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Let $m$ be an ideal of a ring $A$. Set $B:=A/m$. Then, $m$ and $m/m^2$ have natural structures of $A$ and $B$ respectively.

For any $B$-module T which also has a natural structure of an $A$-module, I want to prove that $$Hom_B(m/m^2, T)\cong Hom_A(m,T)$$ as abelian groups.

Consider a homomorphism of albelian goups $H$: $Hom_B(m/m^2, T)\rightarrow Hom_A(m,T)$ by $H(f)=f\circ \pi$

where $\pi:$ $A\rightarrow A/m^2$ is the natural projection.

I can prove the injectivity part but get stuck on the surjectivity part. So, my questions are : Am I on the right track ? and how to prove it ?

Anyone can give me some hints or a reference for that ?

Thank you in advance.

An Khuong Doan
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1 Answers1

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Consider an $A$-linear map $f:\mathfrak m\to T$. Since $T$ is originally a $B$-module, this means $\mathfrak m$ acts trivally on it, so if $ab\in\mathfrak m^2$, we have that $f(ab) = af(b) = 0$. This means $f$ induces a map $\bar{f} : \mathfrak m/\mathfrak m^2\to T$ which you can check is $B$-linear. This defines a map of abelian groups $$F : \hom_A(\mathfrak m,T)\to \hom_B(\mathfrak m/\mathfrak m^2,T)\text { such that } F(f) = \bar{f}.$$

The inverse $G$ is given by precomposing with the projection $\mathfrak m\to\mathfrak m/\mathfrak m^2$. Once you check that $FG$ and $GF$ are the identity, you need not worry about proving injectivity or surjectivity, since you have already proved that these arrows are isomorphisms.

Pedro
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