## 45.10 Further properties

In this section we prove a few more results one obtains if given data (D0), (D1), (D2), and (D3) satisfying (A), (B), and (C) as in Section 45.9.

Lemma 45.10.1. Assume given (D0), (D1), (D2), and (D3) satisfying (A), (B), and (C). Let $X, Y$ be nonempty smooth projective schemes both equidimensional of dimension $d$ over $k$. Then $\int _{X \amalg Y} = \int _ X + \int _ Y$.

Proof. Denote $i : X \to X \amalg Y$ and $j : Y \to X \amalg Y$ be the coprojections. By Lemma 45.9.9 the map $(i^*, j^*) : H^*(X \amalg Y) \to H^*(X) \times H^*(Y)$ is an isomorphism. The statement of the lemma means that under the isomorphism $(i^*, j^*) : H^{2d}(X \amalg Y)(d) \to H^{2d}(X)(d) \oplus H^{2d}(Y)(d)$ the map $\int _ X + \int _ Y$ is tranformed into $\int _{X \amalg Y}$. This is true because

$\int _{X \amalg Y} a = \int _{X \amalg Y} i_*(i^*a) + j_*(j^*a) = \int _ X i^*a + \int _ Y j^*a$

where the equality $a = i_*(i^*a) + j_*(j^*a)$ was shown in the proof of Lemma 45.9.9. $\square$

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

Lemma 45.10.3. Assume given (D0), (D1), (D2), and (D3) satisfying (A), (B), and (C). Let $X$ be a nonempty smooth projective scheme equidimensional of dimension $d$ over $k$. The diagram

$\xymatrix{ \mathop{\mathrm{CH}}\nolimits ^ d(X) \ar[r]_-\gamma \ar@{=}[d] & H^{2d}(X)(d) \ar[d]^{\int _ X} \\ \mathop{\mathrm{CH}}\nolimits _0(X) \ar[r]^\deg & F }$

commutes where $\deg : \mathop{\mathrm{CH}}\nolimits _0(X) \to \mathbf{Z}$ is the degree of zero cycles discussed in Chow Homology, Section 42.40.

Proof. Let $x$ be a closed point of $X$ whose residue field is separable over $k$. View $x$ as a scheme and denote $i : x \to X$ the inclusion morphism. To avoid confusion denote $\gamma ' : \mathop{\mathrm{CH}}\nolimits _0(x) \to H^0(x)$ the cycle class map for $x$. Then we have

$\int _ X \gamma ([x]) = \int _ X \gamma (i_*[x]) = \int _ X i_*\gamma '([x]) = \int _ x \gamma '([x]) = \deg (x \to \mathop{\mathrm{Spec}}(k))$

The second equality is axiom (C)(b) and the third equality is the definition of $i_*$ on cohomology. The final equality is Lemma 45.10.2. This proves the lemma because $\mathop{\mathrm{CH}}\nolimits _0(X)$ is generated by the classes of points $x$ as above by Lemma 45.8.1. $\square$

Lemma 45.10.4. Assume given (D0), (D1), (D2), and (D3) satisfying (A), (B), and (C). Let $X$ be a nonempty smooth projective scheme over $k$ which is equidimensional of dimension $d$. We have

$\sum \nolimits _ i (-1)^ i\dim _ F H^ i(X) = \deg (\Delta \cdot \Delta ) = \deg (c_ d(\mathcal{T}_{X/k}))$

Proof. Equality on the right. We have $[\Delta ] \cdot [\Delta ] = \Delta _*(\Delta ^![\Delta ])$ (Chow Homology, Lemma 42.61.6). Since $\Delta _*$ preserves degrees of $0$-cycles it suffices to compute the degree of $\Delta ^![\Delta ]$. The class $\Delta ^![\Delta ]$ is given by capping $[\Delta ]$ with the top Chern class of the normal sheaf of $\Delta \subset X \times X$ (Chow Homology, Lemma 42.53.5). Since the conormal sheaf of $\Delta$ is $\Omega _{X/k}$ (Morphisms, Lemma 29.32.7) we see that the normal sheaf is equal to the tangent sheaf $\mathcal{T}_{X/k} = \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\Omega _{X/k}, \mathcal{O}_ X)$ as desired.

