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?

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