Isomorphism between injective resolution

Question

When reading homological theory, I confront with a statement as follow.

In an exact sequence of $A$-modules: $0\to M\to U_0\to U_1\to \dots\to U_{n-1}\to C\to0$ with all $U_i$ injective, and let $I$ be an ideal of $A$, we have $\operatorname{Ext}^1_A(A/I, C) \cong \operatorname{Ext}^{n+1}_A(A/I, M)$.

I am not sure how can one get this isomorphism. I guess if we give $C$ an injective resolution, we can expand the original exact sequence, right? Hope someone can explain. Thanks!

Answer 1

There is several way to see this. I will write two of them. The first rely on the notion of hyper-derived functor while the second is more elementary.

First method : Consider the sequence of complexes $$0\to M[0]\to U_\bullet\to C[-n+1]\to 0$$ where $M[0]$ is $M$ placed in degree $0$ and $C[-n+1]$ is just $C$ placed in degree $n-1$. Concretely this is : $$\require{AMScd} \begin{CD} M[0]:@.0@>>> M@>>>0\\ @VVV@.@VVV\\ U_\bullet:@.0@>>>U_0@>>>U_1@>>>...@>>>U_{n-1}@>>>0\\ @VVV@.@.@.@.@VVV\\ C[-n+1]:@.@.@[email protected]@>>>C@>>>0 \end{CD} $$ Now this is not strictly speaking a short exact sequence of complexes, but a so-called distinguished triangle, which means that $\operatorname{Cone}(M[0]\to U_\bullet)\to C[-n+1]$ is a quasi-isomorphism.

This is enough to conclude that there will be a distinguished triangle $$ R\operatorname{Hom}(X,M)\to R\operatorname{Hom}(X,U_\bullet)\to R\operatorname{Hom}(X,C)[-n+1]\to$$ or if you don't know derived categories, a long exact sequences $$ ...\to \operatorname{Ext}^i(X,M)\to \operatorname{\mathbb{Ext}^i}(X,U_\bullet)\to\operatorname{\mathbb{Ext}}^i(X,C[-n+1])\to\operatorname{Ext}^{i+1}(X,M)\to...$$ Now, $\operatorname{\mathbb{Ext}}^i(X,C[-n+1])=\operatorname{Ext}^{i-n+1}(X,C)$ since this is just a complex concentrated in a single degree. Also $\operatorname{\mathbb{Ext}}^i(X,U_\bullet)=H^i(\operatorname{Hom}(X,U_\bullet))$ because $U_\bullet$ is a complex of injectives. It follows that if $i\geq n$, then $ \operatorname{\mathbb{Ext}}^i(X,U_\bullet)=0$.

And that's it. Just take $i=n$, the long exact sequence gives : $$0\to\operatorname{Ext}^1(X,C)\to\operatorname{Ext}^{n+1}(X,A)\to 0$$$$$$$$ so you have the desired isomorphism.

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Second method : More elementarily, you can use the standard trick in homological algebra which consists in splitting the sequence in short exact ones. Here this gives : $$0\to M\to U_0\to Z_1\to 0$$ $$0\to Z_1\to U_1\to Z_2\to 0$$ $$\vdots$$ $$0\to Z_{n-1}\to U_{n-1}\to C\to 0$$ Where $Z_i=\ker(U_i\to U_{i+1})=\operatorname{im}(U_{i-1}\to U_i)$. Now, the long exact sequences associated to each of these short exact sequences give isomorphisms $$\operatorname{Ext}^{n+1}_A(X,M)=\operatorname{Ext}^{n}_A(X,Z_1)=\operatorname{Ext}^{n-1}_A(X,Z_2)=...=\operatorname{Ext}^{2}_A(X,Z_{n-1})=\operatorname{Ext}^{1}_A(X,C)$$ (Use the fact that $U_i$ in injective so $\operatorname{Ext}^k_A(X,U_i)=0$ for any $k>0$).

In particular, with $X=A/I$, we get the desired result $\operatorname{Ext}^{n+1}_A(A/I,M)=\operatorname{Ext}^{1}_A(A/I,C)$.

Answer 2

You can do this by induction on $n$. For $n=0$, the long exact sequence yields $$\DeclareMathOperator{\H}{Hom}\DeclareMathOperator{\E}{Ext} \E_A^1(L,U_0) \to \E_A^1(L,C)\to \E_A^2(L,M)\to \E_A^2(L,U_0) $$ where the first and last terms are zero by injectivity of $U_0$.

For the induction step, split the resolution as $$ 0\to M\to U_0\to\dots\to U_{n-1}\to C'\to 0\\ 0\to C'\to U_n\to C\to 0 $$

Note that the statement holds for any $A$-module $L$, not just for $L=A/I$.