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Say you have a bimodule $M$ over a ring $A$ with maps $f,g:M \rightarrow A^n$ such that $f$ is injective and $g$ is surjective. Can we say that $M \cong A^n$? What if $f$ and $g$ were in the opposite direction? What if $A$ is commutative?

I figured out that if the free module is infinite dimensional, then they need not be isomorphic, but can't figure out the finite dimensional case.

EDIT: I would welcome an answer even if it just answers the commutative case.

Siddharth Gurumurthy
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  • Since $g : M \to A^n$ is surjective and $A^n$ is free (thus projective), the surjection splits and we have $M \cong A^n \oplus \ker(g)$. Thus you need either to find a ring $A$ such that $A^n \oplus T$ injects into $A^n$ for some $A$-module $T \neq 0$ or show that this is not possible. – Matthias Klupsch Jan 09 '20 at 21:09
  • Yes, I did realize that, however, I couldn't proceed further. – Siddharth Gurumurthy Jan 10 '20 at 06:24
  • You want $f$ and $g$ to be bimodule homomorphisms? And for $A^n$ to have the obvious bimodule structure given by left and right multiplication? – Jeremy Rickard Jan 10 '20 at 09:58
  • @JeremyRickard, yes. Although, I would like a partial answer with $M$ being a left module. – Siddharth Gurumurthy Jan 10 '20 at 12:28
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    For left modules over a commutative ring $A$, this is true and follows from Orzech's theorem (since you can use the injective map $f$ to identify $M$ with a submodule of $A^n$). – darij grinberg Jan 12 '20 at 11:22
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    The same holds for bimodules over a commutative ring $A$. Indeed, $A$-$A$-bimodules are "the same as" left modules over the commutative ring $A\otimes_{\mathbb Z}A^{\operatorname{op}}$, and of course you can apply Orzech's theorem over the latter ring as well. (The $A\otimes_{\mathbb Z}A^{\operatorname{op}}$-module $A$ is not usually free, but it is finitely generated, and that's sufficient.) – darij grinberg Jan 12 '20 at 11:24
  • Thanks! That does answer that part of my question. Feel free to add that as an answer. – Siddharth Gurumurthy Jan 12 '20 at 14:18

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