Given an integral domain $R$, we have that $R^n$ embeds in $R^{n+1}$ as free modules over $R$. Can we have a surjective $R$-homomorphism from $R^n$ to $R^{n+1}$? Or more so, given two free $R$-modules $A$ and $B$ where there is an injective homomorphism from $A$ to $B$ and also a surjective homomorphism from $A$ to $B$, can we conclude that $A \cong B$?
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No, and yes. See Corollary 5.11 in http://www.math.uconn.edu/~kconrad/blurbs/linmultialg/extmod.pdf . – darij grinberg Mar 26 '16 at 22:07
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I figured it out. Thanks for the help – JKim Mar 26 '16 at 23:11
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For general (commutative) rings, the notion of rank of a free module is well-behaved : if there is a injective morphism $A\to B$ then $rk(A)\leqslant rk(B)$, and if there is a surjective morphism you have the reverse inequality. So since free modules are determined up to isomorphism by their rank, the answer to your question is yes.
When $R$ is integral, it's even easier to see : just tensor with the field of fraction, which is flat over $R$ so tensoring preserves injectivity and surjectivity.
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