7
$\begingroup$

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

$\endgroup$
2
  • 2
    $\begingroup$ Your best bet is always Eisenstein's criterion. $\endgroup$
    – Ian Coley
    Mar 15, 2014 at 18:42
  • 2
    $\begingroup$ @Ian: how do you use Eisenstein's criterion for the polynomial in the question? $\endgroup$ Mar 16, 2014 at 7:42

1 Answer 1

10
$\begingroup$

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.

General Method

Some discussion of this in the case that the polynomial is irreducible over ANY field, can be found here. There are also some tricks in the case of two variables here.

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$.

$\endgroup$
2
  • $\begingroup$ Yep, that works. But it is not really a general criterion, though a similar Argument might apply in other examples as well. $\endgroup$
    – user39082
    Mar 16, 2014 at 14:07
  • $\begingroup$ @user39082 I've added some details and references which may be useful in the general case. $\endgroup$ Mar 16, 2014 at 16:04

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .