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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"
joro
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  • "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). – Georges Elencwajg Aug 19 '15 at 09:52
  • @GeorgesElencwajg Thank you. So even if my reasoning is wrong, I still suppose it is rational, which implies Magma bug? – joro Aug 19 '15 at 10:11
  • Did you check that the parametrization is correct with Magma? – Georges Elencwajg Aug 19 '15 at 10:39
  • @GeorgesElencwajg Not yet, but I checked it with Sage and Maple. – joro Aug 19 '15 at 10:43
  • That certainly suffices: I can't imagine both systems making a mistake on such a trivial calculation. One possible explanation could be that $f$ is not irreducible and that the parametrization only describes one irreducible component of the surface $f=0$. Did you check the irreducibility of $f$ ? – Georges Elencwajg Aug 19 '15 at 11:00
  • @GeorgesElencwajg Certainly. It is irreducible over the rationals according to both Sage and Maple and absolutely irreducible according to Maple (don't know how to check this in sage). – joro Aug 19 '15 at 11:02
  • A smooth surface in $\mathbb P^3$ of degree $\geq 4$ is guaranteed to be non rational. Did you check smoothness of the closure in $\mathbb P^3$ (which has degree six) of your affine surface? – Georges Elencwajg Aug 19 '15 at 11:55
  • @GeorgesElencwajg No, I didn't check this. Why should I check it? – joro Aug 19 '15 at 12:14
  • Because smooothness would guarantee that your surface is not rational, as I explained in my previous comment. – Georges Elencwajg Aug 19 '15 at 12:26
  • @GeorgesElencwajg Due to ignorance I don't know how to compute it. But am still sceptical that it matters in this case. – joro Aug 19 '15 at 13:29
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    @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. – Schemer Aug 19 '15 at 13:58
  • @Relapsarian So likely this is technical bug in Magma and not in paper they tried to implement? – joro Aug 19 '15 at 15:48
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    Joro: yes, this seems to be a bug in Magma. I am a little surprised, because http://magma.maths.usyd.edu.au/magma/handbook/text/1356#15017 claims explcitly to be able to work with arbitrary surfaces in $\mathbf P^3$. – Schemer Aug 19 '15 at 16:33
  • @Relapsarian To my surprise Magma does some singular rational surfaces right... – joro Aug 20 '15 at 07:00

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