Extension and Self Injective Ring 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.
 A: 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.
A: 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 $R^n$.
