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I know that for a commutative ring $A$ and $A$-modules $M$ and $N$, the set $E_A(M, N)$ of extensions of $M$ by $N$ can be equipped with the Baer sum which gives it an additive group structure. Apparently $E_A(M, N)$ is also isomorphic to $\mathrm{Ext}^1 (M, N)$ as an $A$-module. But how does one define the $A$-module structure on $E_A(M, N)$?

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up vote 4 down vote accepted

Let $0 \to N \to X \to M \to 0$ be a representative of an element $\zeta$ of $E_A(M,N)$. Let $f$ be any $A$-module endomorphism of $N$. We have the diagram $$ \begin{array}{lllll} N & \stackrel\iota\hookrightarrow & X & \twoheadrightarrow & M \\ \downarrow f& &\downarrow & & \downarrow \\ N & \hookrightarrow & P & \twoheadrightarrow & M \end{array} $$ where $P$ is the pushout of $f$ and $\iota$. This is just part of functoriality of $E_A$: the bottow row is a representative of $E_A(f, N) (\zeta)$.

Now let $a \in A$, and let $f_a:N\to N$ be the module homomorphism $n\mapsto an$ (here we really use commutativity). Setting $f=f_a$ in the above diagram, the class of the bottom row represents the image of the class of the top under the action of $a$.

There's a lot to check here: that the bottom row is really exact, that this induced a well-defined map on the set of equivalence classes $E_A(M,N)$,... Note that this construction actually turns $E_A(M,N)$ into an $\operatorname{End}(M)$-$\operatorname{End}(N)$-bimodule, even for $A$ non-commutative. In the commutative case we just happen to have ring homomorphisms $A \to \operatorname{End}(M)$ and $A \to \operatorname{End}(N)$ given by $a \mapsto f_a$.

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Excellent answer, 1+. – Martin Brandenburg Apr 5 '12 at 8:17

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