An example of a non-reduced Picard scheme

Let $X$ be a smooth projective connected scheme over an algebraically closed field $k$ (experts will notice that several of these hypotheses can be weakened in what follows). Attached to $X$ is the Picard scheme $\mathrm{Pic}_{X/k}$ , a locally finite type $k$ -scheme defined functorially as sending a $k$ -scheme $T$ to the group $\mathrm{Pic}(X_T)/\mathrm{Pic}(T)$ . (This uses the fact that there exists a $k$ -point; in general one needs to sheafify with respect to a suitable Grothendieck topology.) Using smoothness of $X$ , one can verify from the functorial definition that $\mathrm{Pic}_{X/k}$ satisfies the valuative criterion of properness, so its identity component $\mathrm{Pic}_{X/k}^0$ is a proper $k$ -group scheme. If $X$ is dimension 1, then using dimensional vanishing for coherent cohomology one can furthermore verify the infinitesimal criterion of smoothness for $\mathrm{Pic}_{X/k}$ , so that $\mathrm{Pic}_{X/k}^0$ is an abelian variety. Smoothness also holds if $k$ is of characteristic 0, since in this case it is a theorem of Cartier that every $k$ -group scheme is smooth. However, there are several examples which show that $\mathrm{Pic}_{X/k}$ need not be reduced even if $X$ is a surface in positive characteristic. In this post I will describe one such example due to Serre.

First, how can one get a handle on smoothness of the Picard scheme? There are two fundamental inequalities which we will describe concerning the dimension of the Picard scheme. First,

$\dim \mathrm{Pic}_{X/k} \leq \dim H^1(X, \mathcal{O}_X),$

immediate from the following more precise Lemma.

Lemma. There is a natural isomorphism $T_0 \mathrm{Pic}_{X/k} \cong H^1(X, \mathcal{O}_X)$ , where $T_0$ denotes the tangent space at the identity.

Proof. Recall the natural isomorphism $T_0 \mathrm{Pic}_{X/k} = \mathrm{ker}(\mathrm{Pic}(X_{k[\epsilon]/(\epsilon^2)}) \to \mathrm{Pic}(X))$ . We have a split short exact sequence of abelian sheaves $0 \to \mathcal{O}_X \to \mathcal{O}_{X_{k[\epsilon]}}^* \to \mathcal{O}_X^* \to 0$ , giving rise to the exact sequence of (Cech) cohomology groups

$H^0(X, \mathcal{O}_{X_{k[\epsilon]}}^*) \to H^0(X, \mathcal{O}_X^*) \to H^1(X, \mathcal{O}_X) \to H^1(X, \mathcal{O}_{X_{k[\epsilon]}}^*) \to H^1(X, \mathcal{O}_X^*).$

Since the short exact sequence is split, the leftmost map in this sequence is surjective and thus the lemma follows from the fact that $\mathrm{Pic}(X) \cong H^1(X, \mathcal{O}_X^*)$ and similarly for $\mathrm{Pic}(X_{k[\epsilon]})$ . QED

If $k$ is characteristic 0 (resp. $X$ is a curve), then Hodge symmetry (resp. Serre duality) implies that $\dim H^1(X, \mathcal{O}_X) = \dim H^0(X, \Omega_{X/k}^1)$ , and it follows that

$\dim \mathrm{Pic}_{X/k} \leq \dim H^0(X, \Omega_{X/k}^1).$

Although Hodge symmetry can fail in positive characteristic (as we will see later in our example), this inequality is always true. This is a result due to Igusa, see A Fundamental Inequality in the Theory of Picard Varieties. In fact, Igusa shows that the Albanese morphism $X \to \mathrm{Alb}(X)$ induces an injective map $H^0(\mathrm{Alb}(X), \Omega_{\mathrm{Alb}(X)/k}^1) \to H^0(X, \Omega_{X/k}^1)$ . Since $\mathrm{Alb}(X) \cong ((\mathrm{Pic}_{X/k}^0)_{\mathrm{red}})^\wedge$ , it follows that $H^0(\mathrm{Alb}(X), \Omega_{\mathrm{Alb}(X)/k}^1) = \dim \mathrm{Pic}_{X/k}$ , yielding the above inequality. This is a deep theorem; it relies on a serious understanding of Albanese varieties to reduce to the case that $X$ is a surface, and then resolution of singularities for surfaces to reduce to the case that $X$ is a smooth surface; it is here that some real geometry takes place. In any case, the combination of these two results shows that it would be enough to find a smooth projective integral $k$ -scheme $X$ such that

