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.
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any group of order $4$ is Abelian – J. W. Tanner May 01 '20 at 16:17
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1non abelian 2_group – Math1 May 01 '20 at 16:18
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3@J.W.Tanner But what makes you think the Galois group of this polynomial has order $4$? – Alex Kruckman May 01 '20 at 16:19
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@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. – Luke Poeppel May 01 '20 at 16:30
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3@LukePoeppel that is true, but $\mathbf{Q}(\sqrt{7},i)$ is not the splitting field. – rae306 May 01 '20 at 16:44
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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 $\mathbf{Q}$.
rae306
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1Thank you..but can you explain more for me $\alpha_1 \alpha_2=\sqrt{7}\not\in \mathbf{Q}(\sqrt{2})$, so $\mathbf{Q}(\alpha_1)\neq \mathbf{Q}(\alpha_2)$ – Math1 May 01 '20 at 17:00
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2This 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 $\alpha_1\alpha_2=\sqrt{7}\in K$. – rae306 May 01 '20 at 17:04
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Did you also see that the answer to the hint is already included in the answer..? :-) – rae306 May 01 '20 at 17:11
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"Since $f$ is irreducible, $\alpha_1$ has degree $4$ over $\mathbf{Q}$"..in this part I think – Math1 May 01 '20 at 17:15
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1Yes, exactly. $\mathbf{Q}(\alpha_1)/\mathbf{Q}$ is clearly not normal, since it does not contain its conjugate $\alpha_2$ by the argument above. – rae306 May 01 '20 at 17:19
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@rae306 I don't think I understand the lemma from your comment. Does quadratic mean something other than "degree 2"? What about ]$K = \mathbb{Q}$, $\alpha = i$, and $\beta = 1+i$? Then $K(\alpha) = K(\beta)$ but $\alpha\beta \not\in K$. Is there a fix which still solves the original question? – CJD Sep 12 '20 at 13:13
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