How do I understand the module structure on Yoneda Ext?

Question

Suppose $R$ is a (commutative) ring, and $M$ and $N$ are (finitely generated) $R$-modules. Then I know each $\mathrm{Ext}_R^i(M, N)$ has the structure of an $R$-module. On the other hand, via Yoneda's description of Ext, each $\varepsilon \in \mathrm{Ext}_R^i(M, N)$ corresponds to an equivalence class of exact sequences starting with $N$ and ending with $M$. My question is this: suppose $r \in R$ and $\varepsilon \in \mathrm{Ext}_R^i(M, N)$. How can I understand $r \varepsilon$ in terms of $\varepsilon$? To what extension does $r \varepsilon$ correspond?

Answer 1

we can just see the case $i=1$: In general,if $M$ is $A-B$ bimodule(i.e.left $A$ module and right $B$ module and $(am)b=a(mb)$),N is $A-C$ bimodule,then $Ext^1(M,N)$ is a $B-C$ bimodule.the structure of left $B$ and right $C$ module as follows:

if $\varepsilon:0\rightarrow N\xrightarrow f X\xrightarrow g M\rightarrow 0$ is a short exact sequence in left $A$-modules.$\forall b\in B$,there is a left $A$-module homomorphism $\varphi_b:M\rightarrow M$ by senting $m$ to $mb$.take pullback with $\varphi_b$ and $g$,we get an element in $Ext^1(N,M)$ this is $b\cdot \varepsilon$.

similarly,if $\forall c\in C$,take pushout with natural right multiplication $\phi_c:N\rightarrow N$ and $f$,we get an element in $Ext^1(N,M)$ this is $\varepsilon\cdot c$. it is easy to Check this: structre of left $B$ module and right $C$ module has associativity. i.e. $(b\cdot \varepsilon)\cdot c=b\cdot (\varepsilon\cdot c)$. As follows: Then $\varphi_a=\alpha^-$ is using the unique map induced by kernel.

Now we consider the commutative case,we only need to check $r\cdot \varepsilon=\varepsilon\cdot r$,suppose $\varepsilon:0\rightarrow N\xrightarrow f X\xrightarrow g M\rightarrow 0$ is a short exact sequence in $R-Mod$. Then consider the following diagram: The $\beta^\prime$ make all the diagram commute. $\beta^\prime$ is also the induced map between kernel, so it is unique. It is also induced by $\phi_r:X\rightarrow X$, hence $\beta^\prime=\phi_r:N\rightarrow N$. By the diagram we have:$r\cdot \varepsilon=\varepsilon\cdot r$.

for $i>1$,the same.