Irreducibility of a polynomial in $\mathbb{Q}(Y,Z)[X]$ I am trying to prove that the polynomial
$$X^3+(Z^3+Z^2+Y^2Z+Y-1)X^2+(Z^3+Z^2+Y^2Z+Y-1)Y^2\in\mathbb{Q}(Y,Z)[X]$$
is irreducible.
Thanks to Eisenstein's criterion, it would be enough to prove that
$$ZY^2+Y+Z^3+Z^2-1\in\mathbb{Q}[Y][Z]$$ is irreducible. However, I don't see how to attack this problem. Could someone give me a piece of advice ?
 A: As a Corollary of Gauss's Lemma, we know that if $D$ is a UFD, and $f(x)\in D[x]$ is a polynomial that is primitive (a gcd of the coefficients is $1$), then $f(x)$ is irreducible in $D[x]$ if and only if $f(x)$ is irreducible in $K[x]$, where $K$ is the field of fractions of $D$.
You can view $ZY^2+Y+Z^3+Z^2-1$ as an element of $(\mathbb{Q}[Z])[Y]$. Since $\mathbb{Q}[Z]$ is a UFD, and this polynomial is primitive as a polynomial in $Y$ ($\gcd(Z,Z^2-1) = 1$), then it is irreducible in $\mathbb{Q}[Y,Z]$ if and only if it is irreducible in $\mathbb{Q}(Z)[Y]$, where the polynomial is quadratic. In order to be reducible there, it has to have a root, and the roots are given by the quadratic formula. The discriminant is
$$1 - 4Z(Z^3+Z^2-1) = -4Z^4 -4Z^3 + 4Z + 1$$
which would have to be a perfect square in $\mathbb{Q}(Z)$, hence (by the rational root theorem) a perfect square in $\mathbb{Q}[Z]$. 
But in $\mathbb{Q}[Z]$, any perfect square has positive leading coefficient; so this element is not a square in $\mathbb{Q}[Z]$; since the discriminant of the polynomial is not a perfect square, the polynomial has no roots in $\mathbb{Q}(Z)$, hence is irreducible in $\mathbb{Q}(Z)[Y]$, and therefore is irreducible in $\mathbb{Q}[Y,Z]$.
A: The general strategy is to consider the degrees of the terms of your polynomial. If it factors as $fg$, then each of $f,g$ must be of smaller degree than your polynomial, i.e they must be of degree 2. In fact, one of them must be linear since you have a degree 3 polynomial.
So assume $f=aZ^2+bY^2+cZY+dZ+eY+k$ and that $g=lZ+mY+n$ and expand $fg$. You see at once that, for example, either $b=0$ or $m=0$ because otherwise $fg$ would contain a $Y^3$-term.
You should arrive at a contradiction.
