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

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


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

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
2 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)).

David Benjamin Lim
2 years ago

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 2 years ago by David Benjamin Lim
Bogdan Zavyalov
2 years ago
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

    \[ Z \subset \mathbf{P}\left(\mathcal O\left(\left(G^\vee\right)^3\right)^*\right) \]

of dimension at least 3. Then Z/G^\vee does the job.

Of course, I skipped many details. I think I’ve worked out something similar a couple of years ago, so I can send you the notes if you want to.

Last edited 2 years ago by Bogdan Zavyalov