$x^2 + y^2 + z^2 - 2xy - 2xz - 2yz$ is irreducible over $\mathbb{Z}[x, y, z]$

Question

I am trying to prove that $f(x,y,z) = x^2 + y^2 + z^2 - 2xy - 2xz - 2yz$ is irreducible over $\mathbb{Z}[x, y, z]$ (I am sure it is) and maybe over $\mathbb{R}[x, y, z]$ (almost sure).

I checked that it is not divisible by $x+y+z$ so factors are not symmetric. And that's where I get confused. Suppose $f = g \cdot h$, then some permutations must swap $g$ and $h$ and maybe there are some that preserve them. How do I make sure I checked all the possibilities? Is it sufficient to check what happens if $x,y \mapsto y,x$ swaps them and $(123) \in S_3$ swaps them?

Answer 1

If the form factors: let it be $$ (ax+by+cz) \cdot (dx+ey+fz) $$ in which $a,b,c,d,e,f$ are permitted to be real or complex.

Multiply it out, take its second partial derivatives and arrange in the Hessian matrix,

$$ H= \left( \begin{array}{ccc} 2ad & ae+bd & af+cd \\ ae+bd & 2be & bf+ce \\ af+cd & bf+ce & 2cf \\ \end{array} \right) $$

Finally, calculate the determinant, which is $$ \det H = 0 $$

So, if the form factors, the determinant of the Hessian is zero. Contrapositive, if that determinant is nonzero, the form does not factor.

Your Hessian matrix, rather half of it, is $$ \left( \begin{array}{rrr} 1 & -1 & -1 \\ -1 & 1 & -1 \\ -1 & -1 & 1 \\ \end{array} \right) $$ and its determinant is nonzero

Answer 2

We observe that, if

$$A(x,y,z):=x^2+y^2+z^2-2xy-2xz-2yz$$

is factorable over $\mathbb R$, then :

$$B(a,b):=1+a^2+b^2-2a-2b-2ab$$

is also factorable over $\mathbb R$ , where $a=\dfrac yx$ and $b=\dfrac zx$ .

Define the variables $(a,b)$ as $a=m+n,\thinspace b=m-n$ . Thus, we can deduce that, if $B(a,b)$ is factorable over $\mathbb R$ then

$$C(m,n):=4n^2-4m+1$$

is factorable too .

But, $C(m,n)=0$ yields $m=n^2+\frac 14$, which makes $a$ and $b$ quadratic factors . A contradiction .