The Stacks project

44.7 Some remarks on Picard groups

This section continues the discussion in Varieties, Section 33.30 and will be continued in Algebraic Curves, Section 53.17.

Lemma 44.7.1. Let $k$ be a field. Let $X$ be a quasi-compact and quasi-separated scheme over $k$ with $H^0(X, \mathcal{O}_ X) = k$. If $X$ has a $k$-rational point, then for any Galois extension $k'/k$ we have

\[ \mathop{\mathrm{Pic}}\nolimits (X) = \mathop{\mathrm{Pic}}\nolimits (X_{k'})^{\text{Gal}(k'/k)} \]

Moreover the action of $\text{Gal}(k'/k)$ on $\mathop{\mathrm{Pic}}\nolimits (X_{k'})$ is continuous.

Proof. Since $\text{Gal}(k'/k) = \text{Aut}(k'/k)$ it acts (from the right) on $\mathop{\mathrm{Spec}}(k')$, hence it acts (from the right) on $X_{k'} = X \times _{\mathop{\mathrm{Spec}}(k)} \mathop{\mathrm{Spec}}(k')$, and since $\mathop{\mathrm{Pic}}\nolimits (-)$ is a contravariant functor, it acts (from the left) on $\mathop{\mathrm{Pic}}\nolimits (X_{k'})$. If $k'/k$ is an infinite Galois extension, then we write $k' = \mathop{\mathrm{colim}}\nolimits k'_\lambda $ as a filtered colimit of finite Galois extensions, see Fields, Lemma 9.22.3. Then $X_{k'} = \mathop{\mathrm{lim}}\nolimits X_{k_\lambda }$ (as in Limits, Section 32.2) and we obtain

\[ \mathop{\mathrm{Pic}}\nolimits (X_{k'}) = \mathop{\mathrm{colim}}\nolimits \mathop{\mathrm{Pic}}\nolimits (X_{k_\lambda }) \]

by Limits, Lemma 32.10.3. Moreover, the transition maps in this system of abelian groups are injective by Varieties, Lemma 33.30.3. It follows that every element of $\mathop{\mathrm{Pic}}\nolimits (X_{k'})$ is fixed by one of the open subgroups $\text{Gal}(k'/k_\lambda )$, which exactly means that the action is continuous. Injectivity of the transition maps implies that it suffices to prove the statement on fixed points in the case that $k'/k$ is finite Galois.

Assume $k'/k$ is finite Galois with Galois group $G = \text{Gal}(k'/k)$. Let $\mathcal{L}$ be an element of $\mathop{\mathrm{Pic}}\nolimits (X_{k'})$ fixed by $G$. We will use Galois descent (Descent, Lemma 35.6.1) to prove that $\mathcal{L}$ is the pullback of an invertible sheaf on $X$. Recall that $f_\sigma = \text{id}_ X \times \mathop{\mathrm{Spec}}(\sigma ) : X_{k'} \to X_{k'}$ and that $\sigma $ acts on $\mathop{\mathrm{Pic}}\nolimits (X_{k'})$ by pulling back by $f_\sigma $. Hence for each $\sigma \in G$ we can choose an isomorphism $\varphi _\sigma : \mathcal{L} \to f_\sigma ^*\mathcal{L}$ because $\mathcal{L}$ is a fixed by the $G$-action. The trouble is that we don't know if we can choose $\varphi _\sigma $ such that the cocycle condition $\varphi _{\sigma \tau } = f_\sigma ^*\varphi _\tau \circ \varphi _\sigma $ holds. To see that this is possible we use that $X$ has a $k$-rational point $x \in X(k)$. Of course, $x$ similarly determines a $k'$-rational point $x' \in X_{k'}$ which is fixed by $f_\sigma $ for all $\sigma $. Pick a nonzero element $s$ in the fibre of $\mathcal{L}$ at $x'$; the fibre is the $1$-dimensional $k' = \kappa (x')$-vector space

\[ \mathcal{L}_{x'} \otimes _{\mathcal{O}_{X_{k'}, x'}} \kappa (x'). \]

Then $f_\sigma ^*s$ is a nonzero element of the fibre of $f_\sigma ^*\mathcal{L}$ at $x'$. Since we can multiply $\varphi _\sigma $ by an element of $(k')^*$ we may assume that $\varphi _\sigma $ sends $s$ to $f_\sigma ^*s$. Then we see that both $\varphi _{\sigma \tau }$ and $f_\sigma ^*\varphi _\tau \circ \varphi _\sigma $ send $s$ to $f_{\sigma \tau }^*s = f_\tau ^*f_\sigma ^*s$. Since $H^0(X_{k'}, \mathcal{O}_{X_{k'}}) = k'$ these two isomorphisms have to be the same (as one is a global unit times the other and they agree in $x'$) and the proof is complete. $\square$

Lemma 44.7.2. Let $k$ be a field of characteristic $p > 0$. Let $X$ be a quasi-compact and quasi-separated scheme over $k$ with $H^0(X, \mathcal{O}_ X) = k$. Let $n$ be an integer prime to $p$. Then the map

\[ \mathop{\mathrm{Pic}}\nolimits (X)[n] \longrightarrow \mathop{\mathrm{Pic}}\nolimits (X_{k'})[n] \]

is bijective for any purely inseparable extension $k'/k$.