Equality on the left. By Lemma 45.10.3 we have

\begin{align*} \deg ([\Delta ] \cdot [\Delta ]) & = \int _{X \times X} \gamma ([\Delta ]) \cup \gamma ([\Delta ]) \\ & = \int _{X \times X} \Delta _*1 \cup \gamma ([\Delta ]) \\ & = \int _{X \times X} \Delta _*(\Delta ^*\gamma ([\Delta ])) \\ & = \int _ X \Delta ^*\gamma ([\Delta ]) \end{align*}

We have used Lemmas 45.9.6 and 45.9.1. Write $\gamma ([\Delta ]) = \sum e_{i, j} \otimes e'_{2d - i , j}$ as in Lemma 45.9.7. Recalling that $\Delta ^*$ is given by cup product (Remark 45.9.3) we obtain

$\int _ X \sum \nolimits _{i, j} e_{i, j} \cup e'_{2d - i, j} = \sum \nolimits _{i, j} \int _ X e_{i, j} \cup e'_{2d - i, j} = \sum \nolimits _{i, j} (-1)^ i = \sum (-1)^ i\beta _ i$

as desired. $\square$

Lemma 45.10.5. Let $F$ be a field of characteristic $0$. Let $F'$ and $F_ i$, $i = 1, \ldots , r$ be finite separable $F$-algebras. Let $A$ be a finite $F$-algebra. Let $\sigma , \sigma ' : A \to F'$ and $\sigma _ i : A \to F_ i$ be $F$-algebra maps. Assume $\sigma$ and $\sigma '$ surjective. If there is a relation

$\text{Tr}_{F'/F} \circ \sigma - \text{Tr}_{F'/F} \circ \sigma ' = n(\sum m_ i \text{Tr}_{F_ i/F} \circ \sigma _ i)$

where $n > 1$ and $m_ i$ are integers, then $\sigma = \sigma '$.

Proof. We may write $A = \prod A_ j$ as a finite product of local Artinian $F$-algebras $(A_ j, \mathfrak m_ j, \kappa _ j)$, see Algebra, Lemma 10.53.2 and Proposition 10.60.6. Denote $A' = \prod \kappa _ j$ where the product is over those $j$ such that $\kappa _ j/k$ is separable. Then each of the maps $\sigma , \sigma ', \sigma _ i$ factors over the map $A \to A'$. After replacing $A$ by $A'$ we may assume $A$ is a finite separable $F$-algebra.

Choose an algebraic closure $\overline{F}$. Set $\overline{A} = A \otimes _ F \overline{F}$, $\overline{F}' = F' \otimes _ F \overline{F}$, and $\overline{F}_ i = F_ i \otimes _ F \overline{F}$. We can base change $\sigma$, $\sigma '$, $\sigma _ i$ to get $\overline{F}$ algebra maps $\overline{A} \to \overline{F}'$ and $\overline{A} \to \overline{F}_ i$. Moreover $\text{Tr}_{\overline{F}'/\overline{F}}$ is the base change of $\text{Tr}_{F'/F}$ and similarly for $\text{Tr}_{F_ i/F}$. Thus we may replace $F$ by $\overline{F}$ and we reduce to the case discussed in the next paragraph.

Assume $F$ is algebraically closed and $A$ a finite separable $F$-algebra. Then each of $A$, $F'$, $F_ i$ is a product of copies of $F$. Let us say an element $e$ of a product $F \times \ldots \times F$ of copies of $F$ is a minimal idempotent if it generates one of the factors, i.e., if $e = (0, \ldots , 0, 1, 0, \ldots , 0)$. Let $e \in A$ be a minimal idempotent. Since $\sigma$ and $\sigma '$ are surjective, we see that $\sigma (e)$ and $\sigma '(e)$ are minimal idempotents or zero. If $\sigma \not= \sigma '$, then we can choose a minimal idempotent $e \in A$ such that $\sigma (e) = 0$ and $\sigma '(e) \not= 0$ or vice versa. Then $\text{Tr}_{F'/F}(\sigma (e)) = 0$ and $\text{Tr}_{F'/F}(\sigma '(e)) = 1$ or vice versa. On the other hand, $\sigma _ i(e)$ is an idempotent and hence $\text{Tr}_{F_ i/F}(\sigma _ i(e)) = r_ i$ is an integer. We conclude that

