Let $M$ be a finitely generated module over a commutative ring $A$. Is it true that if there exists a positive integer $n$ and a pair of homomorphisms $\pi:A^n\rightarrow M$ and $\phi:A^n\rightarrow M$ such that $\pi$ is surjective and $\phi$ is injective then $M\cong A^n$?
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Yes, it is.
Let $N=\phi(A^n)$. Then $N$ is a submodule of $M$ isomorphic to $A^n$. Now define $f:N\to M$ by $f(\phi(x))=\pi(x)$. Then $f$ is surjective, and therefore injective. We thus get that $M$ is free of rank $n$.
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Why does $f$ exist? – Martin Brandenburg Nov 15 '14 at 19:49
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@MartinBrandenburg $f$ is this composition: $\phi(A^n)\simeq A^n\stackrel{\pi}\to M$. – user26857 Nov 15 '14 at 19:52
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Alright, but $f$ is not an endomorphism, and therefore the result doesn't apply. – Martin Brandenburg Nov 15 '14 at 19:53
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6@MartinBrandenburg M. Orzech proved that it is not necessary to be an endomorphism: http://www.jstor.org/discover/10.2307/2316897?uid=3737528&uid=2129&uid=2&uid=70&uid=4&sid=21104550594211 – user26857 Nov 15 '14 at 19:54
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I didn't know it, thank you! – Martin Brandenburg Nov 15 '14 at 19:55