Injection and surjection over free modules.

Question

Let $A$ be a commutative ring and $M$ an $A$-module. Suppose to have both an injection $A^s \to M$ and a surjection $A^s \to M$ of module homomorphisms. Show that $M \simeq A^s$.

This point is the last thing that I need to conclude another homework I started two days ago (An exercise on tensor product in a local integral domain.).

I am curious to know if it is true in that context. Furthermore, also a proof for the easier case $A$ local integral domain (as indicated in my previous exercise) should be absolutely good (because it will lead me to solve the exercise).

Thank you in advance, any suggestion will be appreciated.

I just edited, assuming you meant "both a surjection and an injection", as this makes the most mathematical sense. — Circonflexe, Apr 30 '15 at 08:31
@Circonflexe Thanks, sometimes my English is wrong (I'm not a native speaker). — , Apr 30 '15 at 08:32
Is there any reason you are omitting the "local domain" part of the hypotheses? That might simplify matters considerably. — rschwieb, Apr 30 '15 at 10:14
@rschwieb Yes, the reason is that I am also curious to know if it is true also in the more general case. Thanks for the remark, I'll edit the question. — , Apr 30 '15 at 10:28

score 1 · Accepted Answer · edited Apr 13 '17 at 12:21

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Let $\sigma:R^s\to M$ be an injective homomorphism, and $\pi:R^s\to M$ a surjective homomorphism. Set $N=\sigma(R^s)$ and notice that $\sigma:R^s\to N$ is an isomorphism. Now consider $\pi\sigma^{-1}:N\to M$. This is a surjective homomorphism, hence by Orzech Theorem an isomorphism.

edited Apr 13 '17 at 12:21

Community

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answered Apr 30 '15 at 12:42

user26857

52,094

Thank you very much! I didn't know the Orzech Theorem. – May 01 '15 at 12:01

Injection and surjection over free modules.

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