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 $\int _{X \amalg Y} = \int _ X + \int _ Y$.

## 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.

**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 transformed into $\int _{X \amalg Y}$. This is true because

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

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

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

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

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

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

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

**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

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

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

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

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

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

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

commutes where $\deg : \mathop{\mathrm{CH}}\nolimits _0(X) \to \mathbf{Z}$ is the degree of zero cycles discussed in Chow Homology, Section 42.41.

**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

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

**Proof.**
Equality on the right. We have $[\Delta ] \cdot [\Delta ] = \Delta _*(\Delta ^![\Delta ])$ (Chow Homology, Lemma 42.62.6). 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 42.54.5). 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

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

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

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 10.60.7. 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

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

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 PoincarĂ© duality in the form of axiom (A)(b) the equation above is dual to

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

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

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

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

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

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