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.
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 $D_4$.