R-module homomorphism surjectivity

**zelda2139** · November 8th 2008, 10:22 PM

Let $I$ be a nilpotent ideal in a commutative ring $R$ , let $M$ and $N$ be $R-$ modules and let $\phi$ : $M \rightarrow N$ be an $R-$ module homomorphism. Show that if the induced map $\bar{\phi} : \frac{M}{IM} \rightarrow \frac{N}{IN}$ is surjective, then $\phi$ is surjective.

**NonCommAlg** · November 9th 2008, 12:32 AM

Originally Posted by zelda2139

Let $I$ be a nilpotent ideal in a commutative ring $R$ , let $M$ and $N$ be $R-$ modules and let $\phi$ : $M \rightarrow N$ be an $R-$ module homomorphism. Show that if the induced map $\bar{\phi} : \frac{M}{IM} \rightarrow \frac{N}{IN}$ is surjective, then $\phi$ is surjective.

so $I^k=0,$ for some $k \geq 1.$ since $\bar{\phi}$ is surjective, we have: $\frac{N}{IN}=\bar{\phi}\left(\frac{M}{IM}\right)=\ frac{\phi(M)+IN}{IN}.$ thus: $N=\phi(M)+IN,$ which gives us: $N=\phi(M)+I(\phi(M)+IN)=\phi(M)+I^2N,$ because

$I\phi(M) \subseteq \phi(M).$ repeating this we'll get: $N=\phi(M)+I^2(\phi(M) + IN)=\phi(M) + I^3N = \cdots = \phi(M) + I^kN=\phi(M). \ \Box$

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