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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.

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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
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I have never heard of "projective homological dimension"... and I collect homological algebra books :D –  Mariano Suárez-Alvarez Apr 25 '12 at 4:17

1 Answer 1

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.

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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

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