$\operatorname{Hom}_{R/I}(I/I^2, M)\cong \operatorname{Hom}_R(I,M)$

Question

In the proof of Proposition 6.1.8(d) of Qing Liu's Algebraic Geometry and Arithmetic Curves, the following identification between two $\operatorname{Hom}$-modules is stated without proof, and I'd like to clarify the details.

Let $R$ be a commutative ring, $I\subseteq R$ be an ideal, and $M$ be an $R$-module. Then the claim is that $$\operatorname{Hom}_{R/I}(I/I^2, M)\cong \operatorname{Hom}_R(I,M).$$

This statement should mean that any $R/I$-linear map $I/I^2\rightarrow M$ comes from a $R$-linear map $I\rightarrow M$. I tried to consider the SES of $R$-modules $$0\rightarrow I^2\rightarrow I\rightarrow I/I^2\rightarrow 0$$ and then apply the left-exact contravariant $\operatorname{Hom}_{R}(-, M)$ to obtain $$0\rightarrow \operatorname{Hom}_{R}(I/I^2, M)\rightarrow \operatorname{Hom}_{R}(I,M)\rightarrow \operatorname{Hom}_{R}(I^2, M)$$

where the image of the second homomorphism is the kernel of the third. I feel like the desired identification should follow from here, but I'm having trouble seeing what's going on.

Answer 1

You can see the answer for this from the short exact sequence you have written down although I believe you need $M$ to be an $R/I$ module. (I also don't see how the statement makes sense if you don't have this) Then note that $\text{Hom}_{R/I}(I/I^2,M)=\text{Hom}_{R}(I/I^2,M)$ so it is enough to show that the image of the last map is 0. But now for any $f \in \text{Hom}_{R}(I,M) $ you have that $f(I^2)$=0. Indeed for any $x \in I^2 $ you have that $x$ is a sum of terms of the form $x_{i}y_{i}$ for $x_{i} \ , y_{i} \in I$ but you have $f(x_{i}y_{i})=x_{i}f(y_{i}) \in IM$ but as $M$ is an $R/I$ module $M = M/IM$ so that you have $f(x_{i}y_{i})=0$.

Answer 2

Following the suggestion from Cranium Clamp's comment to use adjunction, here's another way to see what's going on:

\begin{align*} \operatorname{Hom}_{B/I}(I/I^2, M) &\cong \operatorname{Hom}_{B/I}(I\otimes_B (B/I), M)\\ &\cong \operatorname{Hom}_B(I, \operatorname{Hom}_{B/I}(B/I, M))\\ &\cong \operatorname{Hom}_B(I, M) \end{align*}