The torsion component of the Picard scheme

Proof. We induct on the codimension $c=n-d$ of $X$ inside $P={\mathbf{P}}_k^n$. If $c=0$, then this follows from the familiar computations of the cohomology of projective space. In general, by the definition of a complete intersection there is some complete intersection $Z\subset P$ and hyperplane $H\subset P$ of degree $m$ such that $X=Z\cap H$ and moreover there is a short exact sequence

where $i:X\to Z$ is the natural inclusion. Tensoring by ${\mathcal{O}}_Z(-N)$ and passing to cohomology gives exact sequences

By induction, if $1\le i\le c$ then the two outside terms vanish. If $i=0$, then we have the exact sequence

If $N=0$, this shows $H^0(X,{\mathcal{O}}_X)=k$. If $N\ge 1$, then again $H^0(X,{\mathcal{O}}_X(-N))=0$. Since $H^0(X,{\mathcal{O}}_X)=k$, it follows from the Stein factorization that $X$ is geometrically connected.

Now suppose that $X$ is smooth and $d\ge 2$. To compute $H^0(X,{\Omega }_{X/k}^1)$, recall first the Euler exact sequence for $n$-dimensional projective space:

From this, we see that $H^0(P,{\Omega }_{P/k}^1)=0$. In general, if a smooth $X$ is the intersection of hypersurfaces whose degrees sum to $N$, then we look at the conormal exact sequence

for $i:X\to P$ the natural inclusion. Passing to cohomology gives the short exact sequence

and it suffices by the above to show that $H^0(X,i^{\ast }{\Omega }_{P/k}^1)=0$. This now follows from the short exact sequence

since $H^0(P,{\Omega }_{P/k}^1)=0$ and $H^1(P,{\Omega }_{P/k}^1(-N))=0$. This completes the cohomology calculations. 

Proof. First suppose $S=\mathrm{Spec}k$ for an algebraically closed field $k$. By Theorem 1.2(ii) and Lemma 1.1, ${\mathbf{Pic}}_{X/k}$ is a quasi-separated algebraic space locally of finite type over $k$. By Lemma 4.2 of Artin's paper cited above, it follows in this case that in fact ${\mathbf{Pic}}_{X/k}$ is a $k$-group scheme. (Note that quasi-separatedness is part of the definition of an algebraic space in Artin's paper.) As $T_0{\mathbf{Pic}}_{X/k}\cong H^1(X,{\mathcal{O}}_X)$, we see that ${\mathbf{Pic}}_{X/k}^0=0$ by Lemma 1.1. So it suffices to show that $\mathrm{Pic}(X)$ has no torsion. If $d\ge 3$, then a Lefschetz theorem (see SGA2, Exp. XII, Cor. 3.7, and note that no smoothness hypotheses on $X$ are necessary) states that $\mathrm{Pic}(X)\cong \mathrm{Pic}({\mathbf{P}}^n)\cong \mathbf{Z}$, so indeed $\mathrm{Pic}(X)$ is torsion-free in this case. Now suppose $X$ is smooth. Recall that $H_{\mathrm{fppf}}^1(X,{\mu }_n)\cong \mathrm{Pic}(X)[n]$ for all $n\in \mathbf{Z}$ since $k$ is algebraically closed. If $p$ does not divide $n$ then ${\mu }_n\cong \mathbf{Z}/n\mathbf{Z}$, so we have

By a Lefschetz theorem (see SGA2, Exp. XII, Cor. 3.5, and note again the lack of smoothness hypotheses), we have ${\pi }_1(X)\cong {\pi }_1(P)=0$, so that this Hom set is trivial. To prove that there is no $p$-torsion, note that there is an exact sequence of Zariski sheaves

where the first map is given by $x\mapsto x^p$ and the second map is given by $f\mapsto \frac{df}{f}$. The $p$th power map is clearly injective since $X$ is reduced, and the composition of the two maps is evidently $0$. Exactness in the middle is more involved: the idea is to use normality and local freeness of ${\Omega }^1$ to reduce to proving that for a finitely generated field extension $L/k$ and $f\in L$, $df=0$ implies that $f$ is a $p$th power. After checking this for purely transcendental extensions, one uses the existence of a separating transcendence basis to deduce the general case. Since the complete argument is rather long, we omit it.

Now let $L\subset {\Omega }_{X/k}^1$ be the image of the map ${\mathcal{O}}_X^{\ast }\to {\Omega }_{X/k}^1$ in the exact sequence above, so there is a corresponding short exact sequence for $L$ and we obtain, passing to cohomology, an exact sequence

where the right map is multiplication by $p$. Since $H^0(X,{\Omega }_{X/k}^1)=0$ by Lemma 1.1, it follows that $\mathrm{Pic}(X)[p]=0$, and indeed ${\mathbf{Pic}}_{X/k}^{\tau }=0$.

