This is an exercise (Exercise 1.23) in Eisenbud. I am not very familiar with free modules, nor resolutions, so I was wondering if this proof was accurate. I have some suspicions, but I figured it would be better to defer to experts.
Proposition: If $M$ is a $(k[x]/\langle x^n \rangle)$-module with finite free resolution, then $M$ is free.
Potential Proof:
Let $R = k[x]/\langle x^n \rangle$, and suppose that $M$ has a finite free resolution:
$$0 \to R^{n_l} \to R^{n_{l-1}} \to \dots \to R^{n_1} \to R^{n_0} \to M \to 0$$
This induces a free resolution $\mathcal{F} : 0 \to R^z \to R^{n_0} \to M \to 0$, where $z = \sum_{i=1}^l (-1)^{i+1} n_i$. Since the map $R^z \to R^{n_0}$ is a monomorphism, after a change of variables we may consider it the inclusion into the first $z$ coordinates of $R^{n_0}$, so it has a retract in the projection sending elements of $R^{n_0}$ to their first $z$ coordinates. As $\mathcal{F}$'s first nontrivial morphism has a retract, $\mathcal{F}$ is split, so:
$$R^{n_0} \cong M \oplus R^z$$
Since $x^n$ annihilates $R^k$ for any $k$, we multiply both sides by $1^{\oplus n_0-z} \oplus (x^n)^{\oplus z}$ to obtain $R^{n_0 - z} \cong M$, as desired. $\square$
Is this proof correct? If not, where is it incorrect?