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We wanted to do this:

Let $R$ be a commutative ring and $I,J$ ideals of $R$. Then we have the $R$-isomorphism (of modules): $$\text{Hom}_R(R/I, R/J)\cong (J:I)/J$$ where $$(J:I)=\{x\in R\mid Ix\subseteq J\}\supseteq J$$ is the quotient of the ideals $J,I$.

We defined the homomorphism: $$h:(J:I)\rightarrow \text{Hom}_R(R/I, R/J)$$ $$x\mapsto h_x$$ where $$h_x:R/I\rightarrow R/J$$ $$r+I\mapsto xr+J$$

We proved that $h_x,h$ are homomorphisms and that $h$ is surjective, but we have trouble trying to prove that $$\ker h=J$$

Until now we tried to do it like this:


Let $x\in \ker h$, then $(\forall r\in R), h_x(r+I)=0+J$, that is $xr+J=J$, and this is $xr\in J$, but we haven't been able to get further. Any idea about this? Thanks.

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$xr\in J$ for all $r\in R$, in particular for $r=1$, so $x\in J$.

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  • $\begingroup$ Oh, that's funny. Thanks. $\endgroup$ – David Molano Sep 19 '15 at 0:57

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