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If $R$ is a ring and $I\subseteq R$ is an ideal, I want to see why $$\textrm{Hom}_R(I, R/I) \cong \textrm{Hom}_R(I/I^2, R/I)$$ I read somewhere that this is true because of the action of any element in $I$. I'm having difficulty seeing what this means explicitly. Can someone spell it out?

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  • $\begingroup$ Prove the more general statement that $\text{Hom}(M, R/I) \cong \text{Hom}(M/I, R/I)$. $\endgroup$ – Qiaochu Yuan Mar 8 '17 at 1:55
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    $\begingroup$ Is $M/I$ short for $M / IM$? $\endgroup$ – Dean Young Mar 8 '17 at 1:56
  • $\begingroup$ You should not delete a question like this. See math.stackexchange.com/questions/2200131/… $\endgroup$ – reuns Mar 23 '17 at 17:33
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$\DeclareMathOperator{\Hom}{Hom}$ Sure. If an $R$-linear map $\alpha : M \rightarrow P$ kills a submodule $N$ of $M$, then there is a unique $\beta : M/N \rightarrow P$ such that $\beta \circ \pi = \alpha$, where $\pi : M \rightarrow M/N$ is the projection map.

Define a map $\Phi : \Hom_R (I, R/I ) \rightarrow \Hom_R (I / I^2, R / I)$ where $\phi : I \rightarrow R/I$ is sent to the unique $\psi : I / I^2 \rightarrow R / I$ such that $\psi \circ \pi = \phi$, where $\pi : I \rightarrow I/I^2$ is the canonical projection map. Note that $\phi : I \rightarrow R /I$ must indeed kill $I^2$ since $\phi(xy) = x\phi(y) =0$ for $x, y \in I$.

There is also a map $\Psi : \Hom_R (I / I^2, R / I) \rightarrow \Hom_R ( I, R/I)$, where $\phi \mapsto \phi \circ \pi$.

All that's left is to show the two maps are inverse to each other. I'll elaborate on this point upon request.

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Its because they are $R$ modules basically you need that if $$f:I\to R/I$$ is a homomorphism then $$I^2\subseteq \ker f.$$ To see this if $a,b\in I$ then $$f(ab)=af(b)=0.$$ Since $ax=0$ for all $x\in R/I$ and $a\in I$.

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