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1

An explicit example is the Jacobian of the hyperelliptic genus-$3$ curve in characteristic two, $Y^2+Y=X^7$. The vertices of the polygon are $(0,0)$, $(3,1)$, and $(6,3)$, giving slopes of $1/3$ and $2/3$.

1

1) Yes. If there is another one, it differs from $L'$ by a line bundle $M$ with $M^{2}\cong \mathcal{O}_Y$. Consider the resolution $\pi :\hat{Y}\rightarrow Y$ obtained by blowing up the double points $p_1,\ldots ,p_{16}$. Since $\hat{Y}$ is simply connected, we have $\pi^* M\cong \mathcal{O}_{\hat{Y}}\$. Thus $M_{|Y\smallsetminus \{p_i\}}$ is trivial, ...

9

No, this is another entry in the list of ways in which elliptic curves can be a poor guide to the higher-dimensional case. The kernel of $F_{A/k}:A \rightarrow A^{(p)}$ is always contained in $A[p]$ since $\ker F_{A/k}$ is an infinitesimal commutative group scheme whose own Frobenius morphism vanishes (and all such are killed by $p$). The $p$-rank being 0 ...

5

In good cases, the ''isomorphism functor'' $\mathrm{Isom}(X,Y)$ is representable by a scheme. Hence, you are asking that if this scheme is non-empty (i.e. contains a geometric point), whether it contains a solvable point. So your problem is more-or-less equivalent to the usual open problem that every variety contains a solvable point. Now for your special ...

2

1) is correct, 2) is not. Indeed, if $i(C)=C$, then the map $C \rightarrow C'$ is a double cover, and $f^*{\mathcal O}_Y(C')={\mathcal O}_X(C)$ since in a neighbourhood of a general point of $C$ the map $X \rightarrow Y$ is biregular. For a double cover $X\rightarrow Y$, in an analogous situation, you get $f^*{\mathcal O}_Y(C')={\mathcal O}_X(2C)$ when $C$ ...

4

We have $X=\text{Jac}(C)$, for $C$ some hyperelliptic curve. By appropriate choice of basepoint, we can arrange that the embedding $C\to X$ passes through $0\in X$ and is that $[-1]$ on $X$ restricts to the hyperelliptic involution on $C$. Let $C'=[2]^{-1}(C)$. I claim that this curve satisfies all of the desired properties. It is (a) smooth, as $[2]$ is ...

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