Zeta function of $X = \mathbb{F}_p \mathbb{P}^1$

Question

I'm trying to produce a toy version of the RH Weil conjecture. Solving this could help me to get a good start at understanding where the $1/2$'s come in here, ideally without having to prove the Hard Lefschetz theorem.

I'm restricting attention to the scheme $X=\mathbb{F}_p \mathbb{P}^1$. Write $x = \text{Spec}(\mathbb{F}_p)$, $y = \text{Spec}(\mathbb{F}_p^{closure})$, and $Y = X \times_{x} y$. My goal is to show the RH Weil conjecture for $Y$, but I suspect that the fact that $Y$ is $1$ dimensional projective space simplifies things a great deal. Meanwhile, I've worked through how to show that

1. $l$-adic cohomology has a poincare duality for $k \mathbb{P}^1$.

2. $l$-adic cohomology has a Kunneth formula and proper base change theorem.

3. There is a Lefschetz fixed point theorem for $X$ and $Y$.

My hope here is that there is to get a proof that

Let $F$ be the Frobenius endomorphism of $Y$. Then the eigenvalues of $F$ acting on $H^r(Y, \mathbb{Q}_l)$ have absolute value $q^{r/2}$.

I have also shown that the inner product $H^*(Y, \mathbb{Q}_l) \otimes H^*(Y, \mathbb{Q}_l) \rightarrow \mathbb{Q}_l(2)$ is fixed by the Frobenius map.

I'm hoping that since $X$ and $Y$ here are pretty simple spaces, that I don't need the Hard Lefschetz theorem. Could this be possible?

TEST_1

Answer 1

Deligne proved the hard Lefschetz theorem as a consequence of the Weil conjectures, not vice versa.

The RH of the Weil conjectures is a statement about the Weil zeta function. For the projective line, this is easy to prove without étale cohomology at all, only by counting points.

But of course you could ask for a proof that the eigenvalues of Frobenius on $H^i$ have size $q^{i/2}$ (Deligne's RH) in this special case. This is indeed much easier than the general case!

Since $Y$ is connected, it's straightforward to check from the definitions that $H^0 (Y, \mathbb Q_\ell) =\mathbb Q_\ell$, with trivial Frobenius action. So this verifies the $i=1$ case since $q^{0/2}=1$.

Next one checks that $H^1(Y,\mathbb Q_\ell)=0$, which vacuously verifies the $i=1$ case. Probably the easiest way to do this is to use the relation between the first étale cohomology and the étale fundamental group, and show the étale fundamental group is trivial by Riemann-Hurwitz.

Finally, using the Poincaré duality $H^0(Y,\mathbb Q_\ell) \times H^2(Y, \mathbb Q_\ell) \to \mathbb Q_\ell(-1)$ (not $\mathbb Q_\ell(2)$), one sees that $H^2(Y,\mathbb Q_\ell) \cong \mathbb Q_\ell(-1)$ and thus the unique Frobenius eigenvalue on it is $q$, verifying the $i=2$ cases since $q^{2/2}=q$.

Whether that explains where the $1/2$ comes in sufficiently well is up to you! If not, I would look at the case of abelian varieties, where $H^1$ is nontrivial, and the RH can be proved by the theory of endomorphisms of abelian varieties.