$x^2-y^2-1$ irreducible in $K[x,y]$

Question

Being $K$ any field, I am trying to figure out if $x^2-y^2-1$ is irreducible in $K[x,y]$.

My approach was to assume there is a decomposition $x^2-y^2-1=(ax+by+c)(ex+fy+g)$ and try to reach a contradiction. At some point doing this, I reach that the equation $b^2+c^2=0$ must be satisfied. If $K$ were, for example, $\mathbb{R}$, from this I could conclude that $b=c=0$ and that allows me to reach later a contradiction. But since $K$ could be any field, I do not think I can do that.

Am I missing something? Is there a way to easily prove whether this polynomial is irreducible?

If $K=\Bbb{F}_2$ then we have $x^2-y^2-1=(x+y+1)^2$. Unless we are in characteristic two, it looks like it is irreducible. — Jyrki Lahtonen, Dec 25 '22 at 11:25
You can use the techniques described here or here, here,... Or even here. — Jyrki Lahtonen, Dec 25 '22 at 11:40

score 1 · Answer 1 · answered Dec 25 '22 at 11:40

1

Let's regard the polynomial is in $y$ and $x$ is in the coefficient ring. Eisenstein's criterion can be applied to the prime $x+1$ as long as the prime ideals $(x+1)$ and $(x-1)$ are distinct. Thus, it is irreducible if char $K$ $\neq 2.$

answered Dec 25 '22 at 11:40

Ayaka

2,722

$x^2-y^2-1$ irreducible in $K[x,y]$

1 Answers1