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!

  • $\begingroup$ 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. $\endgroup$ Nov 16, 2011 at 21:40

1 Answer 1


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.

  • $\begingroup$ 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 $\text{Ext}^2_S(\Omega\otimes_S M,N)\leftarrow\text{Ext}^2_R(M,N)\leftarrow\text{Ext}^2_S(M,N)$? $\endgroup$ Nov 22, 2011 at 7:20
  • $\begingroup$ 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). $\endgroup$ Nov 22, 2011 at 22:47
  • $\begingroup$ Question: If ($S$ is $R$-projective and) the forgetful functor induces injections on Exts, is $S$ not then separable? $\endgroup$ Nov 22, 2011 at 23:31
  • 1
    $\begingroup$ @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. $\endgroup$
    – Ralph
    Nov 23, 2011 at 21:32
  • $\begingroup$ 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. $\endgroup$ Nov 24, 2011 at 2:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.