Map induced by inclusion has image $0$

Question

Let $R$ be a ring and $I\subseteq R$. Then the inclusion $\varphi:I^2\hookrightarrow I$ induces a map $$\varphi^*:Hom_R(I,R/I)\rightarrow Hom_R(I^2,R/I),\hspace{.5cm} \varphi^*(f)=f\circ\varphi$$ I want to show that $\text{im}(\varphi^*)=0$.

My ideas: Let $x\in I^2$ and let $f:I\rightarrow R/I$. Then it suffices to show that $f(\varphi(x))\in I$. I don't see the trick here, but I think that since the inclusion has a left inverse, $\varphi^*$ will be surjective (I've proved this).

Answer 1

Consider the exact sequence $0\to I^2\xrightarrow{\varphi} I\xrightarrow{\pi}I/I^2\to 0$; if you apply the functor $\operatorname{Hom}_R(-,R/I)$, you get the exact sequence $$ 0\to\operatorname{Hom}_R(I/I^2,R/I) \xrightarrow{\pi^*}\operatorname{Hom}_R(I,R/I) \xrightarrow{\varphi^*}\operatorname{Hom}_R(I^2,R/I) $$ and you know from your question Isomorphism of Hom sets that $\pi^*$ is an isomorphism. By exactness, the kernel of $\varphi^*$ contains the image of $\pi^*$. Therefore $\varphi^*$ is the zero map.

Answer 2

Hint:

How do you describe an element of $I^2$?

Answer 3

Let $i,j\in I$, $f(i.j)=i.f(j)$ since $f$ is an homomorphism of $R$-modules. If $y\in R/I, i\in I, i.y=0$, this implies that $i.f(j)=0$ . $I^2$ is generated by $\{i.j; i,j\in I\}$.