Wow. I hate looking at the dates on old posts. I think that maybe a few days have gone by, and I’m horrified to find that 11 or 12 days have passed. It is hard to keep track of time in grad school.

The goal of this post is to prove the theorem: If V is an irreducible projective variety over an algebraically closed field k, then every regular function on V is constant. Note this says that . Also, an exercise is to think about how this relates to Liouville’s Theorem if our field is .

Proof: Let V be an irreducible projective variety in . WLOG V is not contained in a hyperplane, since then we could just eliminate a variable and work in and repeat this until it was not in any hyperplane.

Let . Consider the affine covering from last time . Note that is regular as an affine morphism on . So we can write this as a polynomial in where . i.e. we can factor out the homogeneous part of the denominator variable to get where is homogeneous of degree .

But we assumed V irreducible, so is prime and hence is an integral domain. Let’s take the field of fractions then, . Then , and are all embedded in L. So in L we can multiply by that denominator we had before to get .

Recall that is a graded ring, so I just am denoting to be the -graded part. Thus if we take any integer , then is a finite-dimensional k-vector space. Moreover, the monomials of degree N span the space.

Let be a monomial. Then it is divisible by for some i, so . Thus .

So we have a chain: . So for any . Thus .

But is Noetherian, since it is finitely generated as a -module, so is also finitely generated over . Thus is integral over .

i.e. there are such that . But this shows that is homogeneous of degree 0. i.e. . So f is constant.

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May 26, 2009 at 11:45 am

Therefore, projective varieties can’t secretly be affine varieties. Unless they’re reduced to a point.

Also, couldn’t you prove it like this- P^n(k) is a proper scheme over k, therefore a morphism from P^n -> k must have a closed image, and that image must be proper. But that image is just an affine variety, so it can’t be proper.

I believe you can do something like this for the regular Liouville theorem too. If f: C -> C is bounded an entire, then by Riemann’s theorem on removable singularities it extends to F: S^2 -> C which is holomorphic, and whose image must be compact. But this contradicts the open mapping theorem. (This is, I think, a much more natural proof of Liouville’s theorem than the standard one.)

May 26, 2009 at 4:49 pm

I was trying to avoid the word “scheme,” since I haven’t defined it yet. I also haven’t talked about proper-ness, although that wouldn’t take long. Not that this blog is self-contained, but I’d like to avoid major assumptions of knowledge that aren’t really considered “standard”.