Galois group of $t^4-3t^2+4$

Question

I am trying to understand all the steps for finding the Galois group of the extension $K:\mathbb{Q}$ where $K$ is known to be the splitting field over $\mathbb{Q}$ of $p(t) = t^4-3t^2+4$. We know that $K:\mathbb{Q}$ is normal by the hypothesis but I'm getting a bit confused on finding $[K:\mathbb{Q}]$, so I can determine the order of $\mathrm{Gal}(K:\mathbb{Q})$. Just because we know $K$ is the splitting field of $p(t)$ over $\mathbb{Q}$ doesn't mean that $[K:\mathbb{Q}] = 4$ (without further computation), correct? Furthermore, I checked on Wolfram for the roots and they are $\pm \sqrt{\frac{1}{2}(3\pm i\sqrt{7})}$ so I suspect that $K = \mathbb{Q}(\sqrt{7},i)$ but I'm looking for a fast way to prove this (otherwise, I would use linear algebra and compare terms). Am I going about this the right way or am I missing something?

Answer 1

You are correct to note that knowing $\deg(p) = 4$ only guarantees $4 \le [K: \Bbb Q] \le 4!$. In order to determine the degree of the extension, you're going to need further information about what happens when you adjoin roots of $p(x)$.

I'm skeptical of your guess that $K = \Bbb Q(\sqrt{7}, i) = \Bbb Q(\sqrt{7} + i)$, in part because the only time $\sqrt{7}$ and $i$ show up is as multiples of each other in your formula for the roots of $p$. You can directly see that $\Bbb Q(i\sqrt{7}) = \Bbb Q(\sqrt{-7})$ is a subfield of $K$, however.

Now an extension of $\Bbb Q(\sqrt{-7})$ will allow $p(x)$ to split if and only in square roots of both $\frac{1}{2}(3+\sqrt{-7})$ and $\frac{1}{2}(3-\sqrt{-7})$ exist. Therefore, there are two possible cases. In the first case, adjoining one square root will guarantee the inclusion of the second, like how $\Bbb Q(\sqrt{2}) = \Bbb Q(\sqrt{8})$. In the second case, you will need to perform two quadratic extensions to add both square roots to $\Bbb Q(\sqrt{-7})$. Note that $[K: \Bbb Q]$ will be $4$ in the first case and $8$ in the second case.

Hopefully this is enough to point you in the right direction.