# 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!

• I don't know of any easy proof, without using any big theorems: but your stated question is answered immediately by the Auslander-Buchsbaum formula. Existence of a finite free resolution is a rather strong requirement on a module (if the ring is not regular), and many things can be deduced from it, but these theorems are not so elementary (although they are well worth studying) – zcn Mar 6 '14 at 21:36
• @user121097 In his book, Eisenbud expects an answer without any development of projective dimension, so I guess that we are missing an easy proof for this fact. – Pedro A. Castillejo Mar 6 '14 at 22:17
• @PedroA.Castillejo You can find many ideas here. – Pedro Tamaroff Mar 4 '15 at 10:30

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.

• Duplicate? I wouldn't want to vote to close since my vote is binding. =) – Pedro Tamaroff Mar 4 '15 at 10:30
• Can you be a little bit more explicit in the row and column transformations? – Pedro A. Castillejo Mar 4 '15 at 10:37
• @PedroA.Castillejo: Once you have a unit, you can scale it to be $1$. Exchanging rows and columns you can move this $1$ to the upper left corner. Finally, applying row and column transformations you eliminate all other entries in the first row or column – Hanno Mar 4 '15 at 10:39
• Dear @PedroTamaroff: Thank you for the link, I didn't see that! I certainly won't feel offended if you close, but it could be worth keeping the answer since the restricted case of finitely generated modules might firstly be conceived as simpler due to the use of matrices, and secondly does not need the axiom of choice. – Hanno Mar 4 '15 at 10:40
• @PedroTamaroff: Thanks for the link! Is a very nice complement. This was asked before the other discussion, and I was asking for an elementary proof using the contents of chapter 1 of Eisenbud's book, as the nice one given by Hanno. But now I know the answer, so if you think that this should be closed I'm ok. – Pedro A. Castillejo Mar 4 '15 at 10:52