The Stacks project

Proof. The meaning of the phrase “has a $k$-rational point” is exactly that the structure morphism $f : X \to \mathop{\mathrm{Spec}}(k)$ has a section, which verifies the first condition. By Varieties, Lemma 33.26.2 we see that $k' = H^0(X, \mathcal{O}_ X)$ is a field extension of $k$. Since $X$ has a $k$-rational point there is a $k$-algebra homomorphism $k' \to k$ and we conclude $k' = k$. Since $k$ is a field, any morphism $T \to \mathop{\mathrm{Spec}}(k)$ is flat. Hence we see by cohomology and base change (Cohomology of Schemes, Lemma 30.5.2) that $\mathcal{O}_ T \to f_{T, *}\mathcal{O}_{X_ T}$ is an isomorphism. This finishes the proof. $\square$

Proof. Immediate from Derived Categories of Schemes, Lemma 36.32.3 applied with $i = 0$ and $r = 1$ and Schemes, Definition 26.15.3. $\square$

Proof. In this proof unadorned products are over $\mathop{\mathrm{Spec}}(k)$. By Proposition 44.3.6 the scheme $H = \underline{\mathrm{Hilb}}^ g_{X/k}$ exists. Consider the universal divisor $D_{univ} \subset H \times X$ and the associated invertible sheaf $\mathcal{O}(D_{univ})$, see Remark 44.3.7. We adjust by tensoring with the pullback via $\sigma _ H : H \to H \times X$ to get

By the Yoneda lemma (Categories, Lemma 4.3.5) the invertible sheaf $\mathcal{L}_ H$ defines a natural transformation

Because $F$ is an open subfuctor, there exists a maximal open $W \subset H$ such that $\mathcal{L}_ H|_{W \times X}$ is in $F(W)$. Of course, this open is nothing else than the open subscheme constructed in Derived Categories of Schemes, Lemma 36.32.3 with $i = 0$ and $r = 1$ for the morphism $H \times X \to H$ and the sheaf $\mathcal{F} = \mathcal{O}(D_{univ})$. Applying the Yoneda lemma again we obtain a commutative diagram

To finish the proof we will show that the top horizontal arrow is an isomorphism.

Let $\mathcal{L} \in F(T) \subset \mathrm{Pic}_{X/k, \sigma }(T)$. Let $\mathcal{N}$ be the invertible $\mathcal{O}_ T$-module such that $Rf_{T, *}\mathcal{L} \cong \mathcal{N}[0]$. The adjunction map

on $X_ T$. Claim: The zero scheme of $s$ is a relative effective Cartier divisor $D$ on $(T \times X)/T$ finite locally free of degree $g$ over $T$.

Let us finish the proof of the lemma admitting the claim. Namely, $D$ defines a morphism $m : T \to H$ such that $D$ is the pullback of $D_{univ}$. Then

Hence $(m \times \text{id}_ X)^*\mathcal{L}_ H$ and $\mathcal{O}(D)$ differ by the pullback of an invertible sheaf on $H$. This in particular shows that $m : T \to H$ factors through the open $W \subset H$ above. Moreover, it follows that these invertible modules define, after adjusting by pullback via $\sigma _ T$ as above, the same element of $\mathrm{Pic}_{X/k, \sigma }(T)$. Chasing diagrams using Yoneda's lemma we see that $m \in h_ W(T)$ maps to $\mathcal{L} \in F(T)$. We omit the verification that the rule $F(T) \to h_ W(T)$, $\mathcal{L} \mapsto m$ defines an inverse of the transformation of functors above.

Proof of the claim. Since $D$ is a locally principal closed subscheme of $T \times X$, it suffices to show that the fibres of $D$ over $T$ are effective Cartier divisors, see Lemma 44.3.1 and Divisors, Lemma 31.18.9. Because taking cohomology of $\mathcal{L}$ commutes with base change (Derived Categories of Schemes, Lemma 36.30.4) we reduce to $T = \mathop{\mathrm{Spec}}(K)$ where $K/k$ is a field extension. Then $\mathcal{L}$ is an invertible sheaf on $X_ K$ with $H^0(X_ K, \mathcal{L}) = K$ and $H^1(X_ K, \mathcal{L}) = 0$. Thus

See Varieties, Definition 33.44.1. To finish the proof we have to show a nonzero section of $\mathcal{L}$ defines an effective Cartier divisor on $X_ K$. This is clear. $\square$

Proof. This proof is a variant of the proof of Varieties, Lemma 33.44.16. We encourage the reader to read that proof first.

