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 $[k' : k]$ $\square$

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).