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Let $f(x,y,z)$ be the degree $6$ polynomial:

x^6 + 6*x^5 + 15*x^4 - 3*x^2*y^2 + 20*x^3 - 18*x^2*y - 6*x*y^2 - 2*y^3 - 12*x^2 - 36*x*y - 21*y^2 + 4*z^2 - 48*x - 72*y - 80

I am interested if the surface $f(x,y,z)=0$ is rational.

We have $f(u+v-1,u^2+v^2-3,u^3-v^3)=0$, which is rational parametrization.

I believe this is sufficient condition the surface to be rational by definition of rational variety.

Magma claims it is not rational.

Is it rational or not?

Magma online code http://magma.maths.usyd.edu.au/calc/:

K<x,y,z,t>:=ProjectiveSpace(Rationals(),3);
p:=x^6 + 6*x^5*t + 15*x^4*t^2 - 3*x^2*y^2*t^2 + 20*x^3*t^3 - 18*x^2*y*t^3 - 6*x*y^2*t^3 - 2*y^3*t^3 - 12*x^2*t^4 - 36*x*y*t^4 - 21*y^2*t^4 + 4*z^2*t^4 - 48*x*t^5 - 72*y*t^5 - 80*t^6;
S:=Surface(K,p);

IsRational(S : CheckADE := true); //return "false"
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  • $\begingroup$ "I believe this is sufficient condition the surface to be rational by definition of rational variety." No, not by definition but by a theorem of Castelnuovo, which implies that for a surface "unirational" is the same as "rational" (in characteristic zero). $\endgroup$ Commented Aug 19, 2015 at 9:52
  • $\begingroup$ @GeorgesElencwajg Thank you. So even if my reasoning is wrong, I still suppose it is rational, which implies Magma bug? $\endgroup$
    – joro
    Commented Aug 19, 2015 at 10:11
  • $\begingroup$ Did you check that the parametrization is correct with Magma? $\endgroup$ Commented Aug 19, 2015 at 10:39
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    $\begingroup$ @GeorgesElencwajg: it is not smooth --- in fact it has a 1-dimensional locus of singularities. (I used Macaulay2 to check this.) joro: what do you mean you are "sceptical that it matters"? As Georges says, if it were smooth, it would be guaranteed to be nonrational. (It isn't, but that's not the point here.) It is very off-putting to read comments like this. $\endgroup$
    – Schemer
    Commented Aug 19, 2015 at 13:58
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    $\begingroup$ Joro: yes, this seems to be a bug in Magma. I am a little surprised, because magma.maths.usyd.edu.au/magma/handbook/text/1356#15017 claims explcitly to be able to work with arbitrary surfaces in $\mathbf P^3$. $\endgroup$
    – Schemer
    Commented Aug 19, 2015 at 16:33

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