Local artinian ring and module with finite projective dimension

Question

$R=k[x]/(x^n)$, $n>1$. Suppose $\operatorname{pd}_RM<\infty$, $M$ might not be finitely generated and $M$ might be uncountably generated. I want to show $M$ is free irregardless of non-finitely generated (i.e., either countably or uncountably generated).

I am aware that the previous post indicates a solution for finitely ranked free modules in the free resolution as $k[x]/(x^n)$ module with finite free resolution is free. However in Eisenbud, there is no such guarantee.

I have two ideas on this.

I was referring to Eisenbud Commutative Algebra Exercise 4.11b to solve the problem for finitely generated case. However, there is no reason to assume $F_i$'s have finite rank. So this clearly does not work. See answers below Mariano Suárez-Álvarez's comments.
The other way is to see $M$ as a limit of finitely generated modules $N_i$. I hope the free resolution is finite and finite rank for sure as it is artinian ring. However, it might not be guaranteed that $N_i$ being free where Mariano Suárez-Álvarez commented on this as well.

score 2 · Accepted Answer · answered Mar 19 '17 at 04:25

2

That algebra is self-injective, and from that it follows that a module is either projective or of infinite projective dimension. As your module M has finite projective dimension, it must be in fact projective. Now, your algebra is local, so the module is free because it is projctive.

answered Mar 19 '17 at 04:25

Mariano Suárez-Álvarez

139,300

Not really. ${}{}$ – Mariano Suárez-Álvarez Mar 19 '17 at 04:37
http://stacks.math.columbia.edu/tag/058Z for example. – Mariano Suárez-Álvarez Mar 19 '17 at 04:40
Alternatively: that algebra has finite epresentation type, so every module is a direct sum of indecomposable finite dimensional modules, and the only indecomposable finite dimensional module which is projective is the free one – Mariano Suárez-Álvarez Mar 19 '17 at 04:41
It is true. If it is not intuitively clear to you then you just have to adjust your intuition. – Mariano Suárez-Álvarez Mar 19 '17 at 04:43
Huh? I have no idea what you are talking about. A projctive module over a local ring is free. That is what you have to interpret. – Mariano Suárez-Álvarez Mar 19 '17 at 04:44
Why don't you just read the proof? (you will see immediately that it is not based on taking limits of Finitely generated modules...) It is quite not true that a direct limit of free modules is free, so I have no idea why you expect it to be true. – Mariano Suárez-Álvarez Mar 19 '17 at 04:51
OK. Thanks. I will check the proof in stacks now. – user45765 Mar 19 '17 at 04:52
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Please do not delete comments to which I have responded, as it turns the comment thread into an incomprehensible mess. – Mariano Suárez-Álvarez Mar 19 '17 at 04:54
Sigh. As you do not seem to care about that I'll just start not caring about you. "Saving face" deleting comments in which you make entirely natural mistakes is below the standards I ask of people who want my help. – Mariano Suárez-Álvarez Mar 19 '17 at 04:57
♦ I am not trying to save face as I do not want to confuse people by my mistake. I will comment in my question part. – user45765 Mar 19 '17 at 05:02
The thread of comments is designed to unconfuse others which might make the same mistake, just as it helped unconfuse you. Now all they will see is a sequence of unconnected and meaningless comments on my answer written by myself. – Mariano Suárez-Álvarez Mar 19 '17 at 05:20
On the other hand, you seem to think that my answer somehow does not answer your question and that there is something else to be elucidated. There isn't, really. – Mariano Suárez-Álvarez Mar 19 '17 at 05:22

Local artinian ring and module with finite projective dimension

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