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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. –  user641 Feb 26 '12 at 23:07
    
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 :) –  Mariano Suárez-Alvarez Feb 29 '12 at 4:51
    
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 mathoverflow.net/questions/33294/… . So if someone can just do this with A being self-injective... –  Harry Altman Mar 5 '12 at 2:21
    
Actually, if one uses the theorem quoted at mathoverflow.net/questions/30066/… , 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. –  Harry Altman Mar 5 '12 at 2:33

2 Answers 2

The question is also true without any commutativity for quasi-Frobenius rings.

Recall that a quasi-Frobenius ring is a ring which is one-sided self injective and one-sided Noetherian. They also happen to be two-sided self-injective and two-sided Artinian.

For every finitely generated projective module $P$ over a quasi-Frobenius ring $R$, a well-known fact is that isomorphisms of submodules of $P$ extend to automorphisms of $P$. (You can find this on page 415 of Lam's Lectures on Modules and Rings.)

Obviously your $P=R^n$ is f.g. projective, and injecting $M$ into $P$ just results in an isomorphism between $M$ and its image, so there you have it!

In fact, this result seems a bit overkill for your original question, so I would not be surprised if a class properly containing the QF rings and satisfying your condition exists.

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Well, this is true if $R$ is commutative and noetherian; I don't know whether that's good enough for what you want. (This solution may be overcomplicated; I do not actually know how to prove all the facts I am using.)

If $R$ is commutative, noetherian, and self-injective, then it's Artinian, it's a finite product of local Artinian rings, hence we can reduce to the local case.

So say $R$ is commutative, noetherian, and local (and hence Artinian, but I won't use that). Take an injective hull of $M$ inside $R^n$; call it $Q$. So $f$ extends injectively to $f:Q\rightarrow R^n$, and we now need to extend it from $Q$ to all of $R^n$. Since $Q$ is injective, it is a direct summand of $R^n$, and hence is also projective. But we assumed $R$ was local, and hence $Q$ is free; say it is isomorphic to $R^m$, $m\le n$.

Then an injective function $R^k \rightarrow R^n$ is the same as an (ordered) linearly independent subset of $R^n$, of $k$ elements. So we have $m$ linearly independent elements of $R^n$ and we want to extend it to $n$ such. We can extend it to a maximal linearly independent set, certainly; the question then is just if such a set necesssarily has $n$ elements.

Now, since we assumed $R$ was commutative and noetherian, we can apply the theorem of Lazarus quoted here, and say yes, a maximal linearly independent set of $R^n$ necessarily has $n$ elements, and so having extended $f$ to $Q\cong R^m$, we can further extend it to $R^n$.

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My understanding of the theorem of Lazarus that you quoted does not guarantee that we can extend $Q$ to $R^n$. It only says if we start we an $R$-module $M$ all maximal linear independent sets in $M$ has the same cardinality. Do I miss something? –  user9077 Mar 6 '12 at 20:59
    
Again, consider what this means when $M$ is free, say isomorphic to $R^n$. It means that all maximal linearly independent sets have size $n$, and thus (since any linearly independent set can be extended to a maximal one) that all linearly independent sets can be extended to one of size $n$. But an (ordered) linearly independent set of size $k$ is exactly the same as an injective homomorphism from $R^k$. So every injective homomorphism from some $R^k$ can be extended to one from $R^n$; and $Q\cong R^m$. –  Harry Altman Mar 6 '12 at 22:56
    
Thank you for your explanation. I am not very good at ring and module theory, so please be patient with me. If we take our $M$ to be the $R^n$ the Lazarus theorem says that every maximal linear independent set in $M$ is of sixe $n$. I understand that. But doesn't that we are looking for is maximal linearly independent set that contains $Q$. Is this extra requirement (containing $Q$) does not make any difference at all? –  user9077 Mar 6 '12 at 23:12
    
Remember, I'm using the fact that $Q$ is free, since I assumed $R$ was local. So, if you prefer to think of it that way, we are taking a free basis for $Q$ and extending it to a maximal linearly independent set. Which then must have size $n$. –  Harry Altman Mar 6 '12 at 23:39

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