# Irreducibility of $x^2+y^2+z^2-xyz-2$

Are there some general criteria for deciding the irreducibility of polynomials? For example the one in the title?

• Your best bet is always Eisenstein's criterion. Mar 15, 2014 at 18:42
• @Ian: how do you use Eisenstein's criterion for the polynomial in the question? Mar 16, 2014 at 7:42

If this polynomial were to factor, it would have to factor into a polynomial of degree 2 and a polynomial of degree 1. \begin{align*} x^2 + y^2 + z^2 - xyz - 2 &= \big([\text{terms of degree } 2] + [\text{terms of degree } \le 1] \big) \\ &\quad \big([\text{terms of degree } 1] + [\text{terms of degree } \le 0] \big) \\ \end{align*} The terms of degree $2$ in the first factor and the terms of degree $1$ in the second factor must multiply to $-xyz$. By symmetry in $x,y,$ and $z$, we may assume they are therefore $-xy$ and $z$, respectively. $$x^2 + y^2 + z^2 - xyz - 2= (-xy + ax + by + cz + d)(z + e)$$ But there is no $x^2$ or $y^2$ term on the RHS, so this is impossible.
One possible approach is to consider any polynomial $p(x,y,z)$ (in general, any number of variables) as a polynomial in $x$ over the ring $\mathbb{Z}[y,z]$ (in general, $\mathbb{Z}$ could be any unique factorization domain). Note that $\mathbb{Z}[y,z]$ is a unique factorization domain itself, so things like Gauss's lemma and Eisenstein apply. In this case, we would have to have $$x^2 + (-yz) x + (y^2 + z^2 - 2) = (x + q_1(y,z))(x + q_2(y,z))$$ for polynomials $q_1$ and $q_2$, which reduces to the problem of showing $y^2 + z^2 - 2$ is irreducible over $\mathbb{Z}[y,z]$. Then in turn we could consider $y^2 + z^2 - 2$ as a polynomial in $y$ over the ring $\mathbb{Z}[z]$, to get $$y^2 + z^2 - 2 = (y + r_1(z))(y + r_2(z))$$ for polynomials $r_1$ and $r_2$ in one variable. Then we have that $r_1(z) + r_2(z) = 0$, so $r_1(z) = -r_2(z)$, so $z^2 - 2 = -r_1(z)^2$, so $2 - z^2$ is a perfect square. So we have $2 - z^2 = (az + b)^2$, and we conclude $a^2 = -1$, $ab = 0$, and $b^2 = 2$. This is of course impossible, not just over $\mathbb{Q}$ but over any field of characteristic $\ne 2$.