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Crossposted from MSE.

I am interested what is the type of the surfaces over the rationals $$ x^5 - y^5 + z^2 + x=0$$

and

$$ x^5 - y^5 + z^2 + x+1=0$$

Magma's KodairaEnriquesType(S : CheckADE:=true); fails to compute it.

According to Magma they are not rational.

Partial answers (e.g. it is not $X$) or approaches how to compute with a CAS are welcome.

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  • $\begingroup$ I would say that both are $K3$: they are double planes ramified at sextics (the order $5$ curves in $x$, $y$ and the line at infinity). One should only check carefully that these sextics have simple singulrities. $\endgroup$
    – Alex Degtyarev
    Commented Aug 19, 2015 at 8:59
  • $\begingroup$ @AlexDegtyarev Thank you, my wild guess was this. If you don't plan to do it would you please give the sextics to check them? $\endgroup$
    – joro
    Commented Aug 19, 2015 at 9:01
  • $\begingroup$ Just drop $z$ from the equations, and add the line at infinity to the resulting affine quintic. Say, the first one is obviously nonsingular in the affine part, but you should also check the singularities at infinity. $\endgroup$
    – Alex Degtyarev
    Commented Aug 19, 2015 at 9:04
  • $\begingroup$ The first one would be $z_0z_1^5-z_0z_2^5+z_0^5z_1=0$ in homogeneous coordinates. $\endgroup$
    – Alex Degtyarev
    Commented Aug 19, 2015 at 9:05
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    $\begingroup$ For Magma, somehow example H119E12 should give the idea of how to resolve singularities, but it is beyond my ability currently. magma.maths.usyd.edu.au/magma/handbook/text/1355#14997 $\endgroup$
    – kantelope
    Commented Aug 19, 2015 at 11:19

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Both are $K3$: they are double planes ramified at sextics (the order $5$ curves in $x$, $y$ and the line at infinity). One should only check carefully that these sextics have simple singulrities. I use the coordinates $x=z_1/z_0$, $y=z_2/z_0$ and multiply by $z_0$ (the line at infinity).

The former: $z_0z_1^5-z_0z_2^5+z_0^5z_1=0$. Obviously, there are no singular points in the affine part ($z_0\ne0$), and the residual quintic intersects the component $z_0=0$ at five distinct points, thus forming five $A_1$, which are all simple.

The latter: $z_0z_1^5-z_0z_2^5+z_0^5z_1+z_0^6=0$. The same conclusion applies literally.

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  • $\begingroup$ Isn't $(0:1:1)$ double singularity on your first curve? $\endgroup$
    – joro
    Commented Aug 19, 2015 at 13:25
  • $\begingroup$ Yes, it is. Once again, the double plane ramified at a sextic is a $K3$ iff all singular points of the sextic, if any, are simple (= $ADE$ = $0$-modal =?= rational double points + 12 more definitions/names). $A_1$ is the simplest simple singularity :) $\endgroup$
    – Alex Degtyarev
    Commented Aug 19, 2015 at 13:45
  • $\begingroup$ Of course, I presume that you compactify your surface and resolve the singularities. Since, in your case, the affine part chosen seems nonsingular, the compactification can also be chosen nonsingular. $\endgroup$
    – Alex Degtyarev
    Commented Aug 19, 2015 at 13:47

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