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I have a polynomial of degree 8 in 6 variables given explicitly by

$$ (\sqrt{1+(x_1+x_2+x_3)^2+(y_1+y_2+y_3)^2}+\sqrt{1+x_1^2+y_1^2}+\sqrt{1+x_2^2+y_2^2}+\sqrt{1+x_3^2+y_3^2})\times\text{the other seven of its Galois conjugates}. $$

I fed it into Maple and it shows it's irreducible. But being unsure of what is going on behind the scene, I am asking here for a less computation-intensive and more conceptual proof of this fact, in the sense that any computation involved in the proof can be checked by hand in a reasonable amount of time.

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  • $\begingroup$ irreducibility follows from smoothness, i.e. a smoothness certificate would do. I'd also try checking irreducibility on a couple of other computer algebra systems, just in case... $\endgroup$
    – Dima Pasechnik
    Commented Feb 27, 2017 at 7:52
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    $\begingroup$ I am not totally convinced. What happens to $x(x-1)$? Also could you please explain what a smooth certificate is? I am not well versed in Algebraic geometry. $\endgroup$
    – Fan Zheng
    Commented Feb 27, 2017 at 15:00
  • $\begingroup$ univariate polynomials always factor in linear factors (over $\mathbb{C}$). As usual in algebraic geometry you have to homogenise: $f(x,y,z)=x(x-y)$ defines a plane projective curve. In a singularity you must have $f'_x=f'_y=f'_z=0$, giving you a linear system with nonzero solution $(0:0:1)$. In general you will have such a system of equations for singular points, which might have no nonzero solutions - then your hypersurface is irreducible. $\endgroup$
    – Dima Pasechnik
    Commented Feb 27, 2017 at 17:30
  • $\begingroup$ @DimaPasechnik A check by Maple using your Criterion shows that this polynomial is singular on a subvariety of dimension 4, so it doesn't quite solve the problem. $\endgroup$
    – Fan Zheng
    Commented Feb 27, 2017 at 18:32
  • $\begingroup$ One can reduce the complexity of the problem as follows: Set $y_3=x_2$, $y_2=0$, $y_1=x_3=1$. Then the resulting bivariate polynomial in $x_1$ and $x_2$ still has degree $8$. One can show that it is irreducible, so the original polynomial is irreducible as well. $\endgroup$
    – Peter Mueller
    Commented Feb 27, 2017 at 19:16

1 Answer 1

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For new variables $x$ and $y$ set $x_1=x_2=x_3=x$ and $y_1=y$, $y_2=2y$, $y_3=3y$. Then $$ \prod\left(\sqrt{1+9x^2+36y^2}\pm\sqrt{1+x^2+y^2}\pm\sqrt{1+x^2+4y^2}\pm\sqrt{1+x^2+9y^2}\right)=-1024\left((3y^2 + 4)x^6 + (23y^4 + 47y^2)x^4 + (45y^6 + 180y^4)x^2 + 225y^6\right). $$ This polynomial has degree $8$, and if it is irreducible, then the original polynomial is irreducible even more. Replacing $x$ with $xy$ and dividing by $y^6$ afterwards reduces to show that $$ (3x^6 + 23x^4 + 45x^2)y^2 + 4x^6 + 47x^4 + 180x^2 + 225 $$ is irreducible. That holds, because the coefficients of $y^0$ and $y^2$ are relatively prime and non-squares.

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