# R-module homomorphism surjectivity

• Nov 8th 2008, 10:22 PM
zelda2139
R-module homomorphism surjectivity
Let $\displaystyle I$ be a nilpotent ideal in a commutative ring $\displaystyle R$, let $\displaystyle M$ and $\displaystyle N$ be $\displaystyle R-$modules and let $\displaystyle \phi$ : $\displaystyle M \rightarrow N$ be an $\displaystyle R-$module homomorphism. Show that if the induced map $\displaystyle \bar{\phi} : \frac{M}{IM} \rightarrow \frac{N}{IN}$ is surjective, then $\displaystyle \phi$ is surjective.
• Nov 9th 2008, 12:32 AM
NonCommAlg
Quote:

Originally Posted by zelda2139
Let $\displaystyle I$ be a nilpotent ideal in a commutative ring $\displaystyle R$, let $\displaystyle M$ and $\displaystyle N$ be $\displaystyle R-$modules and let $\displaystyle \phi$ : $\displaystyle M \rightarrow N$ be an $\displaystyle R-$module homomorphism. Show that if the induced map $\displaystyle \bar{\phi} : \frac{M}{IM} \rightarrow \frac{N}{IN}$ is surjective, then $\displaystyle \phi$ is surjective.

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

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