First we pick an ample invertible sheaf $\mathcal{L}_0$ and we replace $\mathcal{L}$ by $\mathcal{L} \otimes _{\mathcal{O}_{X_ K}} \mathcal{L}_0^{\otimes n}|_{X_ K}$ for some $n \gg 0$. The result will be that we may assume that $H^0(X_ K, \mathcal{L}) \not= 0$ and $H^1(X_ K, \mathcal{L}) = 0$. Namely, we will get the vanishing by Cohomology of Schemes, Lemma 30.17.1 and the nonvanishing because the degree of the tensor product is $\gg 0$. We will finish the proof by descending induction on $t = \dim _ K H^0(X_ K, \mathcal{L})$. The base case $t = 1$ is trivial. Assume $t > 1$.

Observe that for a $k$-rational point $x$ of $X$, the inverse image $x_ K$ is a $K$-rational point of $X_ K$. Moreover, there are infinitely many $k$-rational points by Varieties, Lemma 33.25.6. Therefore the points $x_ K$ form a Zariski dense collection of points of $X_ K$.

Let $s \in H^0(X_ K, \mathcal{L})$ be nonzero. From the previous paragraph we deduce there exists a $k$-rational point $x$ such that $s$ does not vanish in $x_ K$. Let $\mathcal{I}$ be the ideal sheaf of $i : x_ K \to X_ K$ as in Varieties, Lemma 33.43.8. Look at the short exact sequence

Observe that $H^0(X_ K, i_*i^*\mathcal{L}) = H^0(x_ K, i^*\mathcal{L})$ has dimension $1$ over $K$. Since $s$ does not vanish at $x$ we conclude that

is surjective. Hence $\dim _ K H^0(X_ K, \mathcal{I} \otimes _{\mathcal{O}_{X_ K}} \mathcal{L}) = t - 1$. Finally, the long exact sequence of cohomology also shows that $H^1(X_ K, \mathcal{I} \otimes _{\mathcal{O}_{X_ K}} \mathcal{L}) = 0$ thereby finishing the proof of the induction step. $\square$

Proof. Since $k$ is separably closed there exists a $k$-rational point $\sigma $ of $X$, see Varieties, Lemma 33.25.6. As discussed above, it suffices to show that the functor $\mathrm{Pic}_{X/k, \sigma }$ classifying invertible modules trivial along $\sigma $ is representable. To do this we will check conditions (1), (2)(a), (2)(b), and (2)(c) of Lemma 44.5.1.

The functor $\mathrm{Pic}_{X/k, \sigma }$ satisfies the sheaf condition for the fppf topology because it is isomorphic to $\mathrm{Pic}_{X/k}$. It would be more correct to say that we've shown the sheaf condition for $\mathrm{Pic}_{X/k, \sigma }$ in the proof of Lemma 44.4.3 which applies by Lemma 44.6.1. This proves condition (1)

As our subfunctor we use $F$ as defined in Lemma 44.6.2. Condition (2)(b) follows. Condition (2)(a) is Lemma 44.6.4. Condition (2)(c) is Lemma 44.6.5. $\square$

Proof. Pick a $k$-rational point $\sigma $ of $X$. Recall that $\mathrm{Pic}_{X/k}$ is isomorphic to the functor $\mathrm{Pic}_{X/k, \sigma }$. By Derived Categories of Schemes, Lemma 36.32.2 for every $d \in \mathbf{Z}$ there is an open subfunctor

whose value on a scheme $T$ over $k$ consists of those $\mathcal{L} \in \mathrm{Pic}_{X/k, \sigma }(T)$ such that $\chi (X_ t, \mathcal{L}_ t) = d + 1 - g$ and moreover we have

as fppf sheaves. It follows that the scheme $\underline{\mathrm{Pic}}_{X/k}$ (which exists by Proposition 44.6.6) has a corresponding decomposition

where the points of $\underline{\mathrm{Pic}}^ d_{X/k, \sigma }$ correspond to isomorphism classes of invertible modules of degree $d$ on $X$.

