Irreducibily of polynomials in two variables Let $\mathbb{F}_q$ be a finite field where $q$ is odd. Let $f \in \mathbb{F}_q[x,y]$ be the following polynomial
$$f:=(x^2y^2 - 2x^2y - 2xy^2 - x^2 + xy - y^2 - 2x - 2y + 1).$$
How to prove that $f$ is irreducible or absolutely irreducible over $\mathbb{F}_q$? Is there any general criterion to deal with such polynomials? (Absolutely irreducible means irreducible over every algebraic extension of the base field.)
 A: Since this $f$ has a unique top (total) degree term $x^2y^2$, we can just examine all possible factorization and show they don't work.
Any factorization $f=gh$ also give $f_{top}=g_{top}h_{top}$, so the top-degree piece of possible least-degree factors are $x,y,x^2,xy,y^2$.  So we have to examine the four cases


*

*$(x-c)(xy^2+\dots)$

*$(y-c)(x^2y+\dots)$

*$(x^2+\dots)(y^2+\dots)$

*$(xy+\dots)(xy+\dots)$
For example, if $x-c$ is a factor of $f$, then
$$
\begin{cases}
0=c^2-2c-1 & (\text{coeffs of }y^2)\\
0=-2c^2+c-2 & (\text{coeffs of }y)\\
0=-c^2-2c+1 & (\text{coeffs of }1)\\
\end{cases}
$$
which is impossible (if $2\mid q$ then the second equation implies $c=0$ but that doesn't satisfy the other two.  Otherwise, adding the first and third equation gives $c=0$, which again doesn't work).  Same reason works for $y-c$.
For the quadratic factors, note that case (3) must completely separate $x$ and $y$ (because there are no $x^3$ or $y^3$), so it reduces to case 1 if we work over a quadratic extension.  For case (4), use the $x\leftrightarrow y$ symmetry of $f$ to further break into (4a) $(xy+a(x+y)+b)(xy-(2+a)(x+y)+b')$, or (4b) $(xy+ax+by+c)(xy+bx+ay+c)$, $a\neq b$, etc.