$-1 = \sum n m_ i r_ i = n (\sum m_ i r_ i) \quad \text{or}\quad 1 = \sum n m_ i r_ i = n (\sum m_ i r_ i)$

which is impossible. $\square$

Lemma 45.10.6. Assume given (D0), (D1), (D2), and (D3) satisfying (A), (B), and (C). Let $k'/k$ be a finite separable extension. Let $X$ be a smooth projective scheme over $k'$. Let $x, x' \in X$ be $k'$-rational points. If $\gamma (x) \not= \gamma (x')$, then $[x] - [x']$ is not divisible by any integer $n > 1$ in $\mathop{\mathrm{CH}}\nolimits _0(X)$.

Proof. If $x$ and $x'$ lie on distinct irreducible components of $X$, then the result is obvious. Thus we may $X$ irreducible of dimension $d$. Say $[x] - [x']$ is divisible by $n > 1$ in $\mathop{\mathrm{CH}}\nolimits _0(X)$. We may write $[x] - [x'] = n(\sum m_ i [x_ i])$ in $\mathop{\mathrm{CH}}\nolimits _0(X)$ for some $x_ i \in X$ closed points whose residue fields are separable over $k$ by Lemma 45.8.1. Then

$\gamma ([x]) - \gamma ([x']) = n (\sum m_ i \gamma ([x_ i]))$

in $H^{2d}(X)(d)$. Denote $i^*, (i')^*, i_ i^*$ the pullback maps $H^0(X) \to H^0(x)$, $H^0(X) \to H^0(x')$, $H^0(X) \to H^0(x_ i)$. Recall that $H^0(x)$ is a finite separable $F$-algebra and that $\int _ x : H^0(x) \to F$ is the trace map (Lemma 45.10.2) which we will denote $\text{Tr}_ x$. Similarly for $x'$ and $x_ i$. Then by Poincaré duality in the form of axiom (A)(b) the equation above is dual to

$\text{Tr}_ x \circ i^* - \text{Tr}_{x'} \circ (i')^* = n(\sum m_ i \text{Tr}_{x_ i} \circ i_ i^*)$

which takes place in $\mathop{\mathrm{Hom}}\nolimits _ F(H^0(X), F)$. Finally, observe that $i^*$ and $(i')^*$ are surjective as $x$ and $x'$ are $k'$-rational points and hence the compositions $H^0(\mathop{\mathrm{Spec}}(k')) \to H^0(X) \to H^0(x)$ and $H^0(\mathop{\mathrm{Spec}}(k')) \to H^0(X) \to H^0(x')$ are isomorphisms. By Lemma 45.10.5 we conclude that $i^* = (i')^*$ which contradicts the assumption that $\gamma ([x]) \not= \gamma ([x'])$. $\square$

Lemma 45.10.7. Assume given (D0), (D1), (D2), and (D3) satisfying (A), (B), and (C). Let $k'/k$ be a finite separable extension. Let $X$ be a geometrically irreducible smooth projective scheme over $k'$ of dimension $d$. Then $\gamma : \mathop{\mathrm{CH}}\nolimits _0(X) \to H^{2d}(X)(d)$ factors through $\deg : \mathop{\mathrm{CH}}\nolimits _0(X) \to \mathbf{Z}$.

Proof. By Lemma 45.8.1 it suffices to show: given closed points $x, x' \in X$ whose residue fields are separable over $k$ we have $\deg (x') \gamma ([x]) = \deg (x) \gamma ([x'])$.

