59.69 Picard groups of curves
Our next step is to use the Kummer sequence to deduce some information about the cohomology group of a curve with finite coefficients. In order to get vanishing in the long exact sequence, we review some facts about Picard groups.
Let $X$ be a smooth projective curve over an algebraically closed field $k$. Let $g = \dim _ k H^1(X, \mathcal{O}_ X)$ be the genus of $X$. There exists a short exact sequence
\[ 0 \to \mathop{\mathrm{Pic}}\nolimits ^0(X) \to \mathop{\mathrm{Pic}}\nolimits (X) \xrightarrow {\deg } \mathbf{Z} \to 0. \]
The abelian group $\mathop{\mathrm{Pic}}\nolimits ^0(X)$ can be identified with $\mathop{\mathrm{Pic}}\nolimits ^0(X) = \underline{\mathrm{Pic}}^0_{X/k}(k)$, i.e., the $k$-valued points of an abelian variety $\underline{\mathrm{Pic}}^0_{X/k}$ over $k$ of dimension $g$. Consequently, if $n \in k^*$ then $\mathop{\mathrm{Pic}}\nolimits ^0(X)[n] \cong (\mathbf{Z}/n\mathbf{Z})^{2g}$ as abelian groups. See Picard Schemes of Curves, Section 44.6 and Groupoids, Section 39.9. This key fact, namely the description of the torsion in the Picard group of a smooth projective curve over an algebraically closed field does not appear to have an elementary proof.
Lemma 59.69.1. Let $X$ be a smooth projective curve of genus $g$ over an algebraically closed field $k$ and let $n \geq 1$ be invertible in $k$. Then there are canonical identifications
\[ H_{\acute{e}tale}^ q(X, \mu _ n) = \left\{ \begin{matrix} \mu _ n(k)
& \text{ if }q = 0,
\\ \mathop{\mathrm{Pic}}\nolimits ^0(X)[n]
& \text{ if }q = 1,
\\ \mathbf{Z}/n\mathbf{Z}
& \text{ if }q = 2,
\\ 0
& \text{ if }q \geq 3.
\end{matrix} \right. \]
Since $\mu _ n \cong \underline{\mathbf{Z}/n\mathbf{Z}}$, this gives (noncanonical) identifications
\[ H_{\acute{e}tale}^ q(X, \underline{\mathbf{Z}/n\mathbf{Z}}) \cong \left\{ \begin{matrix} \mathbf{Z}/n\mathbf{Z}
& \text{ if }q = 0,
\\ (\mathbf{Z}/n\mathbf{Z})^{2g}
& \text{ if }q = 1,
\\ \mathbf{Z}/n\mathbf{Z}
& \text{ if }q = 2,
\\ 0
& \text{ if }q \geq 3.
\end{matrix} \right. \]
Proof.
Theorems 59.24.1 and 59.68.5 determine the étale cohomology of $\mathbf{G}_ m$ on $X$ in terms of the Picard group of $X$. The Kummer sequence $0\to \mu _{n, X} \to \mathbf{G}_{m, X} \to \mathbf{G}_{m, X}\to 0$ (Lemma 59.28.1) then gives us the long exact cohomology sequence
\[ \xymatrix{ 0 \ar[r] & \mu _ n(k) \ar[r] & k^* \ar[r]^{(\cdot )^ n} & k^* \ar@(rd, ul)[rdllllr] \\ & H_{\acute{e}tale}^1(X, \mu _ n) \ar[r] & \mathop{\mathrm{Pic}}\nolimits (X) \ar[r]^{(\cdot )^ n} & \mathop{\mathrm{Pic}}\nolimits (X) \ar@(rd, ul)[rdllllr] \\ & H_{\acute{e}tale}^2(X, \mu _ n) \ar[r] & 0 \ar[r] & 0 \ldots } \]
The $n$th power map $k^* \to k^*$ is surjective since $k$ is algebraically closed. So we need to compute the kernel and cokernel of the map $n : \mathop{\mathrm{Pic}}\nolimits (X) \to \mathop{\mathrm{Pic}}\nolimits (X)$. Consider the commutative diagram with exact rows
\[ \xymatrix{ 0 \ar[r] & \mathop{\mathrm{Pic}}\nolimits ^0(X) \ar[r] \ar@{>>}[d]^{(\cdot )^ n} & \mathop{\mathrm{Pic}}\nolimits (X) \ar[r]_-\deg \ar[d]^{(\cdot )^ n} & \mathbf{Z} \ar[r] \ar@{^{(}->}[d]^ n & 0 \\ 0 \ar[r] & \mathop{\mathrm{Pic}}\nolimits ^0(X) \ar[r] & \mathop{\mathrm{Pic}}\nolimits (X) \ar[r]^-\deg & \mathbf{Z} \ar[r] & 0 } \]
The group $\mathop{\mathrm{Pic}}\nolimits ^0(X)$ is the $k$-points of the group scheme $\underline{\mathrm{Pic}}^0_{X/k}$, see Picard Schemes of Curves, Lemma 44.6.7. The same lemma tells us that $\underline{\mathrm{Pic}}^0_{X/k}$ is a $g$-dimensional abelian variety over $k$ as defined in Groupoids, Definition 39.9.1. Hence the left vertical map is surjective by Groupoids, Proposition 39.9.11. Applying the snake lemma gives canonical identifications as stated in the lemma.
