Is $x^2+y^2+z^2-xy-xz-yz$ reducible in $\mathbb{C}[x,y,z]$?

Question

In an exercise, I have the polynomial $p(x,y,z)=x^2+y^2+z^2-xy-xz-yz$, I have to decide whether is reducible or not in $\mathbb{C}[x,y,z]$.

I think it's reducible, and thinking of it in $\mathbb{C}[y,z][x]$ I have $p(x,y,z)=x^2-x(y+z)+y^2+z^2-yz$.

Applying Bhaskara I get

$\begin{align*} x_{\pm}&=\frac{y+z\pm \sqrt{y^2+2yz+z^2-4y^2-4z^2+4yz}}{2}\\ &=\frac{y+z\pm \sqrt{-3(y-z)^2}}{2}\\&=\frac{y+z\pm i\sqrt{3}(y-z)}{2}\end{align*}$.

But when I try to verify $(x-x_+)(x-x_-)$ (with Wolfram Alpha) I don't get the original polynomial.

Is it wrong using Bhaskara to try to reduce the polynomial?

Thank you in advance

Answer 1

You are correct.

\begin{align*} (x-x_+)(x-x_-)&=\left(\frac{2x-y-z}2\right)^2-\left(\frac{i\sqrt3(y-z)}2\right)^2\\ &=\frac{4x^2+y^2+z^2-4xy-4xz+2yz+3(y^2+z^2-2yz)}4\\ &=x^2+y^2+z^2-xy-yz-zx \end{align*}

Answer 2

$$(x-\frac{y+z+i\sqrt{3}(y-z)}{2})(x-\frac{y+z-i\sqrt{3}(y-z)}{2})\\=x^2-x(\frac{y+z+z i\sqrt{3}(y-z)}{2}+\frac{y+z-i\sqrt{3}(y-z)}{2})+\frac{y+z+i\sqrt{3}(y-z)}{2}\frac{y+z- i\sqrt{3}(y-z)}{2} \\ =x^2-x(y+z)+(\frac{y+z}{2})^2-\left(\frac{i\sqrt{3}(y-z)}{2} \right)^2\\ =x^2-x(y+z)+ \frac{1}{4} \left( (y+z)^2+3(y-z)^2 \right)\\ =x^2-x(y+z)+ \frac{1}{4} \left( 4y^2+4z^2-4yz \right) $$

Which is your polynomial.

Did you plug in $$x_+=\frac{y+z+i\sqrt{3}(y-z)}{2} \\ x_-=\frac{y+z-i\sqrt{3}(y-z)}{2} $$ or did you wrongly put them with $\pm$?

Answer 3

The theorem, going back to the time of Ludwig Otto Hesse, is that a quadratic form in three variables factors as the product of two linears over the complexes if and only if the determinant of the Hessian matrix is zero.

If you start with that attitude, you can just demand that the coefficients of $x$ are both $1$ and see what happens. $$ \left(x + A y + B z \right) \left(x + C y + D z \right)$$ To get the squares correct, $$ \left(x + A y + B z \right) \left(x + \frac{1}{A} y + \frac{1}{B} z \right)$$ Now we need the mixed terms to have coefficients $-1,$ so $$ A + \frac{1}{A} = -1. $$ Stay with this, $$ A^2 + 1 = -A $$ $$ A^2 + A + 1 = 0 $$ So $A$ is a nontrivial cube root of unity. Naming $$ \omega = \frac{-1 + i \sqrt 3}{2} $$ so $$ \omega^2 = \frac{-1 - i \sqrt 3}{2} $$ we reach $$ \left(x + y \omega + z \omega^2 \right) \left(x + y \omega^2 + z \omega \right) = x^2 + y^2 + z^2 - yz - zx - xy $$

Page from a relevant article by Brookfield:

Answer 4

This following was inspired by Will Jagy's answer.

It is well known that over the real $x^3+y^3+z^3-3xyz$ can be factored as $$(x+y+z)(x^2+y^2+z^2-xy-yz-zx ).$$

Note that $$F:=x^3+y^3+z^3-3xyz$$ $$=x^3+(\omega y)^3+(\omega^2 z)^3-3x(\omega y)(\omega^2 z)$$ $$=x^3+(\omega^2 y)^3+(\omega z)^3-3x(\omega^2 y)(\omega z),$$ where $\omega$ is a primitive third root of unity.

It follows that $F$ also has factors $x+\omega y+\omega^2 z$ and $x+\omega^2 y+\omega z$. Since ${\mathbb C}[x,y,z]$ is a UFD, one has $$F=(x+y+z)(x+\omega y+\omega^2 z)(x+\omega^2 y+\omega z),$$ as required. QED