# Minimal polynomial reducible modulo every prime $p$

Suppose $K = \mathbb{Q}(\alpha)$ with $\alpha = a + b\sqrt{D_1}+c\sqrt{D_2}+d\sqrt{D_1D_2}$ with $D_1,D_2 \in \mathbb{Z}$. Prove that the minimal polynomial $m_\alpha(x)$ for $\alpha$ over $\mathbb{Q}$ is irreducible of degree 4 over $\mathbb{Q}$ but is reducible modulo every prime $p$. In particular show that the polynomial $x^4 - 10x^2 +1$ is irreducible in $\mathbb{Z}[x]$ but is reducible modulo every prime. [Use the fact that there are no biquadratic extensions over finite fields.]

So far I have established the following:

$[\mathbb{Q}(\alpha):\mathbb{Q}]=\deg(m_\alpha(x))=4$

$Gal(\mathbb{F}_{p^n}/\mathbb{F}_p)$ is cyclic, hence no biquadratic extension (which is iso to $V_4$) exists over finite fields.

I'm having a problem proving the reducibility mod every prime though. Any hints?

• If it were irreducible, what would the quotient field be?
– user14972
Jun 12, 2014 at 6:51
• This answer by Qiaochu explains the theory. Here is an elementary discussion of a special case. Jun 12, 2014 at 6:56
• Also $x^4-10x^2+1$ has $\pm\sqrt2\pm\sqrt3$ as its roots. Thus its splitting field over $\Bbb{F}_p$ is $\Bbb{F}_p[\sqrt2,\sqrt3]$, which is ... Jun 12, 2014 at 6:58
• Thank you all, I believe I got it. Jun 12, 2014 at 7:07
• In that case, foaly, I'd encourage you to write it up and post it as an answer. That's encouraged here. Jun 12, 2014 at 7:21

Turns out it's rather simple. As I already observed in the question, $Gal(\mathbb{F}_{p^n}/\mathbb{F}_p)\cong \mathbb{Z}/n\Bbb{Z}$, hence cyclic.
If $f(x)$ of degree $n$ is irreducible over $\Bbb{F}_p$, its splitting field over $\Bbb{F}_p$ is $\Bbb{F}_{p^n}$.
$\because$ the splitting field of $m_\alpha(x)$ would be $\Bbb{F}_p(\sqrt{D_1},\sqrt{D_2})$, which has no element of order $p^4$, i.e. which is not cyclic,
$\therefore$ $m_\alpha(x)$ cannot be irreducible.
The field $\mathbf{F}_{q^2}$ contains the square roots of all of the elements of $\mathbf{F}_q$, so the splitting field of $f$ over $\mathbf{F}_q$ is $\mathbf{F}_{q^2}$, a degree 2 extension.
Thus, the factorization of $f$ consists only of linear and quadratic factors -- in particular, no quartic factors -- thus it is not irreducible over $\mathbf{F}_q$.