# Galois group of splitting field of $X^4-6X^2+7$ is non-abelian

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 group of order $4$ is Abelian May 1 '20 at 16:17
• non abelian 2_group May 1 '20 at 16:18
• @J.W.Tanner But what makes you think the Galois group of this polynomial has order $4$? May 1 '20 at 16:19
• @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. May 1 '20 at 16:30
• @LukePoeppel that is true, but $\mathbf{Q}(\sqrt{7},i)$ is not the splitting field. May 1 '20 at 16:44

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 $$\mathbf{Q}$$.
• Thank you..but can you explain more for me $\alpha_1 \alpha_2=\sqrt{7}\not\in \mathbf{Q}(\sqrt{2})$, so $\mathbf{Q}(\alpha_1)\neq \mathbf{Q}(\alpha_2)$ May 1 '20 at 17:00
• 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 $\alpha_1\alpha_2=\sqrt{7}\in K$. May 1 '20 at 17:04
• 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. May 1 '20 at 17:19
• @CJD I see I forgot to add the condition $\alpha^2,\beta^2\in K$. Sep 12 '20 at 14:03