Now we work in the case of general $S$. Using Lemma 1.1, a simple argument using cohomology and base change shows that $X\to S$ is cohomologically flat in dimension $0$, so ${\mathbf{Pic}}_{X/S}$ is an algebraic space by Theorem 1.2(ii) and ${\mathbf{Pic}}_{X/S}^{\tau }$ is also an algebraic space since ${\mathbf{Pic}}_{X/S}^{\tau }\to {\mathbf{Pic}}_{X/S}$ is representable by open immersions. Since formation of ${\mathbf{Pic}}_{X/S}^{\tau }$ commutes with arbitrary base change, we see that $({\mathbf{Pic}}_{X/S}^{\tau }{)}_s=0$ for all $s\in S$ by the first paragraph. Thus the identity section $S\to {\mathbf{Pic}}_{X/S}^{\tau }$ is an isomorphism on fibers, and since $S$ is trivially flat over itself, the fibral isomorphism theorem (see EGA IV, Cor. 17.9.5) implies that ${\mathbf{Pic}}_{X/S}^{\tau }=0$. (Note that the fibral isomorphism criterion holds for a morphism from a scheme to an algebraic space: morphisms may be checked to be isomorphisms after etale base change, so this follows immediately from the fact that algebraic spaces admit etale covers which are schemes and the relative diagonal of an algebraic space is representable.) This completes the proof. 

Proof. We work with the presheaf ${\mathrm{Pic}}_{X/S}$ defined by ${\mathrm{Pic}}_{X/S}(T)=\mathrm{Pic}(X_T)$ for all $S$-schemes $T$. There is a pullback map ${\pi }^{\ast }:{\mathrm{Pic}}_{X/S}\to {\mathrm{Pic}}_{Y/S}$, and we claim that it has kernel $G^{\vee }$. After fppf sheafification, this easily gives the result. To prove this claim, it suffices by base change on $S$ to show that the kernel of the map $\mathrm{Pic}(X)\to \mathrm{Pic}(Y)$ can be canonically identified with the group $G^{\vee }(S)$, and we will do this below.

Recall that fppf descent for quasicoherent sheaves says that a line bundle on $X$ is equivalent to the data of a line bundle $\mathcal{L}$ on $Y$ along with an isomorphism $\phi :{\pi }_1^{\ast }\mathcal{L}\to {\pi }_2^{\ast }\mathcal{L}$, where ${\pi }_1,{\pi }_2:Y{\times }_{X}Y\to Y$ are the two canonical projections, satisfying the cocycle condition ${\pi }_{23}^{\ast }\phi \circ {\pi }_{12}^{\ast }\phi ={\pi }_{13}^{\ast }\phi$, where ${\pi }_{ij}:Y{\times }_{X}Y{\times }_{X}Y\to Y{\times }_{X}Y$ is the projection onto the $i$ and $j$ coordinates. Given a line bundle $\mathcal{M}$ on $X$, we obtain a datum of this form (a descent datum) via taking $\mathcal{L}={\pi }^{\ast }\mathcal{M}$ and using the canonical isomorphism of functors ${\pi }_1^{\ast }\circ {\pi }^{\ast }\cong {\pi }_2^{\ast }\circ {\pi }^{\ast }$. So a line bundle on $X$ which becomes trivial after pullback to $Y$ is the same as the datum of an isomorphism ${\pi }_1^{\ast }{\mathcal{O}}_Y\to {\pi }_2^{\ast }{\mathcal{O}}_Y$ satisfying the cocycle condition. Since ${\pi }_1^{\ast }{\mathcal{O}}_Y\cong {\pi }_2^{\ast }{\mathcal{O}}_Y\cong {\mathcal{O}}_{Y{\times }_{X}Y}$ canonically, this is the same as an automorphism of ${\mathcal{O}}_{Y{\times }_{X}Y}$, i.e., an element of ${\mathbf{G}}_m(Y{\times }_{X}Y)$, satisfying the cocycle condition. We will show below that the cocycle condition can be described more concretely in terms of $G$.

Since $\pi$ is a $G$-torsor, there is a canonical isomorphism $G{\times }_{S}Y\to Y{\times }_{X}Y$ given functorially by $(g,y)\mapsto (y,gy)$. There is also an identification $G{\times }_{S}G{\times }_{S}Y\cong Y{\times }_{X}Y{\times }_{X}Y$ given functorially by $(g,h,y)\mapsto (y,hy,ghy)$. Under these identifications, the maps ${\pi }_{12}$, ${\pi }_{23}$, and ${\pi }_{13}$ are identified with maps $G{\times }_{S}G{\times }_{S}Y\to G{\times }_{S}Y$ via

Suppose that $\phi \in {\mathbf{G}}_m(G{\times }_{S}Y)$ is such that the cocycle condition holds. Functorially, this means

for all $g,h\in G(T)$ and $y\in Y(T)$, where $T$ ranges over the category of $S$-schemes. Since $Y$ is cohomologically flat in dimension $0$, we have ${\mathbf{G}}_m(G{\times }_{S}Y)\cong {\mathbf{G}}_m(G)$ naturally via pullback of units, so $\phi$ may be regarded as a morphism $\phi :G\to {\mathbf{G}}_m$, and the cocycle condition is precisely saying that $\phi$ is a group homomorphism, i.e., $\phi \in G^{\vee }(S)$. This completes the proof. 

