Recall the following theorem, due to Hilbert. Let be a field, and considered as a graded ring, and take any finitely generated graded -module. Then there is a free resolution of of length at most . This theorem is important because applying it in the case for a homogeneous ideal (i.e. to a projective variety), it tells us that the Hilbert polynomial exists and gives an explicit bound on its degree.

I had always assumed this theorem was in some sense fairly deep or hard, but in Chan’s scheme theory course today we saw that it’s actually a small amount of some fairly basic homological algebra.

Step 1: Do the case , where each acts as the zero homomorphism. I would rather not do the details of this in a blog post, but it isn’t very hard, and you if you try to write down a resolution that’s as efficient as possible, you should end up with a resolution of length , which usually seems to be given the intimidating name `the Koszul complex’.

Step 2: Let now be arbitrary, and consider a minimal free resolution (in the sense that the first map has the smallest possible number of generators, then the second one does, and so on). By minimality, for all a basis element of must be mapped to an element with no constant term in any coefficient. Indeed, if is mapped to , each is homogeneous (since the maps are maps of graded modules) and if some is constant, since is a field and is in the kernel of , we see that the basis vector maps to an element that is expressible as a linear combination of the image of the other s, contradicting minimality. What this observation buys us is that if we take such a complex and tensor it with the -module , we get a complex

and all of the maps save the first one are the zero map.

Step 3: Use the fact that , or more generally that . This really is the key ingredient. Knowing that the above maps are zero, we see that the rank of is just the rank of . And this is the rank of , which can be computed by as the homology of the complex obtained by tensoring the Koszul complex with . In particular, we see that it vanishes for , so Hilbert’s theorem is proved.

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