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I am trying to prove that the polynomial


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 ?

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Suppose $f(Y,Z) = ZY^2 + Y + Z^3 + Z^2 - 1$ is reducible. What does that imply about $f(0, Z)$? (warning: double check your reply to make sure you're not overlooking something) (disclaimer: I have not checked that this comment easily leads to a solution to your question) – Hurkyl Mar 3 '12 at 20:10
@Hurkyl Do you mean using the fact that $\mathbb{Q}[Z]$ is an UFD? Since $f(0,Z)=z^2(z+1)$, if $f=f_1f_2$, then $f_1(0,Z)=z+1$ and $f_2(0,Z)=z^2$ OR $f_1(0,Z)=z(z+1)$ and $f_2(0,Z)=z$, etc. ? – Fred Mar 3 '12 at 20:54
You made a typo somewhere; with the polynomial you write above, $f(0,Z) = Z^3 + Z^2 - 1 \neq Z^2 (Z+1)$. – Hurkyl Mar 3 '12 at 20:56
...which is irreducible over $\mathbb{Q}[Z]$ and from here we get that $f_1$ is a constant. Thanks, it's was very good idea! – Fred Mar 3 '12 at 21:24
That's what I was warning about: - f1 is not necessarily a constant. Instead, it is a constant modulo $y$: $f_1(Y,Z) = 1 + Y g(Y, Z)$ – Hurkyl Mar 4 '12 at 5:04
up vote 6 down vote accepted

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 $\mathbb{Q}[Y,Z]$.

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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.

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Thanks! You mean "would contain a $Y^3$-term, don't you ? Then $b=0$ or $m=0$. By distinguishing the cases $b=0$/$m\not=0$, $m=0$/$b\not=0$ and $m=b=0$, I get $n=0$, which is impossible. So it works, but it is quite tedious! If someone gets a more elegant solution ? – Fred Mar 3 '12 at 20:39
Ah, yes. One has to be a bit careful when checking the cases. I'll slightly edit the answer now. I don't know of more elegant ways to do this, however there are lots of algorithms for factoring polynomials - a Google search should give you many of them. – Fredrik Meyer Mar 3 '12 at 21:24

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