The Stacks project

Lemma 45.10.2. Assume given (D0), (D1), (D2), and (D3) satisfying (A), (B), and (C). Let $X$ be a smooth projective scheme of dimension zero over $k$. Then

  1. $H^ i(X) = 0$ for $i \not= 0$,

  2. $H^0(X)$ is a finite separable algebra over $F$,

  3. $\dim _ F H^0(X) = \deg (X \to \mathop{\mathrm{Spec}}(F))$,

  4. $\int _ X : H^0(X) \to F$ is the trace map,

  5. $\gamma ([X]) = 1$, and

  6. $\int _ X \gamma ([X]) = \deg (X \to \mathop{\mathrm{Spec}}(k))$.

Proof. We can write $X = \mathop{\mathrm{Spec}}(k')$ where $k'$ is a finite separable algebra over $k$. Observe that $\deg (X \to \mathop{\mathrm{Spec}}(k)) = [k' : k]$. Choose a finite Galois extension $k''/k$ containing each of the factors of $k'$. (Recall that a finite separable $k$-algebra is a product of finite separable field extension of $k$.) Set $\Sigma = \mathop{\mathrm{Hom}}\nolimits _ k(k', k'')$. Then we get

\[ k' \otimes _ k k'' = \prod \nolimits _{\sigma \in \Sigma } k'' \]

Setting $Y = \mathop{\mathrm{Spec}}(k'')$ axioms (B)(a) and Lemma 45.9.9 give

\[ H^*(X) \otimes _ F H^*(Y) = \prod \nolimits _{\sigma \in \Sigma } H^*(Y) \]

as graded commutative $F$-algebras. By Lemma 45.9.5 the $F$-algebra $H^*(Y)$ is nonzero. Comparing dimensions on either side of the displayed equation we conclude that $H^*(X)$ sits only in degree $0$ and $\dim _ F H^0(X) = [k' : k]$. Applying this to $Y$ we get $H^*(Y) = H^0(Y)$. Since

\[ H^0(X) \otimes _ F H^0(Y) = H^0(Y) \times \ldots \times H^0(Y) \]

as $F$-algebras, it follows that $H^0(X)$ is a separable $F$-algebra because we may check this after the faithfully flat base change $F \to H^0(Y)$.

The displayed isomorphism above is given by the map

\[ H^0(X) \otimes _ F H^0(Y) \longrightarrow \prod \nolimits _{\sigma \in \Sigma } H^0(Y),\quad a \otimes b \longmapsto \prod \nolimits _\sigma \mathop{\mathrm{Spec}}(\sigma )^*a \cup b \]

Via this isomorphism we have $\int _{X \times Y} = \sum _\sigma \int _ Y$ by Lemma 45.10.1. Thus

\[ \int _ X a = \text{pr}_{1, *}(a \otimes 1) = \sum \mathop{\mathrm{Spec}}(\sigma )^*a \]

in $H^0(Y)$; the first equality by Lemma 45.9.2 and the second by the observation we just made. Choose an algebraic closure $\overline{F}$ and a $F$-algebra map $\tau : H^0(Y) \to \overline{F}$. The isomorphism above base changes to the isomorphism

\[ H^0(X) \otimes _ F \overline{F} \longrightarrow \prod \nolimits _{\sigma \in \Sigma } \overline{F},\quad a \otimes b \longmapsto \prod \nolimits _\sigma \tau (\mathop{\mathrm{Spec}}(\sigma )^*a) b \]

It follows that $a \mapsto \tau (\mathop{\mathrm{Spec}}(\sigma )^*a)$ is a full set of embeddings of $H^0(X)$ into $\overline{F}$. Applying $\tau $ to the formula for $\int _ X a$ obtained above we conclude that $\int _ X$ is the trace map. By Lemma 45.9.5 we have $\gamma ([X]) = 1$. Finally, we have $\int _ X \gamma ([X]) = \deg (X \to \mathop{\mathrm{Spec}}(k))$ because $\gamma ([X]) = 1$ and the trace of $1$ is equal to $[k' : k]$ $\square$

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 0FHQ. Beware of the difference between the letter 'O' and the digit '0'.