Proof. First we observe that the map $\mathop{\mathrm{Pic}}\nolimits (X) \to \mathop{\mathrm{Pic}}\nolimits (X_{k'})$ is injective by Varieties, Lemma 33.30.3. Hence we have to show the map in the lemma is surjective. Let $\mathcal{L}$ be an invertible $\mathcal{O}_{X_{k'}}$-module which has order dividing $n$ in $\mathop{\mathrm{Pic}}\nolimits (X_{k'})$. Choose an isomorphism $\alpha : \mathcal{L}^{\otimes n} \to \mathcal{O}_{X_{k'}}$ of invertible modules. We will prove that we can descend the pair $(\mathcal{L}, \alpha )$ to $X$.

Set $A = k' \otimes _ k k'$. Since $k'/k$ is purely inseparable, the kernel of the multiplication map $A \to k'$ is a locally nilpotent ideal $I$ of $A$. Observe that

\[ X_ A = X \times _{\mathop{\mathrm{Spec}}(k)} \mathop{\mathrm{Spec}}(A) = X_{k'} \times _ X X_{k'} \]

comes with two projections $\text{pr}_ i : X_ A \to X_{k'}$, $i = 0, 1$ which agree over $A/I$. Hence the invertible modules $\mathcal{L}_ i = \text{pr}_ i^*\mathcal{L}$ agree over the closed subscheme $X_{A/I} = X_{k'}$. Since $X_{A/I} \to X_ A$ is a thickening and since $\mathcal{L}_ i$ are $n$-torsion, we see that there exists an isomorphism $\varphi : \mathcal{L}_0 \to \mathcal{L}_1$ by More on Morphisms, Lemma 37.4.2. We may pick $\varphi $ to reduce to the identity modulo $I$. Namely, $H^0(X, \mathcal{O}_ X) = k$ implies $H^0(X_{k'}, \mathcal{O}_{X_{k'}}) = k'$ by Cohomology of Schemes, Lemma 30.5.2 and $A \to k'$ is surjective hence we can adjust $\varphi $ by multiplying by a suitable element of $A$. Consider the map

\[ \lambda : \mathcal{O}_{X_ A} \xrightarrow {\text{pr}_0^*\alpha ^{-1}} \mathcal{L}_0^{\otimes n} \xrightarrow {\varphi ^{\otimes n}} \mathcal{L}_1^{\otimes n} \xrightarrow {\text{pr}_0^*\alpha } \mathcal{O}_{X_ A} \]

We can view $\lambda $ as an element of $A$ because $H^0(X_ A, \mathcal{O}_{X_ A}) = A$ (same reference as above). Since $\varphi $ reduces to the identity modulo $I$ we see that $\lambda = 1 \bmod I$. Then there is a unique $n$th root of $\lambda $ in $1 + I$ (Algebra, Lemma 10.32.8) and after multiplying $\varphi $ by its inverse we get $\lambda = 1$. We claim that $(\mathcal{L}, \varphi )$ is a descent datum for the fpqc covering $\{ X_{k'} \to X\} $ (Descent, Definition 35.2.1). If true, then $\mathcal{L}$ is the pullback of an invertible $\mathcal{O}_ X$-module $\mathcal{N}$ by Descent, Proposition 35.5.2. Injectivity of the map on Picard groups shows that $\mathcal{N}$ is a torsion element of $\mathop{\mathrm{Pic}}\nolimits (X)$ of the same order as $\mathcal{L}$.

Proof of the claim. To see this we have to verify that

\[ \text{pr}_{12}^*\varphi \circ \text{pr}_{01}^*\varphi = \text{pr}_{02}^*\varphi \quad \text{on}\quad X_{k'} \times _ X X_{k'} \times _ X X_{k'} = X_{k' \otimes _ k k' \otimes _ k k'} \]

As before the diagonal morphism $\Delta : X_{k'} \to X_{k' \otimes _ k k' \otimes _ k k'}$ is a thickening. The left and right hand sides of the equality signs are maps $a, b : p_0^*\mathcal{L} \to p_2^*\mathcal{L}$ compatible with $p_0^*\alpha $ and $p_2^*\alpha $ where $p_ i : X_{k' \otimes _ k k' \otimes _ k k'} \to X_{k'}$ are the projection morphisms. Finally, $a, b$ pull back to the same map under $\Delta $. Affine locally (in local trivializations) this means that $a, b$ are given by multiplication by invertible functions which reduce to the same function modulo a locally nilpotent ideal and which have the same $n$th powers. Then it follows from Algebra, Lemma 10.32.8 that these functions are the same. $\square$

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