The Stacks project

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 $F_ i$. Poincaré 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 $H^*$?


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0FI3. Beware of the difference between the letter 'O' and the digit '0'.