$\mathbb k[x]$ / $(x^n)$ is a self-injective algebra

Question

I read that $\mathbb k[x]$ / $(x^n)$ is a self-injective algebra. But I can't find a proof of this fact.

Can someone explain me the proof?

Answer 1

You probably mean self injective ! I'll also assume that $\newcommand{\k}{\mathbb k}\k$ is a field

In this case, one way to prove this is to use Baer's criterion, recall what it says :

Let $R$ be a ring, and $I$ an $R$-module. Then $I$ is injective if and only if for all ideals $a$ of $R$ and maps $a\to I$, there is a map $R\to I$ extending it.

Now we can use this with $R=I=\k[x]/(x^n)$. Now an ideal of $R$ is of the form $a/(x^n)$ for some ideal $a$ of $\k[x]$ containing $(x^n)$. Since $\k$ is a field, $\k[x]$ is principal and so $a$ is of the form $(p)$ for some polynomial $p$, and $(x^n)\subset (p)$ means that $p$ is a divisor of $x^n$ : $a=(x^k)$ for some $0\leq k\leq n$.

Then $a/(x^n) \simeq \k[x]/(x^{n-k})$ as $R$-modules, so a map $a/(x^n)\to R$ amounts to choosing $q\in R$ such that $x^{n-k}q= 0$, which is easily seen to mean $q=x^kr$ for some $r$. Therefore we may extend it to $R$ by sending $1\mapsto r$ (note that $r$ is not unique, and so the extension isn't, but injectivity doesn't require any uniqueness)

By Baer's criterion, $R$ is then an injective $R$-module, which is exactly to say that $\k[x]/(x^n)$ is self-injective.