irreducibility of a bivariate polyonimal over a finite field Let $\mathbb{F}_q$ be the finite field with $q$ elements. Consider the bivariate polynomial
$$P(x,y)=y^2- x(x-1)(x-a)(x-b),$$
where $a\neq b$, and $a,b \neq 0,1$ are some arbitrary elements of $\mathbb{F}_q$. Is $P$ absolutely irreducible (i.e. irreducible over the algebraic closure of $\mathbb{F}_q$)?
 A: Apply Eisenstein's criterion to $P$ in $\bar{\mathbf F}_q[x][y]$ for the "prime" $x$. 
More explicitly, you can show that $P$ is irreducible in $\bar{\mathbf F}_q(x)[y]$ with Eisenstein's criterion, and then apply Gauss' Lemma to conclude that $P$ is irreducible in $\bar{\mathbf F}_q[x][y]=\bar{\mathbf F}_q[x,y]$.
More economically you can argue immediately using 
this variant of Eisenstein's criterion: let $A$ be a UFD, $f\in A[x]$ be of degree $\geq1$, and $p\in A$ a prime element, or equivalently, an irreducible element. Write
$$
f=a_nx^n+\cdots+a_0
$$
with $a_i\in A$ and $a_n\neq0$. Suppose that
$$
\gcd(a_n,\ldots,a_0)=1,
$$
and
$$
p\not\mid a_n,\
p|a_{n-1},\ldots,a_0,\ \text{and}\
p^2\not\mid a_0.
$$
Then $f$ is irreducible in $A[x]$. Indeed, suppose that $f$ is reducible and write $f=gh$ with both $g$ and $h$ nonunits in $A[x]$. Since the coefficients of $f$ are relatively prime, both $g$ and $h$ have degree at least $1$. Reduce modulo $p$ in order to get
$\bar f=\bar g\bar h$ in $A/p[x]$. Now, $\bar f=\bar a_nx^n$ in $A/p[x]$, with $\bar a_n\neq0$. Since $A/p$ is a domain, this means that $\bar g=bx^m$ and $h=cx^\ell$, for some $b,c\in A/p$. Since the leading coefficient of $f$ is the product of the leading coefficients of $g$ and $h$, the leading coefficients of $g$ and $h$ are not divisible by $p$. Since the reductions of $g$ and $h$ are monomials, $m=\deg(g)\geq1$ and $\ell=\deg(h)\geq1$. It follows that both trailing coefficients of $g$ and $h$ are divisible by $p$. Now, the trailing coefficient of $f$ is the product of the trailing coefficients of $g$ and $h$ which are both divisble by $p$. Hence, the trailing coefficient of $f$ is divisible by $p^2$. Contradiction.
As said, one applies this to $A=\mathbf F_q[x]$, $f=P\in A[y]$, and $p=x$.
A: Let us first prove that $P$ is irreducible in the PID $\bar{\mathbf F}_q(x)[y]$.
Since $P$ is a polynomial of degree $2$ in $y$ over the field $\bar{\mathbf F}_q(x)$, it suffices to prove that $P$ has no roots in the field $\bar{\mathbf F}_q(x)$. But this is clear, if $P$ would have a root in $\bar{\mathbf F}_q(x)$, then there would be a rational function $R\in\bar{\mathbf F}_q(x)$ with
$$
R^2=x(x−1)(x−a)(x−b),
$$
which is absurd since $a\neq0,1$. Here, one uses the fact that any nonzero rational function $F\in \bar{\mathbf F}_q(x)$ can be uniquely written as
$$
F=\lambda(x-a_1)^{e_1}(x-a_2)^{e_2}\cdots(x-a_n)^{e_n},
$$
with $\lambda\in\bar{\mathbf F}_q$, $e_1,\ldots,e_n\in\mathbf Z\setminus\{0\}$ and $a_1\ldots,a_n\in\bar{\mathbf F}_q$. In particular, $F$ is a square in $\bar{\mathbf F}_q(x)$ if and only if all $e_i$ are even. In any case, it follows that $P$ is irreducible in $\bar{\mathbf F}_q(x)[y]$. Since $P\in \bar{\mathbf F}_q[x,y]$ and $P$ is monic in $y$, the polynomial $P$ is irreducible in $\bar{\mathbf F}_q[x,y]$ by Gauss' Lemma.