$\dim H^0(X, \Omega_{X/k}^1) < \dim H^1(X, \mathcal{O}_X).$

Our work is now to find such an $X$ . In fact, with (substantially) more work we will show that the $X$ we construct has the property that $\mathrm{Pic}_{X/k}^0 \cong \mu_p$ , see the Remark below.

The example

With these preparations, we are ready to describe Serre’s example. We will show that if $p \geq 5$ is the characteristic of $k$ , then there is a smooth surface $Y$ in $\mathbf{P}_k^3$ with a free action of the group $G = \mathbf{Z}/p\mathbf{Z}$ . Grant this for now; we will construct $Y$ at the very end. Note that $X = Y/G$ exists in this case as a smooth projective variety (see SGA3, Exp. V, Thm. 4.1). We will show $\dim H^1(X, \mathcal{O}_X) = 1$ and $H^0(X, \Omega_{X/k}^1) = 0$ , and it will follow immediately that $\mathrm{Pic}_{X/k}^0$ is a nontrivial finite connected $k$ -group scheme, giving the promised example.

Lemma. $H^0(X, \Omega_{X/k}^1) = 0$ .

Proof. Note that the etale cover $Y \to X$ induces an injective map $H^0(X, \Omega_{X/k}^1) \to H^0(Y, \Omega_{Y/k}^1)$ , so it suffices to show that $H^0(Y, \Omega_{Y/k}^1) = 0$ . Indeed, we will show that this is the case whenever $Y$ is a hypersurface inside a projective space $P$ of dimension $n \geq 3$ . Recall the conormal exact sequence

$0 \to \mathcal{I}_Y/\mathcal{I}_Y^2 \to i^* \Omega_{P/k}^1 \to \Omega_{Y/k}^1 \to 0$

where $i: Y \to P = \mathbf{P}_k^n$ is the natural closed embedding and $\mathcal{I}_Y$ is the quasicoherent sheaf defining it. Note that if $Y$ is a hypersurface of degree $d$ , then $\mathcal{I}_Y = \mathcal{O}_Y(-d)$ and thus $\mathcal{I}_Y/\mathcal{I}_Y^2 = \mathcal{O}_Y(-d)$ . Thus we have an exact sequence

$H^0(Y, i^* \Omega_{P/k}^1) \to H^0(Y, \Omega_{Y/k}^1) \to H^1(Y, \mathcal{O}_Y(-d)),$

and we see that it is enough to show $H^0(Y, i^* \Omega_{P/k}^1) = 0$ and $H^1(Y, \mathcal{O}_Y(-d)) = 0$ . We’ll take these equalities in turns.

1. $H^0(Y, i^* \Omega_{P/k}^1) = 0$ .

Recall the Euler exact sequence for $P$

$0 \to \Omega_{P/k}^1 \to \mathcal{O}_P(-1)^{n+1} \to \mathcal{O}_P \to 0.$

Taking global sections shows that $H^0(P, \Omega_{P/k}^1) = 0$ . Moreover, tensoring by $\mathcal{O}_P(-d)$ and passing again to cohomology shows that $H^1(P, \Omega_{P/k}^1(-d)) = 0$ since $H^0(P, \mathcal{O}_P(-d) = 0$ and $H^1(P, \mathcal{O}_P(-d-1)) = 0$ .

