Simplest example of jumping of cohomology of structure sheaf in smooth families? Using Hodge theory (and the ill-defined Lefschetz principle), one can show that in characteristic 0, given a proper smooth family $X \rightarrow B$, the cohomology groups of the structure sheaf of the fibers are locally constant (as a function on $B$).  I'm aware of the existence of a number of counterexamples to the corresponding statement in positive characteristic.  Given the success of this question, I want to ask:  

What are the simplest examples of a proper smooth family exhibiting jumping of some cohomology group of the structure sheaf?

By the above discussion, such an example will necessarily be in positive characteristic.
By "simplest", I mean by one of the following measures.
(best) An example whose proof is as elementary as possible, and ideally short. 
An example with a simple conceptual underpinning.  (Well, the best answer would do well by both of the first two measures.)
A known example that is simple to state, but may have a complicated proof. (Ideally there should be a reference.)
An expected, folklore, or conjectured example.
 A: This is an attempt to realise Sándor's program of getting an example based on
Kodaira vanishing or non-vanishing varying in a family. It will be done by
keeping the surface fixed but varying the line bundle.
I shall start not with the examples of Raynaud but rather a variation given by
Raynad-Szpiro (for details on it see Szpiro's article in Astérisque 64 or
Flexors exposé in Astérisque 86). Recall that a Tango-Raynaud structure on a
relative smooth and proper curve $f\colon X\to C$ consists of a line bundle
$\mathcal L$ on $X^{(p)}$ together with a map $\mathcal L\to B^1_{X/C}$, where
$B^1_{X/C}\subseteq \alpha_\ast\Omega^1_{X^{(p)}/C}$ is the image of
$d\colon\alpha_\ast\mathcal O_{X}\rightarrow\alpha_\ast\Omega^1_{X/C}$ (and
$\alpha\colon X\to X^{(p)}$ is the relative Frobenius), such that the adjoint
map $\alpha^\ast\mathcal L\rightarrow \Omega^1_{X/C}$ is an isomorpism. If $C$
is a proper smooth curve and the fibration is non-isotrivial, then such an
$\mathcal L$ is ample as $\Omega^1_{X/C}$ is by Szpiro (and $\alpha$ is finite)
and on the other hand, by looking at the exact sequence
$0\rightarrow\mathcal O_X\rightarrow\alpha_\ast\mathcal O_{X^{(p)}}\rightarrow
B^1_{X/C}\rightarrow0$ tensored with $\mathcal L^{-1}$ we get an embedding
$H^0(X,B^1\bigotimes\mathcal L^{-1})\hookrightarrow
H^1(X^{(p)},\mathcal L^{-1})$ and thus the given map $\mathcal L\to B^1_{X/C}$
gives a non-zero element of $H^1(X^{(p)},\mathcal L^{-1})$. The map
$X\rightarrow C$ is constructed by the Kodaira procedure (using that if the
second starting curve $D$ is Tango-Raynaud, then so is $f$).
The aim is now to show that there exists an
$\mathcal{M}\in\mathrm{Pic}^0(X^{(p)})$ such that $H^1(X^{(p)},\mathcal
L\bigotimes \mathcal{M})=0$. Of course, for every $\mathcal{M}$, $\mathcal
L\bigotimes \mathcal{M}$ is ample and $\mathrm{Pic}^0(X^{(p)}$ is connected so
we would be finished if we could do this. Note that $\mathrm{Pic}^0(X^{(p)})$ is positive
dimensional (as it contains a subgroup isogenous to $\mathrm{Pic}^0(C)\times
\mathrm{Pic}^0(D)$) and I shall in fact show that the vanishing is true for all
but a finite number of $\mathcal{M}$'s.
I shall use $\mathcal L'$ to denote a general element of the form
$\mathcal L\bigotimes\mathcal{M}$, where
$\mathcal{M}\in\mathrm{Pic}^0(X^{(p)})$.  


