# Minimal vs characteristic polynomial of geometric Frobenius

Assume $$X$$ is a smooth projective variety over $$\overline{\mathbf{F}}_p$$ and fix a prime $$\ell\neq p$$.

Let $$F_i$$ be the geometric Frobenius on $$\ell$$-adic cohomology

$$H^i_{\rm ét}(X,\overline{\mathbf{Q}}_{\ell})$$ for fixed $$i\ge 0$$.

• What relation is expected to hold between the minimal polynomial $$m_{F_i}(T)$$ of $$F_i$$ and the characteristic polynomial $$P_i(T)$$ of $$F_i$$? (apart from $$m_{F_i}\mid P_i$$)
• Are $$m_{F_i}(T)$$ and $$P_i(T)$$ conjectured to agree? If so, is this a known result?
• Does $$m_{F_i}(T)$$ depend on the Weil cohomology theory chosen to compute $$P_i(T)$$?

We know from Deligne's work on the Weil conjectures, that we have $$P_i(T)\in\mathbf{Z}[T]$$, $$P_i(T)$$ does not depend on the chosen Weil cohomology, and its roots are of the form $$q^{-i/2}\rho$$ for $$\rho$$ an algebraic number whose complex absolute value, for any complex embedding $$\overline{\mathbf{Q}}\subset\mathbf{C}$$, is one.

I'm mostly interested to understand to what extent $$m_{F_i}(T)$$ is, or expected to be, intrinsic.

It's conjectured -- see e.g. this question -- that the Frobenius is always semisimple, so its minimal polynomial is the radical of its characteristic polynomial (the product of its distinct linear factors, each with multiplicity 1). So $$m_{F_i}$$ should be independent of $$\ell$$.
This also shows that $$m_{F_i}$$ is different from $$P_i$$ iff $$P_i$$ has a root of multiplicity $$> 1$$. This can certainly occur, e.g. consider a supersingular elliptic curve over $$\mathbf{F}_{p^2}$$.
• Thank you! In particular, the "semisimplicity conjecture" depends on $\ell$. Am I understanding this correctly? If one proves Frobenius is semisimple on $H^i(X,\overline{\mathbf{Q}}_{\ell})$, then $m_{F_i} = \text{rad}(P_i)$, the radical, so I guess for every $\ell\neq \ell',p$, this $m_{F_i}$ will divide the minimal polynomial of Frobenius on $H^i(X,\overline{\mathbf{Q}}_{\ell'})$, but nothing more can be said. Or is it somehow known by some nontrivial argument that the semisimplicity conjecture is "independent of $\ell$"? Jul 10 at 11:46
• When I wrote "should be", I meant that if semisimplicity holds for $\ell$-adic cohomology for all $\ell$ (as is conjectured), then it implies $m_{F_i}$ is independent of $\ell$. But I don't think this is known unconditionally; as far as I'm aware, semisimplicity for one $\ell$ doesn't imply it for other $\ell$. Jul 10 at 12:38