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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  • 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].

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