## If I is nilpotent and an induced R-module homomorphism M/IM → N/IN is surjective, then M → N is surjective

Let $R$ be a commutative ring and let $I \subseteq R$ be a nilpotent ideal, say with $I^k = 0$. Let $M$ and $N$ be left $R$-modules and suppose $\varphi : M \rightarrow N$ is an $R$-module homomorphism. Prove that if the induced mapping $\psi : M/IM \rightarrow N/IN$ is surjective, then $\varphi$ is surjective.

[I am ashamed to admit that I had to refer to this forum post to see how to prove this one.]

The induced mapping $\psi$ is given by $\psi(m + IM) = \varphi(m) + IN$.

Note that since $\psi$ is surjective, we have $N/IN = \psi[M/IM] = (\varphi[M] + IN)/IN$. By the Lattice Isomorphism theorem for modules, we have $N = \varphi[M] + IN$.

We claim that $N = \varphi[M] + I^tN$ for all $t \geq 1$. We prove this by induction, with the base case already shown. Suppose now that the equation holds for some $t \geq 1$; then $N = \varphi[M] + I^tN$ $= \varphi[M] + I^t(\varphi[M] + IN)$ $= \varphi[M] + I^t\varphi[M] + I^{t+1}N$ $= \varphi[M] + I^{t+1}N$ as desired, since $I^t\varphi[M] \subseteq \varphi[M]$.

Since $I^k = 0$, we have $N = \varphi[M] + I^kN = \varphi[M]$. Thus $\varphi$ is surjective.

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### Comments

• Becca  On May 30, 2011 at 11:15 am

Please may you explain the last step of the induction procedure. Is it supposed to be f[M] + I^(t+1} N, and if so, how do you reach this conclusion?
Thank you.

• nbloomf  On May 30, 2011 at 9:05 pm

Sorry- there was a typo. ($t$ should have been $t+1$.) Thanks for the heads up.

The final step follows because $I^t \varphi[M] \subseteq \varphi[M]$.