Consider now the exact sequence

$0 \to \Omega_{P/k}^1(-d) \to \Omega_{P/k}^1 \to i_* i^* \Omega_{P/k}^1 \to 0$

coming from tensoring $\Omega_{P/k}^1$ against the exact sequence defining the closed subscheme $Y \subset P$ . Taking cohomology gives the exact sequence

$H^0(P, \Omega_{P/k}^1) \to H^0(P, i_* i^* \Omega_{P/k}^1) \to H^1(P, \Omega_{P/k}^1(-d)).$

The two outside terms are $0$ by the above, so we see $H^0(Y, i^* \Omega_{P/k}^1) = H^0(P, i_* i^* \Omega_{P/k}^1) = 0$ .

2. $H^1(Y, \mathcal{O}_Y(-d)) = 0$ .

We have an exact sequence

$0 \to \mathcal{O}_P(-2d) \to \mathcal{O}_P(-d) \to i_* \mathcal{O}_Y(-d) \to 0$

coming from tensoring $\mathcal{O}_P(-d)$ against the exact sequence defining the closed subscheme $Y \subset P$ . This gives the exact sequence on cohomology

$H^1(P, \mathcal{O}_P(-d)) \to H^1(P, i_* \mathcal{O}_Y(-d)) \to H^2(P, \mathcal{O}_P(-2d)).$

As $P$ is a projective space of dimension at least 3, the two outside terms vanish and thus so does the inner term. Since $i$ is a closed embedding (hence in particular affine), we find

$H^1(Y, \mathcal{O}_Y(-d)) = H^1(P, i_*\mathcal{O}_Y(-d)) = 0$

as desired. This completes the proof of the Lemma.

Lemma. $\dim H^1(X, \mathcal{O}_X) = 1$ .

Proof. We will first prove the weaker claim that $H^1(X, \mathcal{O}_X)$ is nonzero, in order to introduce some cool results. This is not a necessary step, but the ideas will be used later to deduce that $\mathrm{Pic}_{X/k}^0 \cong \mu_p$ . Recall first the following amazing isomorphism. If $G$ is a finite commutative $k$ -group scheme and $X$ is an arbitrary proper flat connected $k$ -scheme, then

$H^1_{\mathrm{fppf}}(X, G) \cong \mathrm{Hom}_{k-\mathrm{gp}}(\mathbf{D}(G), \mathrm{Pic}_{X/k}),$

where $\mathbf{D}(G)$ denotes the Cartier dual of $G$ . There is an easy-to-define natural map from left to right, but it is not obvious that it is an isomorphism; for this see Milne’s Etale Cohomology, III, Prop. 4.16, which proves a more general statement. Taking $G = \mathbf{Z}/p\mathbf{Z}$ , we see that

$\mathrm{Hom}_{k-\mathrm{gp}}(\mu_p, \mathrm{Pic}_{X/k}) \cong H^1_{\mathrm{fppf}}(X, \mathbf{Z}/p\mathbf{Z}) \cong \mathrm{Hom}(\pi_1(X), \mathbf{Z}/p\mathbf{Z}).$

Note that this latter group is nonzero because $X$ admits a covering by $\mathbf{Z}/p\mathbf{Z}$ (namely, $Y$ !). In fact, by a Lefschetz theorem (see SGA2, Exp. XII, Cor. 3.5), since $Y$ is a hypersurface in the $3$ -dimensional simply connected variety $P$ , it follows that $Y$ is simply connected and thus $\pi_1(X) \cong \mathbf{Z}/p\mathbf{Z}$ .

Consider now the Artin-Schreier exact sequence of etale sheaves

$0 \to \mathbf{Z}/p\mathbf{Z} \to \mathcal{O}_X \to \mathcal{O}_X \to 0$

where the rightmost nonzero map is given by $t \mapsto t^p - t$ . Passing to etale cohomology gives

$H^0(X, \mathcal{O}_X) \to H^0(X, \mathcal{O}_X) \to H^1(X, \mathbf{Z}/p\mathbf{Z}) \to H^1(X, \mathcal{O}_X) \to H^1(X, \mathcal{O}_X).$

The leftmost map is surjective because $H^0(X, \mathcal{O}_X) = k$ and $k$ is algebraically closed. Thus we have the equality $\mathbf{Z}/p\mathbf{Z} = H^1(X, \mathbf{Z}/p\mathbf{Z}) = H^1(X, \mathcal{O}_X)^F$ (the superscript $F$ denoting Frobenius-invariants). In particular, $H^1(X, \mathcal{O}_X) \neq 0$ .

