## 33.30 Picard groups of varieties

In this section we collect some elementary results on Picard groups of algebraic varieties.

Lemma 33.30.1. Let $A \to B$ be a faithfully flat ring map. Let $X$ be a quasi-compact and quasi-separated scheme over $A$. Let $\mathcal{L}$ be an invertible $\mathcal{O}_ X$-module whose pullback to $X_ B$ is trivial. Then $H^0(X, \mathcal{L})$ and $H^0(X, \mathcal{L}^{\otimes -1})$ are invertible $H^0(X, \mathcal{O}_ X)$-modules and the multiplication map induces an isomorphism

$H^0(X, \mathcal{L}) \otimes _{H^0(X, \mathcal{O}_ X)} H^0(X, \mathcal{L}^{\otimes -1}) \longrightarrow H^0(X, \mathcal{O}_ X)$

Proof. Denote $\mathcal{L}_ B$ the pullback of $\mathcal{L}$ to $X_ B$. Choose an isomorphism $\mathcal{L}_ B \to \mathcal{O}_{X_ B}$. Set $R = H^0(X, \mathcal{O}_ X)$, $M = H^0(X, \mathcal{L})$ and think of $M$ as an $R$-module. For every quasi-coherent $\mathcal{O}_ X$-module $\mathcal{F}$ with pullback $\mathcal{F}_ B$ on $X_ B$ there is a canonical isomorphism $H^0(X_ B, \mathcal{F}_ B) = H^0(X, \mathcal{F}) \otimes _ A B$, see Cohomology of Schemes, Lemma 30.5.2. Thus we have

$M \otimes _ R (R \otimes _ A B) = M \otimes _ A B = H^0(X_ B, \mathcal{L}_ B) \cong H^0(X_ B, \mathcal{O}_{X_ B}) = R \otimes _ A B$

Since $R \to R \otimes _ A B$ is faithfully flat (as the base change of the faithfully flat map $A \to B$), we conclude that $M$ is an invertible $R$-module by Algebra, Proposition 10.83.3. Similarly $N = H^0(X, \mathcal{L}^{\otimes -1})$ is an invertible $R$-module. To see that the statement on tensor products is true, use that it is true after pulling back to $X_ B$ and faithful flatness of $R \to R \otimes _ A B$. Some details omitted. $\square$

Lemma 33.30.2. Let $A \to B$ be a faithfully flat ring map. Let $X$ be a scheme over $A$ such that

1. $X$ is quasi-compact and quasi-separated, and

2. $R = H^0(X, \mathcal{O}_ X)$ is a semi-local ring.

Then the pullback map $\mathop{\mathrm{Pic}}\nolimits (X) \to \mathop{\mathrm{Pic}}\nolimits (X_ B)$ is injective.

Proof. Let $\mathcal{L}$ be an invertible $\mathcal{O}_ X$-module whose pullback $\mathcal{L}'$ to $X_ B$ is trivial. Set $M = H^0(X, \mathcal{L})$ and $N = H^0(X, \mathcal{L}^{\otimes - 1})$. By Lemma 33.30.1 the $R$-modules $M$ and $N$ are invertible. Since $R$ is semi-local $M \cong R$ and $N \cong R$, see Algebra, Lemma 10.78.7. Choose generators $s \in M$ and $t \in N$. Then $st \in R = H^0(X, \mathcal{O}_ X)$ is a unit by the last part of Lemma 33.30.1. We conclude that $s$ and $t$ define trivializations of $\mathcal{L}$ and $\mathcal{L}^{\otimes -1}$ over $X$. $\square$

Lemma 33.30.3. Let $k'/k$ be a field extension. Let $X$ be a scheme over $k$ such that

1. $X$ is quasi-compact and quasi-separated, and

2. $R = H^0(X, \mathcal{O}_ X)$ is semi-local, e.g., if $\dim _ k R < \infty$.

Then the pullback map $\mathop{\mathrm{Pic}}\nolimits (X) \to \mathop{\mathrm{Pic}}\nolimits (X_{k'})$ is injective.

Proof. Special case of Lemma 33.30.2. If $\dim _ k R < \infty$, then $R$ is Artinian and hence semi-local (Algebra, Lemmas 10.53.2 and 10.53.3). $\square$

Example 33.30.4. Lemma 33.30.3 is not true without some condition on the scheme $X$ over the field $k$. Here is an example. Let $k$ be a field. Let $t \in \mathbf{P}^1_ k$ be a closed point. Set $X = \mathbf{P}^1 \setminus \{ t\}$. Then we have a surjection

$\mathbf{Z} = \mathop{\mathrm{Pic}}\nolimits (\mathbf{P}^1_ k) \longrightarrow \mathop{\mathrm{Pic}}\nolimits (X)$

The first equality by Divisors, Lemma 31.28.5 and surjective by Divisors, Lemma 31.28.3 (as $\mathbf{P}^1_ k$ is smooth of dimension $1$ over $k$ and hence all its local rings are discrete valuation rings). If $\mathcal{L}$ is in the kernel of the displayed map, then $\mathcal{L} \cong \mathcal{O}_{\mathbf{P}^1_ k}(nt)$ for some $n \in \mathbf{Z}$. We leave it to the reader to show that $\mathcal{O}_{\mathbf{P}^1_ k}(t) \cong \mathcal{O}_{\mathbf{P}^1_ k}(d)$ where $d = [\kappa (t) : k]$. Hence

$\mathop{\mathrm{Pic}}\nolimits (X) = \mathbf{Z}/d\mathbf{Z}$

Thus if $t$ is not a $k$-rational point, then $d > 1$ and this Picard group is nonzero. On the other hand, if we extend the ground field $k$ to any field extension $k'$ such that there exists a $k$-embedding $\kappa (t) \to k'$, then $\mathbf{P}^1_{k'} \setminus X_{k'}$ has a $k'$-rational point $t'$. Hence $\mathcal{O}_{\mathbf{P}^1_{k'}}(1) = \mathcal{O}_{\mathbf{P}^1_{k'}}(t')$ will be in the kernel of the map $\mathbf{Z} \to \mathop{\mathrm{Pic}}\nolimits (X_{k'})$ and it will follow in the same manner as above that $\mathop{\mathrm{Pic}}\nolimits (X_{k'}) = 0$.

The following lemma tells us that “rationally equivalence invertible modules” are isomorphic on normal varieties.

Lemma 33.30.5. Let $k$ be a field. Let $X$ be a normal variety over $k$. Let $U \subset \mathbf{A}^ n_ k$ be an open subscheme with $k$-rational points $p, q \in U(k)$. For every invertible module $\mathcal{L}$ on $X \times _{\mathop{\mathrm{Spec}}(k)} U$ the restrictions $\mathcal{L}|_{X \times p}$ and $\mathcal{L}|_{X \times q}$ are isomorphic.

Proof. The fibres of $X \times _{\mathop{\mathrm{Spec}}(k)} U \to X$ are open subschemes of affine $n$-space over fields. Hence these fibres have trivial Picard groups by Divisors, Lemma 31.28.4. Applying Divisors, Lemma 31.28.1 we see that $\mathcal{L}$ is the pullback of an invertible module $\mathcal{N}$ on $X$. $\square$

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).