*

* 
We have that $H^1(X^{(p)},\mathcal L'^{-p^n})=0$ for $n>0$. This is proven by
descending induction on $n$, the statement being true for large $n$ by Serre and
ampleness of $\mathcal L'$. It is therefore enough to show that the $p$'th
power map $H^1(X^{(p)},\mathcal L'^{-p^n})\rightarrow
H^1(X^{(p)},\mathcal L'^{-p^{n+1}})$ is injective. This is the map induced by
adjunction $\mathcal L''\rightarrow F_\ast F^\ast\mathcal L''$, where $F\colon
X\rightarrow X$ is the absolute Frobenius (on both $X$ and $X^{(p)}$) and
$\mathcal L'':=\mathcal L'^{-p^n}$. We prove this by factoring $F$ as
$\alpha\circ\beta$, where $\beta\colon X^{(p)}\rightarrow X$ is the base change
of the Frobenius on $C$. Hence it will be enough to show that
$H^1(X^{(p)},\mathcal L'')\rightarrow H^1(X,\alpha^\ast\mathcal L'')$ and
$H^1(X,\alpha^\ast\mathcal L'')\rightarrow
H^1(X,\beta^\ast\alpha^\ast\mathcal L'')$ are both injective. For the first we
have that its kernel is equal to $\mathrm{Hom}(\mathcal L'',B^1_{X/C})$ and an
element of it gives (by inclusion and adjunction as above) rise to a map
$\alpha^\ast\mathcal L''\rightarrow \Omega^1_{X/C}\cong
\alpha^\ast\mathcal L$. Such a map is zero as because of the ampleness of
$\mathcal L$, $\mathcal L''$ is more positive than $\mathcal L$.

As for the injectivity of $H^1(X,\alpha^\ast\mathcal L'')\rightarrow
H^1(X,\beta^\ast\alpha^\ast\mathcal L'')$ we get similarly that the kernel is
equal to $\mathrm{Hom}(\alpha^\ast\mathcal L'',f^{\ast}B^1_{C})$. However,
$\alpha^\ast\mathcal L''$ has strictly positive degree on the fibres of $f$ so
all such maps are zero.


* 
Hence to show that $H^1(X^{(p)},\mathcal L'^{-1})=0$ for general $\mathcal{M}$
it would be enough to show that for such $\mathcal{M}$ the $p$'th power map
$H^1(X^{(p)},\mathcal L'^{-1})\rightarrow H^1(X^{(p)},\mathcal L'^{-p})$ is
injective. Again we can factor it and the second part of the argument works as
before so it will be enough to show that there are only a finite number of
$\mathcal{M}$'s for which $\mathrm{Hom}(\mathcal L',B^1_{X/C})$ could be
non-zero. As for $\mathcal L$ a non-zero such map would give a non-zero map
$\alpha^\ast\mathcal L'\rightarrow \Omega^1_{X/C}\cong
\alpha^\ast\mathcal L$. However, $\mathcal L'$ is numerically equivalent to
$\mathcal L$ and hence the map would have to be an isomorphism. This implies
that $\mathcal{M}$ would have to lie in the kernel of $\alpha^\ast$ which is
finite as $\alpha$ is finite flat (and surjective).


