Let $R$ be an artinian chain (uniserial) ring. I want to prove that $R$ is self-injective, that is, any $\alpha:I\rightarrow R$ right $R$-module homomorphism can be extended to a homomorphism $\tilde{\alpha}:R\rightarrow R$ where $I$ is a right ideal of $R$.
Let $J$ be the jacobson radical of $R$. We knov that all right ideals of $R$ form a chain $$R\supset J\supset J^2\supset ...\supset J^{m-1}\supset J^m=0$$ and $J^k=p^kR$ for some $p\in J\setminus J^2$ and any $k=1,2,...$ since $R$ is an artinian chain ring. So I need to show that any $\alpha:p^k\rightarrow R$ right $R$-module homomorphism can be extended to a homomorphism $\tilde{\alpha}:R\rightarrow R$.
How can i find the extension. I need some help.
Thanks a lot...
Given that you know that the right ideals are precisely the powers of the Jacobson radical, I'm guessing you also know that they coincide with the left ideals, and that $J^k=p^kR=Rp^k$ for any $p\in J\setminus J^2$?
If $\alpha:p^kR\to R$ is a right $R$-module homomorphism then $\operatorname{im}(\alpha)$ is a right ideal of $R$ and can't have length greater than that of $p^kR$ (since it's a quotient of $p^kR$), so $\operatorname{im}(\alpha)=p^lR=Rp^l$ for some $l\geq k$.
$p^kR$ is generated as a right module by $p^k$, and so $\alpha$ is determined by $\alpha(p^k)$, which is $sp^l$ for some $s\in R$.
Now just define $\bar{\alpha}:R\to R$ by $\bar{\alpha}(r)=sp^{l-k}r$.