# 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 map which is surjective on Exts is called an homological epimorphism (these were studied by Auslander, Platzek and Todorov a while ago). I've never seen the injective variant. Nov 16, 2011 at 21:40

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.

• Do you need that $M$ is an $S$-module, which becomes an $R$-module after restricting along $R\to S$? If so, isn't $\Omega\otimes_S M \to S\otimes_R M\to M$ in your context split as well, as a sequence of left $S$-modules? So that $\text{Ext}^2_S(-,N)$, being an additive functor, maps it to a split short exact sequence $\text{Ext}^2_S(\Omega\otimes_S M,N)\leftarrow\text{Ext}^2_S(S\otimes_R M,N)\leftarrow\text{Ext}^2_S(M,N)$, which amounts to the split short exact sequence $\text{Ext}^2_S(\Omega\otimes_S M,N)\leftarrow\text{Ext}^2_R(M,N)\leftarrow\text{Ext}^2_S(M,N)$? Nov 22, 2011 at 7:20
• Matthias, indeed: the first sentence list the requirements for the spectral sequence, for the rest is in the context of the question; I'll edit so asto make it clear. As for your second point, of course!: that gives injectivity (in all degrees). Nov 22, 2011 at 22:47
• Question: If ($S$ is $R$-projective and) the forgetful functor induces injections on Exts, is $S$ not then separable? Nov 22, 2011 at 23:31
• @Mariano: Concerning your question: This is true at least if $S$ is a supplemented $R$-algebra. For, by $R$-projectivity, there is an isomorphism $Ext^n_{S\otimes_R S}(S,N) \cong Ext^n_S(R;N)$ (Cartan-Eilenberg, X, Prop. 1.1) and the latter embedds into $Ext^n_R(R;N)=0$ if $n>0$. In particular $Ext^1_{S\otimes_R S}(S,N) =0$ and hence $S$ is separable (see Rowen, Ring Theory II, 5.3.7). Thus a supplemented projective $R$-algebra $S$ is separable iff $\operatorname{res}_R^S: Ext_S \to Ext_R$ is injective. Nov 23, 2011 at 21:32
• Ralph, That was the motivation for the question in the general case, actually :) You should probably copy your comment to an answer to the OP. Nov 24, 2011 at 2:20