The Stacks project

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. $\square$