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I am hoping to analyze the Galois group of the splitting field for the irreducible polynomial $$ x^4+3x^2+5 $$ over $\mathbb{Q}$, preferably without using facts about the discriminant.

I have found the roots to be the intimidating $$ \pm \frac1{\sqrt{2}}\sqrt{-3\pm i\sqrt{11}} $$ I know the transitive subgroups of $S_4$ are $S_4$, $A_4$, $D_4$, $C_4$ and $C_2\times C_2$. My hunch is the Galois group for this is $D_4$ but I am unsure of this and less sure how to prove it (maybe a nice degree argument coupled with finding a non-Galois subfield?). Is there a nicer field this field extension is equal to?

Any hints or guidance would be appreciated. If I really just need to memorize the discriminant stuff for problems like these that would also be valuable to know.

Thanks in advance.

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  • $\begingroup$ Many hints here: math.stackexchange.com/questions/45893/… $\endgroup$ – vadim123 May 1 '18 at 3:43
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    $\begingroup$ You already know that the roots are $\alpha, -\alpha, \bar{\alpha}, -\bar{\alpha}$. If the galois groups are either $C_{4}$ or $C_{2}\times C_{2}$, then the extension $K=\mathbb{Q}(\alpha)$ is Galois. Since $\bar{\alpha}=\frac{\sqrt{5}}{\alpha}$ and $\alpha$ is in $K$, then if $\sqrt{5}$ is in $K$ then $K$ is Galois. If not, then $K(\sqrt{5})$ is Galois, which is degree 8 over $Q$. From there, you can narrow down which groups to choose from without determining if $K$ is Galois or not. $\endgroup$ – Dave huff May 1 '18 at 3:51
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If you call $$\pm\sqrt{\frac{3\pm\sqrt{-11}}2}$$ intimidating, then it is just as well that the Galois group isn't $A_4$ or $S_4$. Then any formula for the roots would be much more complex.

The splitting field is $K=\Bbb Q(\sqrt{-11},\alpha,\beta)$ where $\alpha$ and $\beta$ are square roots of $\frac12(3+i\sqrt{11})$ and $\frac12(3-i\sqrt{11})$. Then $|K:\Bbb Q(\sqrt{-11})|$ is a factor of $4$, so the Galois group cannot be $A_4$ or $S_4$. Let's suppose that $\beta=\bar\alpha$. Then $\alpha\beta$ is the positive square root of $\frac14(3+i\sqrt{11})(3-i\sqrt{11})=5$. Therefore $\alpha \beta=\sqrt5\in K$. As $\sqrt5$ is not a square in $\Bbb Q(\sqrt{-11})$ then $|K:\Bbb Q(\sqrt{-11})|=4$ from elementary Kummer theory. Thus $|K:\Bbb Q|=8$ and so the Galois group is $D_4$.

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