A: Here is an idea of how one (let's say I) might try to construct such an example. I think this idea, if it works, might qualify under rule #2. Also, it is possible that the answer Matt gave on MSE help fill the gap in this. Even though this is not a complete answer I think it still might be useful.
So, let's say that $X$ is a smooth projective variety and $\mathscr L$ an ample line bundle on $X$ such that $H^1(X,\mathscr L^{-1})\neq 0$. This can obviously happen only in positive characteristic by Kodaira vanishing, but it can happen there. I think Raynaud was the first to give examples of this (in 1978?) and Ekedahl showed that this can even happen with $\mathscr L=\omega_X$ in characteristic $2$. Others gave various examples for this to happen, I think maybe including Mukai.
Anyway, suppose that such an $X$ comes in a polarized family (that is, there is a relatively ample line bundle on the total space that restricts to $\mathscr L$ on $X$) where the general fiber does not have satisfy this (for any power of the line bundle that's the deformation of $\mathscr L$). I am not sure if this is an outrageous expectation, but I am kind of thinking that the failure of Kodaira vanishing is special and so a general deformation will not be a counter-example to Kodaira vanishing, in other words Kodaira vanishing holds on it. One potential problem I see with this is that it is possible that such an $\mathscr L$ would perhaps not deform, so I could not reasonably take a "general" deformation. Well, actually if one uses $\mathscr L=\omega_X$ as in Torsten's example, then it definitely deforms! I am sure that Torsten will read this and tell me why I am wrong. :)
In any case, what I need is just that $\dim H^1(X,\mathscr L^{-1})$ would jump and this still seems an easier task than the original since $\mathscr L$ is undetermined, so one has a little more freedom.
So, if we have that, then we're in business. Take a general section of a high power of the global line bundle that restricts to $\mathscr L$ and take the finite cover it determines. This finite cover has the property that the cohomology of $\mathscr O$ of the fibers of this cover over the original base is the sum of the cohomologies of the negative powers (up to the degree minus one) of the deformations of $\mathscr L$. The assumption implies that either the original family gave you an example or there will be a jump in this one.
A: By the way, for Enriques this more than just an example - it reflects their classification: the moduli space of Enriques surfaces is connected in any characteristic. All such surfaces arise as desingularizations of quotients $Y/G$, where Y is a (possibly singular) complete intersection of $3$ quadrics in $\mathbb{P}^5$, and $G$ is a finite flat group scheme of length $2$.
In characteristic $p\neq2$, all Enriques surfaces satisfy $h^{01}=h^{10}=0$, and their moduli space is irreducible. Also, over an algebraically closed field of characteristic $p\neq2$, the only possibility for $G$ is $\mathbb{Z}/2\mathbb{Z}$, which is isomorphic to $\mu_2$.
In characteristic $p=2$, we have $h^{01}=0$, $h^{10}=1$ if the surface arises as a quotient by $G=\mu_2$ ("classical"), we have $h^{10}=h^{01}=1$ if $G=\alpha_2$ ("supersingular"), and finally, we have $h^{01}=1, h^{10}=0$ if $G=\mathbb{Z}/2\mathbb{Z}$ ("singular"). The moduli space is connected, but has two irreducible components: one corresponds to $\mu_2$-quotients, the other to $\mathbb{Z}/2\mathbb{Z}$-quotients, and their intersection to $\alpha_2$-quotients. 
Thus, the Hodge numbers reflect the position in the moduli space, and a jumping corresponds to a change in type.
A: One example is given by Enriques surfaces in characteristic $2$. There are three types depending on the value of $\mathrm{Pic}^\tau$ (as a group scheme) which can be either $\mathbb Z/2$, $\mu_2$ or $\alpha_2$. In the first case $\omega_X$ is the generator so in particular it is non-trivial and $H^2(X,\mathcal O_X)=0$ (using of course Serre duality). In the two other cases  $\omega_X$ is trivial (as it is numerically trivial and all numerically trivial line bundles are trivial) so that $h^2(X,\mathcal O_X)=1$. Now, $\alpha_2$ can be deformed to both $\mathbb Z/2$ and $\mu_2$ and such deformations can be lifted to deformations of Enriques surfaces (in fact $\mathrm{Pic}^\tau$ is flat in families of Enriques surfaces and the functor from deformations of the surfaces to those of $\mathrm{Pic}^\tau$ is formally smooth - Liedtke: arXiv:1007.0787, Ekedahl-Shepherd-Barron: unpublished). If we pick a connected family of Enriques surfaces with some special value being an $\alpha_2$-surface and generically $\mathbb Z/2$, then we get an example.
Such an example can be constructed (very) explicitly without deformation theory. Here is a semi-explicit construction which works in any positive characteristic. Fixa a a group scheme $A$ of order $p$ over $\mathbb A^1$ localised at $0$ (say) which is $\alpha_p$ at $0$ and $\mathbb Z/p$ elsewhere. By the Godeaux construction (which Raynaud - Prop. 4.2.3, p-torsion du schema de Picard, Astérisque 64 - showed works for such families) there is a free action of $A$ on a flat complete intersection $Y$ (of any dimension, which we assume is $\ge 2$) such that $X=Y/A$ is smooth (note that contrary to the case of an étale group scheme $Y$ will not be smooth). As $Y$ is a complete intersection we have that $\mathrm{Pic}^\tau_Y=0$ and from that it follows that $\mathrm{Pic}^\tau_X=A$. Now, $H^1(-,\mathcal O_-)$ is the tangent space of  $\mathrm{Pic}^\tau$ so it is zero outside of $0$ and $1$-dimensional at $0$. (By being careful one can get Enriques surfaces for $p=2$, this I guess was the inspiration for the Godeaux construction).