We first reduce to the case of $k'$-rational points. Let $k''/k'$ be a Galois extension such that $\kappa (x)$ and $\kappa (x')$ embed into $k''$ over $k$. Set $Y = X \times _{\mathop{\mathrm{Spec}}(k')} \mathop{\mathrm{Spec}}(k'')$ and denote $p : Y \to X$ the projection. By our choice of $k''/k'$ there exists a $k''$-rational point $y$, resp. $y'$ on $Y$ mapping to $x$, resp. $x'$. Then $p_*[y] = [k'' : \kappa (x)][x]$ and $p_*[y'] = [k'' : \kappa (x')][x']$ in $\mathop{\mathrm{CH}}\nolimits _0(X)$. By compatibility with pushforwards given in axiom (C)(b) it suffices to prove $\gamma ([y]) = \gamma ([y'])$ in $\mathop{\mathrm{CH}}\nolimits ^{2d}(Y)(d)$. This reduces us to the discussion in the next paragraph.

Assume $x$ and $x'$ are $k'$-rational points. By Lemma 45.8.3 there exists a finite separable extension $k''/k'$ of fields such that the pullback $[y] - [y']$ of the difference $[x] - [x']$ becomes divisible by an integer $n > 1$ on $Y = X \times _{\mathop{\mathrm{Spec}}(k')} \mathop{\mathrm{Spec}}(k'')$. (Note that $y, y' \in Y$ are $k''$-rational points.) By Lemma 45.10.6 we have $\gamma ([y]) = \gamma ([y'])$ in $H^{2d}(Y)(d)$. By compatibility with pushforward in axiom (C)(b) we conclude the same for $x$ and $x'$. $\square$

Lemma 45.10.8. Assume given (D0), (D1), (D2), and (D3) satisfying (A), (B), and (C). Let $f : X \to Y$ be a dominant morphism of irreducible smooth projective schemes over $k$. Then $H^*(Y) \to H^*(X)$ is injective.

Proof. There exists an integral closed subscheme $Z \subset X$ of the same dimension as $Y$ mapping onto $Y$. Thus $f_*[Z] = m[Y]$ for some $m > 0$. Then $f_* \gamma ([Z]) = m \gamma ([Y]) = m$ in $H^*(Y)$ because of Lemma 45.9.5. Hence by the projection formula (Lemma 45.9.1) we have $f_*(f^*a \cup \gamma ([Z])) = m a$ and we conclude. $\square$

Lemma 45.10.9. Assume given (D0), (D1), (D2), and (D3) satisfying (A), (B), and (C). Let $k''/k'/k$ be finite separable algebras and let $X$ be a smooth projective scheme over $k'$. Then

$H^*(X) \otimes _{H^0(\mathop{\mathrm{Spec}}(k'))} H^0(\mathop{\mathrm{Spec}}(k'')) = H^*(X \times _{\mathop{\mathrm{Spec}}(k')} \mathop{\mathrm{Spec}}(k''))$

Proof. We will use the results of Lemma 45.10.2 without further mention. Write

$k' \otimes _ k k'' = k'' \times l$

for some finite separable $k'$-algebra $l$. Write $F' = H^0(\mathop{\mathrm{Spec}}(k'))$, $F'' = H^0(\mathop{\mathrm{Spec}}(k''))$, and $G = H^0(\mathop{\mathrm{Spec}}(l))$. Since $\mathop{\mathrm{Spec}}(k') \times \mathop{\mathrm{Spec}}(k'') = \mathop{\mathrm{Spec}}(k'') \amalg \mathop{\mathrm{Spec}}(l)$ we deduce from axiom (B)(a) and Lemma 45.9.9 that we have

$F' \otimes _ F F'' = F'' \times G$

The map from left to right identifies $F''$ with $F' \otimes _{F'} F''$. By the same token we have

$H^*(X) \otimes _ F F'' = H^*(X \times _{\mathop{\mathrm{Spec}}(k')} \mathop{\mathrm{Spec}}(k'')) \times H^*(X \times _{\mathop{\mathrm{Spec}}(k')} \mathop{\mathrm{Spec}}(l))$

as modules over $F' \otimes _ F F'' = F'' \times G$. This proves the lemma. $\square$

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