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Consider the surface $X \subset \mathbb{P}^2_{x, y, z}\times\mathbb{P}^1_{t, s}$: $$ s^3y^2 + t^3yz = (t+s)x^2 + tsxz + t(t^2 +s^2)z^2 $$ over a finite field $k = \mathbb{F}_{2^d}$, $\mathrm{gcd}(d, 6) = 1$.

We have the conic bundle $pr\!: X \to \mathbb{P}^1_{t,s}$. The surface $X$ has the unique singular point, which lays on the unique non-reduced fiber $X_0 = pr^{-1}(0:1)$. If we reduce this fiber, then we'll get a surface $Z$ over $k$ with degenerate fibers for $t = 0, w, w^2, \zeta, \zeta^2, \zeta^4$, where $w^2 + w + 1 = 0$ and $\zeta^3 + \zeta + 1 = 0$. It is easily seen that we can simultaneously blow down fibers $Z_{\zeta}$, $Z_{\zeta^2}$ and $Z_{\zeta^4}$ over $k$. We get a surface $Y$ over $k$ with 3 degenerate fibers for $t = 0, w, w^2$.

Iskovskikh theorem states that $Y$ is the del Pezzo surface of degree 5. In that case the anticanonical linear system $|\!-\!K_Y|$ gives the embedding of $Y$ to $\mathbb{P^5}$.

Questions:

  1. What is a canonical divisor $K_Y$ of $Y$?

  2. How can I find a $k$-basis of $H^0(Y, -K_Y)$ as functions from variables $x$, $y$, $z$, $t$, $s$?

I know the surfaces $X$ and $Y$ are birationally equivalent over $k$ to the Kummer surface of the Jacobian of genus 2 curve $C_b$: $$ y^2 + y = x^5 + x^3 + b, $$ where $b \in \mathbb{F_2}$. This is a motivation of my questions.

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    $\begingroup$ Any del Pezzo surface of degree $5$ with a conic bundle can be embedded as a surface of bidegree $(2,1)$ into $\mathbb{P}^2 \times \mathbb{P}^1$. It might be easier to calculate this model instead. You can then read off the anticanonical bundle using the adjunction formula, from which you can calculate the global sections. $\endgroup$
    – Daniel Loughran
    Commented Oct 14, 2016 at 14:22
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    $\begingroup$ @potentiallydense, the characteristics is even and the curve $C_b$ is supersingular. By Katsura's article On Kummer surfaces in characteristic 2 the Kummer surface is rational. $\endgroup$
    – Dimitri Koshelev
    Commented Oct 14, 2016 at 14:31
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    $\begingroup$ The anticanonical bundle of $\mathbb{P}^2 \times \mathbb{P}^1$ is $\mathcal{O}(3,2)$. Applying the adjunction formula, one finds the anticanonical bundle of $Y$ to be $\mathcal{O}(1,1)$. A basis for the global sections is therefore given by $\{sx,sy,sz,tx,ty,tz\}$. This gives the required embedding in $\mathbb{P}^5$. $\endgroup$
    – Daniel Loughran
    Commented Oct 14, 2016 at 15:15
  • $\begingroup$ I don't understand. You said that a professor already found this embedding for you. What remains to be done? $\endgroup$
    – Daniel Loughran
    Commented Oct 14, 2016 at 15:56
  • $\begingroup$ @DanielLoughran, I am sorry. I misinformed you. I don't know (2,1)-embedding to $\mathbb{P}^2\times\mathbb{P}^1$. Do you have ideas how to compute this embedding? $\endgroup$
    – Dimitri Koshelev
    Commented Oct 16, 2016 at 14:13

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