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:


$Gal(\mathbb{F}_{p^n}/\mathbb{F}_p)$ 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?

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    $\begingroup$ If it were irreducible, what would the quotient field be? $\endgroup$
    – user14972
    Jun 12, 2014 at 6:51
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    $\begingroup$ This answer by Qiaochu explains the theory. Here is an elementary discussion of a special case. $\endgroup$ Jun 12, 2014 at 6:56
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    $\begingroup$ 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 ... $\endgroup$ Jun 12, 2014 at 6:58
  • $\begingroup$ Thank you all, I believe I got it. $\endgroup$
    – rollover
    Jun 12, 2014 at 7:07
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    $\begingroup$ In that case, foaly, I'd encourage you to write it up and post it as an answer. That's encouraged here. $\endgroup$ Jun 12, 2014 at 7:21

2 Answers 2


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 $\mathbf{F}_q$.


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