2
$\begingroup$

Let $E$ be the splitting field of $f(x)=x^{4}-6x^{2}+7$ over $\mathbf{Q}$. Show that $\operatorname{Gal}(E/\mathbf{Q})$ is a non-abelian $2$-group.

Any help please. Thanks.

$\endgroup$
5
  • $\begingroup$ any group of order $4$ is Abelian $\endgroup$ May 1 '20 at 16:17
  • 1
    $\begingroup$ non abelian 2_group $\endgroup$
    – Math1
    May 1 '20 at 16:18
  • 3
    $\begingroup$ @J.W.Tanner But what makes you think the Galois group of this polynomial has order $4$? $\endgroup$ May 1 '20 at 16:19
  • $\begingroup$ @AlexKruckman But it is 4, right? By the tower law, we have that $[\mathbb{Q}(\sqrt{7}, i) : \mathbb{Q}] = [\mathbb{Q} (\sqrt{7}, i): \mathbb{Q} (\sqrt{7})] [\mathbb{Q} (\sqrt{7}) : \mathbb{Q}] = 2 \cdot 2 = 4$. The order of the Galois group is equal to the degree of the extension. $\endgroup$ May 1 '20 at 16:30
  • 3
    $\begingroup$ @LukePoeppel that is true, but $\mathbf{Q}(\sqrt{7},i)$ is not the splitting field. $\endgroup$
    – rae306
    May 1 '20 at 16:44
4
$\begingroup$

Note that $f$ is irreducible since $f(X+1)$ is Eisenstein for $p=2$. The roots of $f$ are $\alpha_1=\sqrt{3+\sqrt{2}}$, $\alpha_2=\sqrt{3-\sqrt{2}}$, $\alpha_3=-\alpha_1$ and $\alpha_4=-\alpha_2$.

The splitting field of $f$ is $\Omega=\mathbf{Q}(\alpha_1,\alpha_2)$. Since $f$ is irreducible, $\alpha_1$ has degree $4$ over $\mathbf{Q}$. Note that $\alpha_1 \alpha_2=\sqrt{7}\not\in \mathbf{Q}(\sqrt{2})$, so $\mathbf{Q}(\alpha_1)\neq \mathbf{Q}(\alpha_2)$. But $\alpha_2$ is a zero of $X^2+\alpha_1^2-6\in \mathbf{Q}(\alpha_1)[X]$. This implies that $\Omega$ has degree $2^3$ over $\mathbf{Q}$. The Galois group is thus of order $2^3$. It remains to show that it is non-abelian..

Hint. If the Galois group were abelian, by the Galois correspondence every intermediate extension would be normal extension over $\mathbf{Q}$.

$\endgroup$
10
  • 1
    $\begingroup$ Thank you..but can you explain more for me $\alpha_1 \alpha_2=\sqrt{7}\not\in \mathbf{Q}(\sqrt{2})$, so $\mathbf{Q}(\alpha_1)\neq \mathbf{Q}(\alpha_2)$ $\endgroup$
    – Math1
    May 1 '20 at 17:00
  • 2
    $\begingroup$ This is the following lemma in Galois theory on quadratic extensions. Let $K$ be an extension of $\mathbf{Q}$ and $K(\alpha),K(\beta)$ be two quadratic extensions of $K$. Then $K(\alpha)=K(\beta) \iff \alpha\beta\in K$. In this particular case we let $K=\mathbf{Q}(\sqrt{2})$. The fields $\mathbf{Q}(\alpha_1)$ and $\mathbf{Q}(\alpha_2)$ are clearly quadratic over $K$ and so they are equal if and only if $\alpha_1\alpha_2=\sqrt{7}\in K$. $\endgroup$
    – rae306
    May 1 '20 at 17:04
  • $\begingroup$ Thank you very much! $\endgroup$
    – Math1
    May 1 '20 at 17:07
  • 1
    $\begingroup$ Yes, exactly. $\mathbf{Q}(\alpha_1)/\mathbf{Q}$ is clearly not normal, since it does not contain its conjugate $\alpha_2$ by the argument above. $\endgroup$
    – rae306
    May 1 '20 at 17:19
  • 1
    $\begingroup$ @CJD I see I forgot to add the condition $\alpha^2,\beta^2\in K$. $\endgroup$
    – rae306
    Sep 12 '20 at 14:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.