Flat base exchange for Ext groups Let $R$ be a unital commutative ring with ideal $I$, let $M$ be a flat $R$-module. Then there is a base exchange isomorphism
\begin{equation} Ext^s_R(M,R/I)\cong Ext^s_{R/I}(M \otimes_R R/I, R/I)  \end{equation}
I looked up for a proof and found this note
http://www1.mate.polimi.it/~enrsch/EXT.pdf
in which they use an argument involving a spectral sequence (lemma 2.5).
The proof is fine but I think there should be an easier argument not involving such machinery. My idea is the following: consider the short exact sequence
\begin{equation} 0 \rightarrow I \rightarrow R \rightarrow R/I \rightarrow 0  \end{equation}
by tensoring with M we get another s.e.s.
\begin{equation} 0 \rightarrow I\otimes M \rightarrow M \rightarrow R/I\otimes M \rightarrow 0. \end{equation}
Now I think you should prove that $Ext^s_R(I \otimes M, R/I)=0$ and so deduce from the induced long exact sequence $Ext^s_R(M, R/I)\cong Ext^s_R(M\otimes R/I, R/I)$, then my hope is $Ext^s_R(M\otimes R/I, R/I)\cong Ext^s_{R/I}(M\otimes R/I, R/I)$. But I do not know how to prove the two claimed isomorphisms.
For $Ext^s_R(I \otimes M, R/I)=0$ I thought that the projective resolution of $I \otimes M$ should still be in the form $I \otimes N$ for some module $N$, thus taking $Hom(-,R/I)$ we get $0$. But I do not know how to actually produce such a resolution. The fact that tensoring with $M$ does not preserve projectives does not help.
 A: Here is an elementary argument by dimension shifting.
Consider a short exact sequence
$$(*): \;\;\;\;\;0 \rightarrow S \rightarrow P \rightarrow M \rightarrow 0$$
where $P$ is projective. Applying $-\otimes R/I$ yields a short exact sequence
$$(*)/I: \;\;\;\;\;0 \rightarrow S/IS \rightarrow P/IP \rightarrow M/IM \rightarrow 0$$
since $\mathrm{Tor}_1^R(M, R/I)=0$. Moreover, note that $S$ is a flat $R$-module since it is a kernel of a map between flat modules, and the same goes for $S/IS$ (as a flat $R/I$-module).
We may apply $\mathrm{Hom}_R(-,R/I)$ and $\mathrm{Hom}_{R/I}(-,R/I)$ to $(*)$ and $(*)/I$, resp. Then we obtain exact sequences
$$0 \rightarrow \mathrm{Hom}_R(M,R/I) \rightarrow \mathrm{Hom}_R(P,R/I) \rightarrow \mathrm{Hom}_R(S,R/I) \rightarrow \mathrm{Ext}^1_R(M, R/I) \rightarrow 0,$$
$$0 \rightarrow \mathrm{Hom}_{R/I}(M/IM,R/I) \rightarrow \mathrm{Hom}_{R/I}(P/IP, R/I) \rightarrow \mathrm{Hom}_{R/I}(S/IS,R/I) \rightarrow \mathrm{Ext}^1_{R/I}(M/IM, R/I) \rightarrow 0,$$
and the first three terms of the respective sequences are naturally identified with one another by elementary considerations (and this is the case $s=0$ of the claim to be proved). This induces an isomorphism on the remaining term, giving $$\mathrm{Ext}^1_{R/I}(M/IM, R/I) \simeq \mathrm{Ext}^1_{R}(M, R/I).$$ So for $s=1$, the claim is proved.
For higher $s$, we proceed by induction. Suppose that the claim was proved for all flat $R$-modules and all $1\leq s'<s$. It remains to observe that from the short exact sequences $(*)$, $(*)/I$, we have
$$\mathrm{Ext}^s_R(M, R/I)\simeq \mathrm{Ext}^{s-1}_R(S, R/I),$$
$$\mathrm{Ext}^s_{R/I}(M/IM, R/I)\simeq \mathrm{Ext}^{s-1}_{R/I}(S/IS, R/I),$$
and since $\mathrm{Ext}^{s-1}_R(S, R/I)\simeq \mathrm{Ext}^{s-1}_{R/I}(S/IS, R/I)$ holds by induction hypothesis, we are done.
Remark: The functor $\mathrm{Hom}(-, R/I)$ can be in fact replaced by $\mathrm{Hom}(-, N)$ for any $R/I$-module $N$, i.e. the argument really shows an isomorphism $\mathrm{Ext}^s_{R/I}(M/IM, N) \simeq \mathrm{Ext}^s_{R}(M, N)$ whenever $N$ is an $R/I$-module.