Proof. The claims about base change are simple from the functorial definition, and will be omitted. For the first representability claim, note that there is a Cartesian diagram

where $\alpha$ denotes the action map $\alpha (g,x)=(x,gx)$. Since $X$ is separated, $\Delta$ is a closed embedding, and it follows by base change that ${\mathrm{Stab}}_{G_X}({\mathrm{id}}_X)$ is a closed $X$-subgroup of $G{\times }_{S}X$, hence in particular a finite $X$-scheme. Now if $x:T\to X$ is any $S$-morphism, we see that there is a Cartesian diagram

so that ${\mathrm{Stab}}_{G_T}(x)$ is a finite $T$-group scheme. The morphism ${\mathrm{Stab}}_{G_T}(x)\to G_T$ is a proper monomorphism, hence a closed embedding, and we obtain the claim.

Recall that if $H\to S$ is a finite morphism, then the function $s\mapsto \mathrm{rk}{H}_s$ is an upper semicontinuous function: this follows directly from Nakayama's lemma. If $H$ is moreover an $S$-group scheme then $\mathrm{rk}{H}_s\ge 1$ for all $s\in S$ (since there exists a section $S\to H$), and $H$ is trivial if and only if the fiber $H_s$ is trivial for all $s\in S$ (apply the fibral isomorphism criterion to the identity section). It follows from these considerations applied to $H={\mathrm{Stab}}_G({\mathrm{id}}_X)$ that if $U=\{x\in X:{\mathrm{Stab}}_{G_{k(x)}}(x)=0\}$ then $U$ is an open subscheme of $X$ and $U$ represents ${\mathrm{Free}}_{G,X/S}$. This proves the lemma. 

Proof. We offer a very brief sketch of the idea and refer to loc. cit. for a detailed argument. First, pass to geometric fibers to assume that $S$ is the spectrum of an algebraically closed field, say of characteristic $p\ge 0$. If $x\in P$ has nontrivial stabilizer, then it contains a subgroup isomorphic to one of $\mathbf{Z}/p\mathbf{Z}$, ${\alpha }_p$, or ${\mu }_{\ell }$ for some prime number $\ell$. One first shows that the locus fixed by any of these subgroups in $P$ has large codimension. In general, $G$ contains only finitely many subgroups of the form $\mathbf{Z}/p\mathbf{Z}$ or ${\mu }_{\ell }$ for $\ell \ne p$, but it may contain infinitely many subgroups of the form ${\alpha }_p$ and ${\mu }_p$. However, such subgroups are determined by their Lie algebras, so the collection of subgroups of each of these forms is parameterized by some projective space. One shows then that this projective space is of relatively small dimension to obtain the lemma. 

Proof. We prove first that $X$ is $S$-flat. Note that the special fiber $X_s$ is integral of dimension $r^n-t-1$, and Chevalley's theorem on upper-semicontinuity of fiber dimension (see EGA IV, 13.1.3 for a proof for general $S$) implies that in fact every fiber of $X$ over $S$ has dimension at most $r^n-t-1$. But every fiber of $P/G^{\vee }$ over $S$ is integral of dimension $r^n-1$, and intersection with a hyperplane can cut the dimension down by at most $1$; this follows from Krull's height theorem. Since every zerodivisor in a noetherian local ring $A$ is contained in a minimal prime ideal, if $A$ is equidimensional and $f$ is a nonunit then $f$ is a nonzerodivisor if and only if $\dim A/f=\dim A-1$. By induction on $t$, using the above considerations and the fact that $P/G^{\vee }$ has integral fibers, we see that $X$ has equidimensional fibers of dimension $r^n-t-1$. If each $H_i$ is defined locally on $P/G^{\vee }$ by some function $f_i$, it follows that $(f_1,\dots ,f_t)$ is a regular sequence since this may be checked on fibers over $S$. So if $U=\mathrm{Spec}A\subset P/G^{\vee }$ is an affine open on which $H_1$ is defined by $f_1$, there is an exact sequence $0\to A\to A\to A/f_1\to 0$, where the first map is given by multiplication by $f_1$. This map is injective after base change to any residue field of $S$ (as we noted above), so that $A/f_1$ is flat over $S$: we see that ${\mathrm{Tor}}_1^S(A/f_1,k(s^{\prime }))=0$ for all $s^{\prime }\in S$, so flatness follows from Stacks Project, Tag 00M5 and a Zorn's lemma-style argument. These same considerations show by induction on $n$ that $X$ is $S$-flat.

Now recall that the $S$-smooth locus in $X$ is open, and $X$ is $S$-smooth at all points of $X_s$ since $X$ is $S$-smooth at a point $x$ if and only if it is flat over $S$ at $x$ and $x$ is smooth in its fiber. Since $X$ is proper, it follows immediately that $X$ is smooth over $S$, completing the proof of the Lemma.