These are some questions concerning Mumford's "Lectures on curves on an algebraic surface".


We concern ourselves with questions of the Picard variety $P$, and its dimension, of a complete nonsingular surface $F$ over an algebraically closed ground field $k$ of arbitrary characteristic. Let $\mathfrak{o}$ be the sheaf of local rings on $F$, $\{C\_\alpha: \alpha \in S\}$ a family of curves on $F$ such that $C\_0 = C$ is nonsingular and $N$ the sheaf of sections of the normal bundle to $C$ on $F$. Then the equality (A): $\dim P = h^1(\mathfrak{o})$ is equivalent to (B): $\{\text{Tangent space to }S \text{ at }\alpha = 0\} \overset{\rho}{\to} H^0(N)$ is surjective for suitable $\{C\_\alpha : \alpha \in S\}$.


I was wondering if someone could give me their explanation/intuitions behind the following, as I am finding the book to be quite terse.


1. The algebraic solution of problem (A) for characteristic $0$, following Grothendieck, using a theorem of Cartier on an algebraic group scheme.
2. Problem (A) in characteristic $p$, i.e. $\dim P$ may not be given in general by $h^1(\mathfrak{o})$ but that the tangent space to $P$ (at a point) corresponds to the subspace of $H^1(\mathfrak{o})$ annihilated by the Bockstein operators.


Thanks in advance!



Intuition behind results in Mumford's "Lectures on curves on an algebraic surface", I
I think I can provide some intuition for (A), both in characteristic $0$ and $p > 0$. What follows below is more or less a proof, but with a lot of omissions (and hopefully not too many lies...).


I believe that everything I state works for all (geometrically) integral projective $k$-schemes. For simplicity, let's assume $k = \bar k$.


**Reference.** A great reference for the Grothendieck-style approach to this theory is *FGA Explained*, chapter 9, or *Néron Models*, chapter 8. (For general discussion of deformation theory and Hilbert and Quot schemes, I would recommend FGA Explained, whereas for the actual construction of the Picard scheme maybe Néron Models is a bit more readable.)




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For a $k$-scheme $X$, we can view the tangent space at a closed point $x \to X$ as the extensions
$$x \to \operatorname{Spec} k[\varepsilon]/\varepsilon^2 \to X.$$
That is, it is the preimage of $x \in X(k)$ under the map $\phi\colon X(k[\varepsilon]/\varepsilon^2) \to X(k)$. If $X$ is a group scheme, then $\phi$ is a group homomorphism, and the tangent space at the identity corresponds to the kernel of $\phi$.


Now apply this to the Picard group scheme $\operatorname{\underline{Pic}}\_X$. This is usually defined as the fppf sheafification of the functor $Y \mapsto \operatorname{Pic}(X\times Y)/\operatorname{Pic}(Y)$, but in the projective case it suffices to take the étale sheafification; the big difficult theorem is that this is representable by a scheme. (I believe that the sheafification takes care of dividing out by $\operatorname{Pic}(Y)$, so you could omit that step if you want to.)


Thus, we have to compute line bundles on $X\_\varepsilon := X \times \operatorname{Spec} k[\varepsilon]/\varepsilon^2$ that are trivial on the central fibre $X$. (Since $k[\varepsilon]/\varepsilon^2$ has no étale covers, we don't need to worry about the sheafification business.) On $X\_\varepsilon$, we have a short exact sequence
$$0 \to \mathcal O\_X \to \mathcal O\_{X\_\varepsilon} \to \mathcal O\_X \to 0,$$
coming from $0 \to k \to k[\varepsilon]/\varepsilon^2 \to k \to 0$. Tensoring with a line bundle $\mathcal L$ trivial on $X$ gives
$$0 \to \mathcal O\_X \to \mathcal L \to \mathcal O\_X \to 0.$$
Conversely, any such extension can be given the structure of line bundle on $X\_\varepsilon$ by having $\varepsilon$ act by the composition $\mathcal L \to \mathcal O\_X \to \mathcal L$. Hence, we are classifying extensions of $\mathcal O\_X$ by itself. These are given by
$$\operatorname{Ext}^1\_X(\mathcal O\_X,\mathcal O\_X) = H^1(X,\mathcal O\_X).$$
Thus, we have proven:



> 
> **Theorem.** The tangent space $T\_0 \operatorname{\underline{Pic}}\_X$ equals $H^1(X,\mathcal O\_X)$. $\square$
> 
> 
> 


In characteristic $0$, any finite type group scheme is smooth, so we get the equality of dimensions. In characteristic $p > 0$, there are non-smooth group schemes, the simplest example being
$$\mu\_p = \operatorname{Spec} k[x]/(x^p-1) \subseteq \mathbb G\_m.$$
In those cases, you actually get the wrong dimension:
$$\dim H^1(X,\mathcal O\_X) \geq \dim \operatorname{\underline{Pic}}\_X,$$
with strict inequality if $\operatorname{\underline{Pic}}\_X$ is singular.


