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Let $R$ be a self injective ring. Then $R^n$ is an injective module. Let $M$ be a submodule of $R^n$ and let $f:M\to R^n$ be an $R$-module homomorphism. By injectivity of $R^n$ we know that we can extend $f$ to $\tilde{f}:R^n\to R^n$.

My question is that if $f$ is injective, can we also find an injective extension $\tilde{f}:R^n\to R^n$?

Thank you in advance for your help.

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    The answer is yes if $M$ is essential, or if $R^n$ is the injective hull of $M$. I don't know what happens in the general case though.2012-02-26
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    If $R$ is Artinian, this implies that $\operatorname{Aut}(R^n)$ operates transitively on each set of isomorphic submodules of $R^n$. It'd be nice :)2012-02-29
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    Note that it's possible to come up with a commutative ring $A$ and a homomorphism $f:A\rightarrow A^2$ such that it can't be extended injectively to $\tilde{f}:A^2 \rightarrow A^2$; see http://mathoverflow.net/questions/33294/linearly-independent-subsets-of-a-free-module . So if someone can just do this with A being self-injective...2012-03-05
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    Actually, if one uses the theorem quoted at http://mathoverflow.net/questions/30066/cardinality-of-maximal-linearly-independent-subset/30369#30369 , it's not too hard to see it's true if we additionally assume R is commutative, noetherian, and local; but that's probably way too many conditions to be helpful. I'll see if I can knock off one or two.2012-03-05

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