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?


1 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:enter image description here 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: enter image description here 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.

  • $\begingroup$ Thanks for your comment, this is very helpful. It is still somewhat unclear to me what happens for higher Exts, though. Is there some way to extend what you have written? I'm happy to go read on my own, if there is some source that describes how to take this structure on Ext^1 and deduce the structure on Ext^i. $\endgroup$ Nov 20, 2018 at 19:33
  • $\begingroup$ @EricCanton GTM4 $\endgroup$
    – Jian
    Nov 21, 2018 at 0:01

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