To get the noncanonical identifications of the lemma we need to show the kernel of $n : \mathop{\mathrm{Pic}}\nolimits ^0(X) \to \mathop{\mathrm{Pic}}\nolimits ^0(X)$ is isomorphic to $(\mathbf{Z}/n\mathbf{Z})^{\oplus 2g}$. This is also part of Groupoids, Proposition 39.9.11.
$\square$
Lemma 59.69.2. Let $\pi : X \to Y$ be a nonconstant morphism of smooth projective curves over an algebraically closed field $k$ and let $n \geq 1$ be invertible in $k$. The map
\[ \pi ^* : H^2_{\acute{e}tale}(Y, \mu _ n) \longrightarrow H^2_{\acute{e}tale}(X, \mu _ n) \]
is given by multiplication by the degree of $\pi $.
Proof.
Observe that the statement makes sense as we have identified both cohomology groups $ H^2_{\acute{e}tale}(Y, \mu _ n)$ and $H^2_{\acute{e}tale}(X, \mu _ n)$ with $\mathbf{Z}/n\mathbf{Z}$ in Lemma 59.69.1. In fact, if $\mathcal{L}$ is a line bundle of degree $1$ on $Y$ with class $[\mathcal{L}] \in H^1_{\acute{e}tale}(Y, \mathbf{G}_ m)$, then the coboundary of $[\mathcal{L}]$ is the generator of $H^2_{\acute{e}tale}(Y, \mu _ n)$. Here the coboundary is the coboundary of the long exact sequence of cohomology associated to the Kummer sequence. Thus the result of the lemma follows from the fact that the degree of the line bundle $\pi ^*\mathcal{L}$ on $X$ is $\deg (\pi )$. Some details omitted.
$\square$
Lemma 59.69.3. Let $X$ be an affine smooth curve over an algebraically closed field $k$ and $n \in k^*$. Let $X \subset \overline{X}$ be a smooth projective compactification (Varieties, Remark 33.43.9). Let $g$ be the genus of $\overline{X}$ and let $r$ be the number of points of $\overline{X} \setminus X$. Then
$H_{\acute{e}tale}^0(X, \mu _ n) = \mu _ n(k)$;
$H_{\acute{e}tale}^1(X, \mu _ n) \cong (\mathbf{Z}/n\mathbf{Z})^{2g+r-1}$, and
$H_{\acute{e}tale}^ q(X, \mu _ n) = 0$ for all $q \geq 2$.
Proof.
Write $X = \overline{X} - \{ x_1, \ldots , x_ r\} $. Then $\mathop{\mathrm{Pic}}\nolimits (X) = \mathop{\mathrm{Pic}}\nolimits (\overline{X})/ R$, where $R$ is the subgroup generated by $\mathcal{O}_{\overline{X}}(x_ i)$, $1 \leq i \leq r$. Since $r \geq 1$, we see that $\mathop{\mathrm{Pic}}\nolimits ^0(\overline{X}) \to \mathop{\mathrm{Pic}}\nolimits (X)$ is surjective, hence $\mathop{\mathrm{Pic}}\nolimits (X)$ is divisible (see discussion in proof of Lemma 59.69.1). Applying the Kummer sequence, we get (1) and (3). For (2), recall that
\begin{align*} H_{\acute{e}tale}^1(X, \mu _ n) & = \{ (\mathcal L, \alpha ) | \mathcal L \in \mathop{\mathrm{Pic}}\nolimits (X), \alpha : \mathcal{L}^{\otimes n} \to \mathcal{O}_ X\} /\cong \\ & = \{ (\bar{\mathcal L},\ D,\ \bar\alpha )\} /\tilde{R} \end{align*}
where $\bar{\mathcal L} \in \mathop{\mathrm{Pic}}\nolimits ^0(\overline{X})$, $D$ is a divisor on $\overline{X}$ supported on $\left\{ x_1, \ldots , x_ r\right\} $ and $ \bar{\alpha }: \bar{\mathcal L}^{\otimes n} \cong \mathcal{O}_{\bar{X}}(D)$ is an isomorphism. Note that $D$ must have degree 0. Further $\tilde{R}$ is the subgroup of triples of the form $(\mathcal{O}_{\overline{X}}(D'), n D', 1^{\otimes n})$ where $D'$ is supported on $\left\{ x_1, \ldots , x_ r\right\} $ and has degree 0. Thus, we get an exact sequence
\[ 0 \longrightarrow H_{\acute{e}tale}^1(\overline{X}, \mu _ n) \longrightarrow H_{\acute{e}tale}^1(X, \mu _ n) \longrightarrow \bigoplus _{i = 1}^ r \mathbf{Z}/n\mathbf{Z} \xrightarrow {\ \sum \ } \mathbf{Z}/n\mathbf{Z} \longrightarrow 0 \]
where the middle map sends the class of a triple $(\bar{ \mathcal L}, D, \bar\alpha )$ with $D = \sum _{i = 1}^ r a_ i (x_ i)$ to the $r$-tuple $(a_ i)_{i = 1}^ r$. It now suffices to use Lemma 59.69.1 to count ranks.
$\square$
Comments (0)