Remark 45.11.5. Let $H^*$ be a Weil cohomology theory (Definition 45.11.4). Let $X$ be a geometrically irreducible smooth projective scheme of dimension $d$ over $k'$ with $k'/k$ a finite separable extension of fields. Suppose that

$H^0(\mathop{\mathrm{Spec}}(k')) = F_1 \times \ldots \times F_ r$

for some fields $F_ i$. Then we accordingly can write

$H^*(X) = \prod \nolimits _{i = 1, \ldots , r} H^*(X) \otimes _{H^0(\mathop{\mathrm{Spec}}(k'))} F_ i$

Now, our final assumption in Definition 45.11.4 tells us that $H^0(X)$ is free of rank $1$ over $\prod F_ i$. In other words, each of the factors $H^0(X) \otimes _{H^0(\mathop{\mathrm{Spec}}(k'))} F_ i$ has dimension $1$ over $F_ i$. PoincarĂ© duality then tells us that the same is true for cohomology in degree $2d$. What isn't clear however is that the same holds in other degrees. Namely, we don't know that given $0 < n < \dim (X)$ the integers

$\dim _{F_ i} H^ n(X) \otimes _{H^0(\mathop{\mathrm{Spec}}(k'))} F_ i$

are independent of $i$! This question is closely related to the following open question: given an algebraically closed base field $\overline{k}$, a field of characteristic zero $F$, a classical Weil cohomology theory $H^*$ over $\overline{k}$ with coefficient field $F$, and a smooth projective variety $X$ over $\overline{k}$ is it true that the betti numbers of $X$

$\beta _ i = \dim _ F H^ i(X)$

are independent of $F$ and the Weil cohomology theory $H^*$?

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