Now we prove the actual statement of the Lemma (using none of the preceding). Recall that the Hochschild-Serre spectral sequence of the $G = \mathbf{Z}/p\mathbf{Z}$ covering $Y \to X$ is given by

$E_2^{pq} = H^p(G, H^q(Y, \mathcal{O}_Y)) \Rightarrow H^{p+q}(X, \mathcal{O}_X).$

The exact sequence of low degree terms is given by

$0 \to H^1(G, H^0(Y, \mathcal{O}_Y)) \to H^1(X, \mathcal{O}_X) \to H^0(G, H^1(Y, \mathcal{O}_Y)).$

We have already seen that $H^1(Y, \mathcal{O}_Y) = 0$ . Moreover, $H^0(Y, \mathcal{O}_Y) = k$ and the action of $G$ on this vector space is trivial since all of the functions in it are constant. Thus by familiar properties of group cohomology we have $H^1(G, H^0(Y, \mathcal{O}_Y)) = \mathrm{Hom}(G, k) \cong k$ , using that $G = \mathbf{Z}/p\mathbf{Z}$ and $k$ is characteristic $p$ . Thus $H^1(X, \mathcal{O}_X) \cong k$ . QED

We have now shown $H^0(X, \Omega_{X/k}^1) = 0$ while $H^1(X, \mathcal{O}_X) \cong k$ . By our introductory remarks, it follows that $\mathrm{Pic}_{X/k}^0$ is $0$ -dimensional but has a nontrivial tangent space at the identity. Thus indeed it is a nontrivial finite connected group scheme, as desired.

Remark. In fact, we have $\mathrm{Pic}_{X/k}^0 \cong \mu_p$ : recall first that a finite connected $k$ -group is scheme-theoretically isomorphic to $\mathrm{Spec} k[x_1, \dots, x_m]/(x_1^{p^{n_1}}, \dots, x_m^{p^{n_m}})$ for some $m \geq 1$ and $n_i \in \mathbf{Z}_{> 0}$ . Since $H^1(X, \mathcal{O}_X)$ is 1-dimensional, it follows that for $\mathrm{Pic}_{X/k}^0$ we have $m = 1$ , and we need to show $n_1 = 1$ . In any case, we have shown that there is a nontrivial $k$ -homomorphism $\mu_p \to \mathrm{Pic}_{X/k}^0$ , so $\mathrm{Pic}_{X/k}^0$ contains $\mu_p$ as a closed $k$ -subgroup. It is clear $H = \mathrm{Pic}_{X/k}^0/\mu_p$ is a finite connected group scheme, and if it is nonzero then it has a $1$ -dimensional tangent space. Suppose for the sake of contradiction that $H$ is nontrivial, so by our above remarks $\mathrm{ker}(F_H)$ is a connected group scheme of order $p$ , hence isomorphic either to $\mu_p$ or $\alpha_p$ by the classification of group schemes of order $p$ over an algebraically closed field of characteristic $p$ . Let $K \subset \mathrm{Pic}_{X/k}^0$ denote the corresponding order $p^2$ closed subgroup, so that $K$ is an extension of either $\mu_p$ or $\alpha_p$ by $\mu_p$ . In the first case, $K$ must be either $\mu_{p^2}$ or $\mu_p \times \mu_p$ (since one can show using the Kummer sequence for $p$ that $\mathrm{Ext}^1(\mu_p, \mu_p)$ is isomorphic to $\mathrm{Hom}(\mu_p, \mathbf{G}_m) \cong \mathbf{Z}/p\mathbf{Z}$ ); in the second, $K$ is $\mu_p \times \alpha_p$ (since one can show via similar methods that $\mathrm{Ext}^1(\alpha_p, \mu_p) \cong \mathrm{Ext}^1(\mathbf{G}_a, \mathbf{G}_m)$ and the latter is trivial), so in particular $\mathrm{Pic}_{X/k}^0$ contains $\alpha_p$ as a closed $k$ -subgroup. But note

