rightarrow R-Mod$ give injections $Ext^i_S(M,N)\rightarrow Ext_R^i(M,N)$?
If this question is too broad, here are the special cases i'm especially interested in:
We start with $R=\mathbb{C}[[x,y]]$, on the algebra side one may choose $S=M_2(R)$ or the subalgebra of the matrices given by:
\begin{pmatrix}R &R \\ xR &R \end{pmatrix}
and one may choose $N=S$ and $M=S/T$ to be an finite length quotient of $S$, i.e there is a sequence $0\rightarrow T\rightarrow S \rightarrow M\rightarrow 0$. Further more the case $i=2$ would be enough. 
So what i'm really interested in is:
Is the map $ Ext_S^2(S/T,S)\rightarrow Ext_R^2(S/T,S)$ injective in these cases?
Inducing injective maps on the $Ext$-groups is, i think, more than being a faithful functor. Maybe this property has already been studied and has its own name? I don't see any sequences which relate these two groups, so that one could see this by doing some kind of diagram chasing. 
Any hints or ideas are welcome!
 A: Given a map or algebras $R\to S$, a left $R$-module $M$ and a left $S$-module $N$, there is a natural first quadrant, cohomologically graded spectral sequence with $$E_2^{p,q}=\mathrm{Ext}^p_S(\mathrm{Tor}_R^q(S,M),N)$$ converging to $\mathrm{Ext}^\bullet_R(M,N)$.
If $S$ is flat as a left $R$-module, this collapses to natural isomorphisms 
$$
\mathrm{Ext}_S^\bullet(S\otimes_RM,N)\cong\mathrm{Ext}^\bullet_R(M,N). \qquad\qquad(\star)
$$
Suppose now moreover that $S$ is a separable $R$-algebra, and go to the situation of the question in which both $M$ and $N$ are a left $S$-modules. If $S\otimes_RS\to S$ is the map induced by multiplication in $S$, which is a map of $S$-bimodules, and $\Omega$ is its kernel, we have a short exact sequence of $S$-bimodules $$0\to\Omega\to S\otimes_RS\to S\to 0$$ which splits—this is separability. Tensoring this with $M$ over $S$ gives then a split short exact sequence $$0\to\Omega\otimes_SM\to S\otimes_RS\otimes_SM\to S\otimes_SM\to 0$$ which can be identified with $$0\to\Omega\otimes_SM\to S\otimes_RM\to M\to 0.$$ The long exact sequence obtained from it by applying $\mathrm{Ext}_S^\bullet(\mathord-,N)$ splits into little exact sequences $$0\to\mathrm{Ext}_S^p(M,N)\to\mathrm{Ext}_S^p(S\otimes_RM,N)\to \mathrm{Ext}^p_S(\Omega_S\otimes_SM,N)\to 0.$$
In particular, the map $\mathrm{Ext}_S^p(M,N)\to\mathrm{Ext}_S^p(S\otimes_RM,N)$ is injective, so when we compose it with $(\star)$ we get an injection
$$\mathrm{Ext}_S^p(M,N)\to\mathrm{Ext}_R^p(M,N).$$ It is not hard to see that this composition is the map you wanted.
This deals with your matrix algebra.
