Prove $F[x]/(x^n)$ is an injective module

Question

Let $F$ be a field and $n\geq1$

(1) Prove $R=F[x]/(x^n)$ is an injective $R$-module.

(2) Give a projective resolution and an injective resolution of the $R$-submodule $M=(x)/(x^n)$

For part (1), I know that by Baer's criterion, it suffices to show for all ideals $(x^k)/(x^n)\ $ ($k\leq n$) that the inclusion

$$\iota:(x^k)/(x^n)\to F[x]/(x^n)$$ has a left inverse. But I don't know how to construct one. I have proven that $\iota(x^k+(x^n))$ has the form $$\iota(x^k)+(x^n)=a_kx^k+a_{k+1}x^{k+1}+\cdots+a_{n-1}x^{n-1}+(x^n),\quad a_k\neq0$$ and tried to work from there.

For part (2) I don't have a clue.

Any help is greatly appreciated. Thanks in advance!

Answer 1

The expansion $R\to R$ can be constructed by mapping $1$ to $a_k+a_{k+1}x+\cdots+a_{n-1}x^{n-1-k}$ (and I think k need not divide n).

The projective resolution is a series of $R$'s and maps are multiplication of $x$ and $x^{n-1}$ alternatively, ending by $x$ to the ideal. $\require{AMScd}$ \begin{CD} ... @>x^{n-1}>> R@>x>>R@>x^{n-1}>>R@>x>>\mathcal{a} \end{CD} The injective one is similar, and begin by inclusion.

Answer 2

My favourite proof for (1) uses the vector space duality $D:=\mathrm{Hom}_F(-,F)$. It is easy to see that a finite dimensional $R$-module $X$ is projective (respectively invective) if and only if $D(X)$ is injective (respectively projective).

Now consider the map $$ \tau \in D(R), \quad a_0+a_1x+\cdots+a_{n-1}x^{n-1} \mapsto a_{n-1}. $$ Then $\tau$ induces an isomorphism of $R$-modules $R\xrightarrow\sim D(R)$, $1\mapsto\tau$. Since it is clear that $R$ is projective, we see that it is also injective. In fact, this shows that $R$ is a symmetric algebra.

For (2) we can construct an exact sequence $$ 0 \to M \to R \xrightarrow{f} R \to M \to 0, $$ where $f\colon 1\mapsto x^{n-1}$. This is then the start of both a projective resolution of $M$, and an injective coresolution. Since it repeats, we can continue this in either direction to get the full (co-)resolution.