$\mathrm{Hom}(\mu_{p^2}, \mathrm{Pic}_{X/k}^0) = H^1(X, \mathbf{Z}/p^2\mathbf{Z}) = \mathbf{Z}/p\mathbf{Z}$

and

$\mathrm{Hom}(\mu_p^2, \mathrm{Pic}_{X/k}^0) = H^1(X, (\mathbf{Z}/p\mathbf{Z})^2) = (\mathbf{Z}/p\mathbf{Z})^2,$

using in both cases that $\pi_1(X) \cong \mathbf{Z}/p\mathbf{Z}$ . On the other hand, we have $\mathrm{Hom}(\mu_{p^2}, \mu_{p^2}) \cong \mathbf{Z}/p^2\mathbf{Z}$ and $\mathrm{Hom}(\mu_p^2, \mu_p^2) \cong \mathrm{GL}_2(\mathbf{Z}/p\mathbf{Z})$ , so $\mathrm{Pic}_{X/k}^0$ cannot contain $\mu_{p^2}$ or $\mu_p^2$ as a closed $k$ -subgroup. Moreover, $H^1(X, \mathcal{O}_X)$ is 1-dimensional and we saw that the Frobenius on $H^1(X, \mathcal{O}_X)$ is not identically 0 (since $\mathbf{Z}/p\mathbf{Z} \cong H^1(X, \mathcal{O}_X)^F$ ), so being semi-linear it must be injective. In particular, using the exact sequence of fppf sheaves $0 \to \alpha_p \to \mathcal{O}_X \to \mathcal{O}_X \to 0$ we deduce $H^1(X, \alpha_p) = 0$ . Because $\alpha_p$ is self-dual it follows that

$\mathrm{Hom}(\alpha_p, \mathrm{Pic}_{X/k}^0) = H^1(X, \alpha_p) = 0.$

This exhausts the possibilities for $K$ , showing that $H$ must be trivial, i.e., $\mathrm{Pic}_{X/k}^0 \cong \mu_p$ .

Question. Is there a good way to get a hold on $H^1(X, \alpha_p)$ in general? E.g., is there a good example of some $X$ such that this is nontrivial?

It remains only to construct the surface $Y$ in $\mathbf{P}_k^3$ with a free $\mathbf{Z}/p\mathbf{Z}$ -action. As before, let $p$ be a prime number greater than or equal to $5$ and let $G = \mathbf{Z}/p\mathbf{Z}$ . Let $V$ be the 4-dimensional $k$ -representation of $G$ defined by letting the canonical generator of $G$ act via the matrix

$\begin{pmatrix} 1 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 \end{pmatrix}.$

Note that this matrix is equal to $1 + N$ , where $N$ is a matrix whose fourth power is $0$ ; it follows that the $p$ th power of this matrix is indeed $1$ since $p \geq 5$ . This representation induces an action of $G$ on $k[x_0, x_1, x_2, x_3]$ respecting the natural grading, and by functoriality of the $\mathrm{Proj}$ construction this gives an action of $G$ on $P = \mathbf{P}_k^3$ whose only fixed point is $y = [0: 0: 0: 1]$ . The quotient $P/G$ in the sense of locally ringed spaces exists as a scheme (see SGA3, Exp. V, Thm. 4.1), and it is projective (see SGA3, Exp. V, Rem. 5.1), hence a closed subscheme of some projective space $Q$ . Moreover, $P/G$ is smooth away from the image $x$ of $y$ . By Bertini’s theorem, there is some hyperplane $H$ in $Q$ not containing $x$ such that $X = H \cap P/G$ is smooth and integral. If $Y$ is the preimage of $X$ in $P$ , then $Y$ is also smooth, being a $G$ -torsor over the smooth $X$ , and $G$ acts freely on $Y$ because $Y$ does not contain the unique fixed point $y$ . This completes the construction.

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David Benjamin Lim

3 years ago

I think there’s a small typo: When you discuss the Hochschild-Serre spectral sequence, the $E_2^{0,1}$ term should be $H^1(G, H^0(Y,\mathcal{O}_Y))$ .

