How to view set of equivalence classes of extensions of M by N as an A-module 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 $E_A(M, N)$?
 A: 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 $a \mapsto f_a$.
