An easy infinite free resolution I'm doing exercise 1.23 on Eisenbud's Commutative algebra, and I have the following situation: let $k$ be a field and $R = k[x]/(x^n)$. They ask for a free resolution of $R/(x^m)$, for some $m \leq n$. We have the following resolution:
$$ \cdots \rightarrow R \rightarrow R  \rightarrow R \rightarrow R \rightarrow R/(x^m) \rightarrow 0 $$
where the penultimate arrow is multiplication by $x^{n-m}$, the one before multiplication by $x^m$, the one before by $x^{n-m}$, and so on.
How can I proof that $R/(x^m)$ has no finite free resolution for $m<n$? I'm looking for an easy proof, without using any big theorem. (That is, this question is supposed to be solvable just after giving the definition of free resolution.)
At best, I have to solve the following question using only the contents of chapter 1 of the book (pages 44 to 46):

Question: Show that the only $k[x]/(x^n)$-modules with finite free resolution are the free modules.

Thank you!
 A: This answer restricts to finitely generated modules for simplicity.
The last non-zero morphism in a bounded and finite rank free resolution of some $R := k[x]/(x^n)$-module is an injective homomorphism of free, finite rank $R$-modules. Choosing bases, such is given by a matrix $A\in\text{Mat}_{a\times b}(R)$, say.
Now, look at the first column of $A$. If all its entries were multiples of $x$, then $A$ wouldn't denote an injective morphism, as the $x^{n-1}$-th multiple of the first basis vector would be sent to zero.
From this we deduce that the first colum contains a unit, and after row and column transformations, we may therefore assume that $A$ has the form $\begin{pmatrix} 1 & 0 \\ 0 & A^{\prime}\end{pmatrix}$ for some $A^{\prime}$. 
Continuing this way, you finally deduce that $A$ denotes a split monomorphism of $R$-modules, or more precisely one whose image has a free complement. This allows for shortening the hypothetical finite free resolution by one. Continuing this over and over, you see that any finite length free resolution can be shortened to a free resolution of length $0$, proving that the module being resolved was actually free.
The point of all this is the self-injectivity of $R$: $R$ is injective as a module over itself.
