Our goal for today is to prove the following theorem:

**Theorem 1:** Let be a projective algebraic curve of genus and an endomorphism of degree . Let be a reasonable cohomology theory. Then the action of on has eigenvalues which are algebraic integers, with norm .

For those who know the term, “reasonable cohomology theory” means “Weil cohomology theory”.

The consequence of this theorem, which does not mention cohomology, is

**Theorem 2:** Let be a projective algebraic curve of genus and an endomorphism of degree . Then there are algebraic integers , , …, with norm such that

.

In a previous post, we established this in characteristic zero, by putting a positive definite hermitian structure on such that became unitary. But, as I discussed last time, we can’t define when has characteristic . Instead, will be defined over some other field of characteristic zero, like . We will therefore need to know that the eigenvalues of are algebraic integers before we can even make sense of the statement that they have norm .

It is possible to take the proof I present here and strip it down to its bare essentials, to give a proof of Theorem 2 which doesn’t even mention cohomology. See Hartshorne Exercise V.1.10. I am going to do the opposite; I will go slowly and focus on what each step is proving about . The essential argument here is Weil’s, although I have modernized the presentation.