With regards to your example of a scheme $X$ with non-trivial $H^1(X,\alpha_p)$ , would the following example work? Take $E/k$ a supersingular elliptic curve. Then $H^1(E, \mathcal{O}_E)$ is one-dimensional by Riemann-Roch. Now $H^0(E, \mathcal{O}_E) \simeq k$ and the frobenius is surjective on $k$ . Therefore $H^1(E, \alpha_p) \simeq H^1(E,\mathcal{O}_E)[\ker F]$ . But now $E$ is supersingular so $F = 0$ on $H^1(E, \mathcal{O}_E)$ , and hence $H^1(E, \alpha_p) \simeq k$ .

Last edited 3 years ago by David Benjamin Lim

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Sean Cotner

Ben, thanks for the correction and the example. Another way to see the equality you say is by noting that elliptic curves are self-dual and supersingular elliptic curves have torsion which is a nontrivial extension of $\alpha_p$ by $\alpha_p$ (this uses some Dieudonne module calculations), so the multiplication-by-p map $E \to E$ admits an intermediate $\alpha_p$ cover.

The question wasn’t very well-posed; it must be very hard, for example, to give a nice characterization of those abelian varieties over $k$ which admit $\alpha_p$ as a closed subgroup. Perhaps a better set of questions: are there any examples of a smooth projective connected variety over $k$ with $\mathrm{Pic}_{X/k}^0 \cong \alpha_p$ ? What about other finite group schemes $G$ of $p$ -power order? Are there obstructions to the existence of such $X$ in terms of $G$ ? If $G$ is of multiplicative type then I suspect (though I haven’t checked carefully) that you can run the same kind of construction above to construct such $X$ (of higher dimension), but the same methods don’t seem to apply to finite unipotent group schemes like $\alpha_p$ , where the fundamental group plays no role in understanding $G$ -covers.

Bogdan Zavyalov

Reply to Sean Cotner

There is a smooth projective connected variety with $\mathrm{Pic}^0_{X/k}\cong G$ for any connected finite commutative group scheme $G/k$ . Or, more generally, there is a smooth projective connected variety with $\mathrm{Pic}^{\tau}_{X/k}\cong G$ for any finite commutative group scheme $G/k$ .

Here is a rough idea of the construction:

1) Firstly consider the stack $X=BG=[*/G]$ and observe that $\mathrm{Pic}^{\tau}_{X/k} \simeq \mathrm{Hom}_{k-gp}(G, \mathbf{G}_m)=G^{\vee}$ by faithfully flat descent. So if we apply the construction to $X=BG^{\vee}$ , we get that $\mathrm{Pic}^{\tau}_{X/k} \simeq G$ .

This already gives you an example of a connected, smooth, proper Artin stack over $k$ with $\mathrm{Pic}^{\tau}\simeq G$ . If $G$ is connected then we get that $\mathrm{Pic}^{\tau}_{X/k} \simeq G$ is already connected, so $\mathrm{Pic}^0_{X/k} \simeq \mathrm{Pic}^{\tau}_{X/k} \simeq G$ .

2) Now the idea is to approximate $BG^{\vee}$ by $Z/G^\vee$ where $Z \subset \mathbf{P}^N$ is a complete intersection with a free action of $G^{\vee}$ . If $\mathrm{dim} Z>2$ then $\mathrm{Pic}^{\tau}_{Z/k} \simeq 0$ . So the same argument with faithfully flat descent would show that $\mathrm{Pic}^{\tau}_{(Z/G)/k} \simeq G$ . Similarly, if $G$ is connected we get

$\mathrm{Pic}^0_{(Z/G^\vee)/k} \simeq \mathrm{Pic}^{\tau}_{(Z/G^\vee)/k} \simeq G$

3) The only thing that is left is to construct $Z$ with an action of $G^\vee$ as above. The construction is similar to the one you explained in your post. One considers the standard action of $G^\vee$ on $\mathbf{P}\left(\mathcal O\left(\left(G^\vee\right)^3\right)^*\right)$ and shows that the locus, where the action is non-free, is of codimension at least $4$ . So you can run a similar argument as in the post to construct the desired