Fix $d \geq 0$. There is a morphism

coming from the invertible sheaf $\mathcal{O}(D_{univ})$ on $\underline{\mathrm{Hilb}}^ d_{X/k} \times _ k X$ (Remark 44.3.7) by the Yoneda lemma (Categories, Lemma 4.3.5). Our proof of the representability of the Picard functor of $X/k$ in Proposition 44.6.6 and Lemma 44.6.4 shows that $\gamma _ g$ induces an open immersion on a nonempty open of $\underline{\mathrm{Hilb}}^ g_{X/k}$. Moreover, the proof shows that the translates of this open by $k$-rational points of the group scheme $\underline{\mathrm{Pic}}_{X/k}$ define an open covering. Since $\underline{\mathrm{Hilb}}^ g_{X/K}$ is smooth of dimension $g$ (Proposition 44.3.6) over $k$, we conclude that the group scheme $\underline{\mathrm{Pic}}_{X/k}$ is smooth of dimension $g$ over $k$.

By Groupoids, Lemma 39.7.3 we see that $\underline{\mathrm{Pic}}_{X/k}$ is separated. Hence, for every $d \geq 0$, the image of $\gamma _ d$ is a proper variety over $k$ (Morphisms, Lemma 29.41.10).

Let $d \geq g$. Then for any field extension $K/k$ and any invertible $\mathcal{O}_{X_ K}$-module $\mathcal{L}$ of degree $d$, we see that $\chi (X_ K, \mathcal{L}) = d + 1 - g > 0$. Hence $\mathcal{L}$ has a nonzero section and we conclude that $\mathcal{L} = \mathcal{O}_{X_ K}(D)$ for some divisor $D \subset X_ K$ of degree $d$. It follows that $\gamma _ d$ is surjective.

Combining the facts mentioned above we see that $\underline{\mathrm{Pic}}^ d_{X/k}$ is proper for $d \geq g$. This finishes the proof of (2) because now we see that $\underline{\mathrm{Pic}}^ d_{X/k}$ is proper for $d \geq g$ but then all $\underline{\mathrm{Pic}}^ d_{X/k}$ are proper by translation.

It remains to prove that $\gamma _ d$ is smooth for $d \geq 2g - 1$. Consider an invertible $\mathcal{O}_ X$-module $\mathcal{L}$ of degree $d$. Then the fibre of the point corresponding to $\mathcal{L}$ is

with its natural scheme structure. Since any isomorphism $\mathcal{O}_ X(D) \to \mathcal{L}$ is well defined up to multiplying by a nonzero scalar, we see that the canonical section $1 \in \mathcal{O}_ X(D)$ is mapped to a section $s \in \Gamma (X, \mathcal{L})$ well defined up to multiplication by a nonzero scalar. In this way we obtain a morphism

(dual because of our conventions). This morphism is an isomorphism, because given an section of $\mathcal{L}$ we can take the associated effective Cartier divisor, in other words we can construct an inverse of the displayed morphism; we omit the precise formulation and proof. Since $\dim H^0(X, \mathcal{L}) = d + 1 - g$ for every $\mathcal{L}$ of degree $d \geq 2g - 1$ by Varieties, Lemma 33.44.17 we see that $\text{Proj}(\text{Sym}(\Gamma (X, \mathcal{L})^*)) \cong \mathbf{P}^{d - g}_ k$. We conclude that $\dim (Z) = \dim (\mathbf{P}^{d - g}_ k) = d - g$. We conclude that the fibres of the morphism $\gamma _ d$ all have dimension equal to the difference of the dimensions of $\underline{\mathrm{Hilb}}^ d_{X/k}$ and $\underline{\mathrm{Pic}}^ d_{X/k}$. It follows that $\gamma _ d$ is flat, see Algebra, Lemma 10.128.1. As moreover the fibres are smooth, we conclude that $\gamma _ d$ is smooth by Morphisms, Lemma 29.34.3. $\square$