Galois group of $t^4-3t^2+4$ 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?
 A: 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.
A: Even though this question was asked a long time ago, I'm leaving an alternative, more elementary approach in case people look it up in the future for homework (like me). 
We can start by factoring $t^4 - 3t^2 + 4$ over $\mathbb{R}$, namely by splitting it into its linear factors the grouping the conjugates together: $(t - \alpha)(t - \beta)(t + \alpha)(t + \beta)$, where $\alpha = \sqrt{\dfrac{3 + i\sqrt{7}}{2}}$, and $\beta = \sqrt{\dfrac{3 - i\sqrt{7}}{2}}$. The key point is that $\alpha + \beta \in \mathbb{R}$, since you are adding 2 conjugates, is real; recall that Re($z$) = $\dfrac{z + \overline{z}}{2}$. Also, $z\overline{z}$ is real, so $(t - \alpha)(t - \beta) \in \mathbb{R}[t]$. We can then apply the quadratic equation and find $\dfrac{\sqrt{7} \pm i}{2}$ are the roots. Apply a similar argument with $(t + \alpha)(t + \beta)$ and we now have the splitting field that can be written in terms of a "better basis," ie $\mathbb{Q}(i, \sqrt{7})$. We then find four possible maps: the identity, one that maps $i \mapsto -i$ and fixes everything else; one mapping $\sqrt{7} \mapsto -\sqrt{7}$, and $i \mapsto -i, \sqrt{7} \mapsto -\sqrt{7}$. Clearly, there are elements of order 2 in the group, so we have to $\Gamma$ must be isomorphic to the Klein-4 group, since the only groups of order 4 are $\mathbb{Z}_2 \times \mathbb{Z}_2$ or the cyclic group.   
