Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

While studying the Koszul complex, I can't properly recall a certain fact. I remember if $\bigwedge V$ is the exterior algebra of a finite dimensional vector space, then $\bigwedge V$ has infinite projective dimension. (I hope I have remembered this correctly.)

This feels weird to me, since I'm more used to considering modules over $\bigwedge V$ than $\bigwedge V$ itself. Essentially, is there an explanation of what goes wrong if $\bigwedge V$ were to admit a finite projective resolution? Of course, I welcome a more direct explanation, but this is just how I was trying to recall it myself. Thank you.

share|cite|improve this question
For every algebra $A$, the module $A$ has finite projective dimension, so you are remembering incorrectly! – Mariano Suárez-Alvarez Apr 24 '12 at 4:55
On the other hand, $\Lambda V$ has infinite global dimension, but that is a rather different statement. – Mariano Suárez-Alvarez Apr 24 '12 at 4:56
@MarianoSuárez-Alvarez Oh dear, that is certainly a problem on my part, sorry! I want to make sure there isn't a miscommunication in the terminology. By "infinite projective dimension," I meant to assert that $\bigwedge(V)$ has an infinite projective homological dimension. Is this the same thing as global dimension which you refer to? – Chelsea Dirks Apr 25 '12 at 4:15
I have never heard of "projective homological dimension"... and I collect homological algebra books :D – Mariano Suárez-Alvarez Apr 25 '12 at 4:17
up vote 4 down vote accepted

Let $M$ be a finitely generated left $\Lambda V$-module with finite projective dimension $n\geq0$.

Since $\Lambda V$ is a finite dimensional algebra, there exists a minimal projective resolution by finitely generated projectives $$0\to P_n\to P_{n-1}\to\cdots\to P_1\to P_0\to M\to 0$$ Suppose $n>0$. The module $P_n$ is projective. Since $\Lambda V$ is local, every f.g. projective is free, so in particular $P_n$ is free. Since $\Lambda V$ is an injective module over itself, $P_n$ —being a finite direct sum of copies of $\Lambda V$—is itself injective. This implies that the injection $P_n\to P_{n-1}$ is split, and this is absurd because the resolution was supposed to be minimal.

It follows that necessarily $n=0$.

We conclude in this way that every finitely generated module over $\Lambda V$ whose projective dimension is finite is in fact projective —we say that $\Lambda V$ has «finitistic global dimension equal to zero».

Now, if $\Lambda V$ were of finite global dimension, then we would have that it is in fact of global dimension zero. But it isn't: for example, if we let $S$ be the simple module $\Lambda V/\operatorname{rad}\Lambda V$, there is a surjective map $\Lambda V\to S$ which is not split, as you can easily check, so that $S$ is not projective.

N.B. I used above the fact that $\Lambda V$ is an injective module over itself. We say that it is therefore a self-injective algebra. Such an algebra, when finite dimensional, is always either semisimple or of infinite global dimension, by the same reasoning as above.

share|cite|improve this answer
Try to find a splitting for that surjection: it is very easy to show directly that there isn't one. – Mariano Suárez-Alvarez Apr 25 '12 at 4:34
Ah you're right. Nevermind. – Chelsea Dirks Apr 25 '12 at 4:40

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.