Is the theta characteristic attached to an etale double cover of a plane quintic arising from a cubic threefold even in characteristic two? We work over an algebraically closed field $k$ of characteristic $2$. Let $X$ be a cubic threefold realized as a conic bundle via $f: X \to \mathbb{P}^2$ (after blowing up some line). Let $C \to \mathbb{P}^2$ be a plane quintic curve describing the locus of degenerate conics of $f$, and let $\pi: \widetilde{C} \to C$ be the etale double cover parametrizing the lines on these degenerate conics. Let $H$ be the hyperplane section on $C$, and consider $M = \pi^*H$. I need to work out the dimension $h^0(M)$. If $k$ had characteristic not $2$ then we know $h^0(M) = 4$, since $h^0(M) = h^0(\pi_*M) = h^0(H) + h^0(H + \eta)$ where $\eta$ is some line bundle satisfying $\eta^2 \cong \mathcal{O}_C$, and we know $h^0(H) = 3$, and $h^0(H + \eta) = 1$ as in Beauville's paper on singularities of the Theta divisor.
However this approach does not work in characteristic $2$. Here we simply get a short exact sequence
$$0 \to \mathcal{O}_C \to \pi_*\mathcal{O}_{\widetilde{C}} \to \mathcal{O}_C \to 0$$
because in characteristic $2$ we deal with Artin-Schreier extensions. After twisting by $H$ and taking long exact sequence, this bounds $3 \leq h^0(M) \leq 6$ for us. It would suffice in my case to know that this quantity is even, so essentially I want to ensure that even in characteristic $2$, the theta divisor is $\textbf{even}$.
I know that $det(\pi_*M) = 2H$ has $h^0(2H) = 11$ since $C$ is genus $6$ and $2H$ is the canonical divisor, but I am not sure if this is of any use. I also tried to manually compute the global sections from cocycle data but that seems like a nightmare.
 A: I'll show the rank is at most 4 in characteristic 2, modulo some claims that I think can be justified.
Your exact sequence
$$0 \to \mathcal{O}_C \to \pi_*\mathcal{O}_{\widetilde{C}} \to \mathcal{O}_C 
\to 0$$
induces a long exact sequence on cohomology
$$0 \to H^0(C, \mathcal{O}_C(1) ) \to  H^0(C, \pi_*\mathcal{O}_{\widetilde{C}}(1) )\to  H^0(C, \mathcal{O}_C(1) ) \to H^1(C ,\mathcal O_C(1))$$
and it suffices to show that the rank of $H^0(C, \mathcal{O}_C(1) ) \to H^1(C ,\mathcal O_C(1))$ is at least $2$.
It's not hard to check that this map is given by cup product with the extension class of the extension $\pi_* \mathcal O_{\tilde{C}} $ of $\mathcal O_C$ with $\mathcal O_C$ in $\operatorname{Ext}^1(C,\mathcal O_C) = H^1 ( C, \mathcal O_C)  = H^0 (C, K_C)^\vee$.
Now by adjunction, $K_C = \mathcal O_C(2)$ and $H^0(C,  K_C) = H^0(C,\mathcal O_C(2) ) = H^0( \mathbb P^2, \mathcal O_{\mathbb P^2}(2))$ is the space of quadratic polynomials in three variables. So the extension class is a linear form on quadratic polynomials in three variables, i.e. a three-by-three symmetric matrix.
The cup product $H^0(C, \mathcal O_C(1)) \times H^0 (C, K_C)^\vee \to H^1(C,\mathcal O_C(1)) = H^0(C, K_C(-1))$ is dual to the product map $H^0(C,\mathcal O_C(1)) \times H^0(C, K_C(-1)) \to H^0(C,K_C)$ so the rank of the cup product map is equal to the rank of this symmetric matrix. So it suffices to prove the symmetric matrix can't have rank $\leq 1$.
The symmetric matrix is certainly nonzero as the extension is nontrivial, so it suffices to prove it can't have rank exactly $1$. By $GL_3$-symmetry, we can rule out a specific $3\times 3$ matrix, i.e. $\begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$, that corresponds to the linear form on quadratic polynomials that extracts the coefficient of $x^2$.
What do we know about this extension class? By the Artin-Schreier exact sequence $H^1(C, \mathbb F_2)  \to H^1(C,\mathcal O_C) \to H^1(C, \mathcal O_C)$, it must be stable under the Frobenius map $H^1(C,\mathcal O_C) \to H^1(C,\mathcal O_C)$, which is dual to the Verschiebung map $H^0(C, K_C) \to H^0(C,K_C)$.
How does Verschiebung act on quadratic polynomials in $x,y,z$? I claim it acts by multiplying by the degree $5$ polynomial $f$ defining your curve, then ignoring all monomials where the exponent of $x,y$, or $z$ is even, then dividing by $xyz$, so all the exponents are even, and then taking the square root. This sends a degree $2$ polynomial to a degree $\frac{ 2+5 - 3}{2} =2$ polynomial.
If the linear form that extracts the coefficient of $x^2$ is stable under Frobenius, then the coefficient of $x^4 \cdot xyz$ in $fq$ must match the coefficient of $x^2$ in $q$ for an arbitrary quadratic polynomial $q$. This can only happen if the coefficients of $x^5, x^4y , x^4z$ in $f$ all vanish, since otherwise we could multiply by $yz, xz, xy$ respectively and get a contradiction. But this forces the vanishing locus of $f$ to have a singularity at $(1:0:0)$, contradicting the assumption that it is smooth.
So indeed the rank of the matrix is $\geq 2$, making the rank of global sections $\leq 4$.
