I am having difficulty finishing this homework problem.

Calculate the Galois group of $f(x)=x^4+x^2+1\in\mathbb{Q}[x]$.

Here's what I've done so far. Denote the roots of $f$ as $\alpha, \overline{\alpha}$, $\beta, \overline{\beta}$. Calculating the roots and relabeling we have that $-\alpha=\beta$, so $\beta\in \mathbb{Q}(\alpha)$, and similarly, we have that $-\overline{\alpha}=\overline{\beta}$, so the splitting field of $f$ is $K=\mathbb{Q}(\alpha,\overline{\alpha})$. The map $\tau\in\mathcal{Gal}(K/\mathbb{Q})$ given by conjugation, and moreover $\tau^2=e$, where $e$ is the identity map. Therefore the Galois group has a subgroup of order $2$. I want to show that this is the entire Galois group. How can I do so?

  • 1
    $\begingroup$ But it's not! If you believe that this polynomial is irreducible, its Galois group has to have order at least $4$. $\endgroup$ – Qiaochu Yuan Dec 6 '15 at 2:31
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    $\begingroup$ Note that if $z,\bar z$ are conjugate, then $(X-z)(X-\bar z)$ lies (at least) in $\Bbb R[X]$, and hopefully in $\Bbb Q[X]$, or even $\Bbb Z[X]$. You're saying your polynomial has two pairs of conjugate complex roots, so it must be reducible at least over $\Bbb R$. $\endgroup$ – Pedro Tamaroff Dec 6 '15 at 2:32
  • $\begingroup$ Note that you can find the roots by letting $u =x^2$ and solving the resulting quadratic. $\endgroup$ – Brandon Thomas Van Over Dec 6 '15 at 2:52
  • $\begingroup$ Ah I see. $K(\alpha)$ must have degree 4, so the Galois group has order at least $4$. I have found the roots. How can I go about finding the Galois group? I know that since it was irreducible, it acts transitively on the roots. I now think it is S4. Let me try it out. $\endgroup$ – Gael Diego Fernandez Dec 6 '15 at 3:16
  • $\begingroup$ It's not $S_4$ either. How about you try figuring out what the roots are, exactly? $\endgroup$ – Qiaochu Yuan Dec 6 '15 at 4:19

Let me take a stab at this.

We know our polynomial factors as $x^4+x^2+1=(x^2+x+1)(x^2-x+1)$, with roots

$$ \alpha= \frac{1+\sqrt{-3}}{2}, \overline{\alpha}=\frac{1-\sqrt{-3}}{2},-\overline{\alpha}=\frac{-1+\sqrt{-3}}{2},-\alpha=\frac{-1-\sqrt{-3}}{2}.$$

In particular, this polynomial is separable, so we define its Galois group to be the Galois group of the splitting field over $\mathbb{Q}$.

Notice that all of the roots are contained in $\mathbb{Q}(\sqrt{-3})$; meanwhile $\mathbb{Q}(\sqrt{-3})$ has minimal polynomial $x^2+3$, so is degree $2$ over $\mathbb{Q}$. But then there are no intermediate fields $\mathbb{Q} \subsetneq k \subsetneq \mathbb{Q}(\sqrt{-3})$, so $\mathbb{Q}(\sqrt{-3})$ really is the splitting field. Finally, $Gal(\mathbb{Q}(\sqrt{-3})/\mathbb{Q})$ is the group of order $2$.

So it looks to me you were right the first time. If I've made a mistake, someone please let me know; my Galois theory is rusty and I've only just started brushing up.


As noted, your polynomial isn't irreducible: $$x^4 + x^2 + 1 = (x^2 + x + 1)(x^2 - x +1) = \Phi_3(x)\Phi_6(x)$$

In particular, the splitting field is $\mathbb Q(\zeta_6)$, because you obtain all of the third roots of unity and the remaining primitive sixth root after adjoining one primitive root. The Galois group of this extension is isomorphic to $\mathbb Z_6^\times$, where its only nontrivial automorphism takes $\zeta_6$ to $\zeta_6^{5} = \zeta_6^{-1}$.


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