Efficiently calculating the Galois group of $p(x)=x^4+4x^2-2$ I need to find the Galois group of $p(x)=x^4+4x^2-2$. Here is what I have done so far:
By Eisenstein's criterion, $p$ is irreducible over $\Bbb Q$. Therefore, the Galois group is a transitive subgroup of $S_4$ so it is either $S_4,A_4,D_4,\Bbb Z_4,$ or $K_4$. $\Delta(x^4+4x-2)^2=-18432 \implies \Delta(x^4+4x-2)\notin\Bbb Q.$ Hence, the Galois group has transpositions so it is either $S_4$ or $D_4$.
At this point, I don't really know how to narrow it down any further without just computing $(E:\Bbb Q)$ in the first place. The roots of $p$ are $\big\{\pm\sqrt{\sqrt6-2}\,,\pm\,\, i\sqrt{\sqrt6+2}\big\}$, so $E=\Bbb Q\big(\sqrt{\sqrt6-2},i\sqrt{\sqrt6+2}\big)$. 
$$\Big(\Bbb Q\big(\sqrt{\sqrt6-2},i\sqrt{\sqrt6+2}\big):\Bbb Q\Big)=\Big(\Bbb Q\big(\sqrt{\sqrt6-2},i\sqrt{\sqrt6+2}\big):\Bbb Q \big(\sqrt{\sqrt6-2}\big)\Big)\cdot\Big(\Bbb Q \big(\sqrt{\sqrt6-2}\big):\Bbb Q\Big)=2\cdot4=8$$
$\implies$ The Galois group is $D_4$. 
But, showing the above calculation requires a lot of work, so my question is: Is there an easier way to do this? In particular, how could I have just continued my argument after calculating the discriminant?
 A: With this particular polynomial, you can see that if $\alpha$ is a root of $p,$ then so is $-\alpha, $ hence $p$ will either split in $\mathbb{Q}(\alpha)$ or it will factor as $(x-\alpha)(x+\alpha)g$ where $g$ is irreducible of degree $2.$ Since $|\text{Gal}(f)|\not=4$ we can conclude that in $\mathbb{Q}(\alpha)$ it has an irreducible factor of degree $2$, so $\text{Gal}(f)$ has order $8,$ and is therefore $D_4.$
A: Denote $$p(x) := x^4 + A x^2 + B.$$ 
First, here's another way to see that $\operatorname{Gal}(p)$ that requires little computation: Using that $p$ is even and that $A > 0 > B$, we conclude that $p$ has two real roots and two conjugate, nonreal roots. In particular, complex conjugation is a field automorphism in $\operatorname{Gal}(p)$ that fixes the real roots and interchanges the complex ones, hence it corresponds to a transposition in $S_4$.
Henceforth we can forget about the signs of $A, B$ and hence treat a general irreducible biquadratic. The roots are $\pm \alpha, \pm \beta$ for some $\alpha, \beta \in \Bbb R \cup i \Bbb R$. Then, Vieta's Formulas show that the particular algebraic combination $(\alpha)(-\alpha) + (\beta)(-\beta) = -(\alpha^2 + \beta^2)$ coincides with $A$, which in particular is in $\Bbb Q$. Thus, any permutation of roots in $\operatorname{Gal}(p)$ fixes that particular algebraic combination but not all permutations in $S_4$ do, leaving as the only possibility $$\color{#df0000}{\boxed{\operatorname{Gal}(p) \cong D_4}} .$$
Remark This argument does not depend on the signs of $A, B$, so by checking which transitive subgroups of $S_4$ preserves such an algebraic combination we conclude that any irreducible biquadratic polynomial over $\Bbb Q$ has Galois group $D_4$, $K_4$, or $\Bbb Z_4$. (A little more work shows that the Galois group is isomorphic to $K_4$ iff $B$ is a square in $\Bbb Q$.)
More generally, if we denote the roots of a quartic polynomial $$y^4 + a_3 y^3 + a_2 y^2 + a_1 y + a_0$$ by $r_1, r_2, r_3, r_4$, we can form the resolvent cubic
$$R(y) = [y - (r_1 r_2 + r_3 r_4)] [y - (r_1 r_3 + r_2 r_4)] [y - (r_1 r_4 + r_2 r_3)],$$
and appealing again to Vieta's formulas gives
$$R(y) = y^3 - a_2 y^2 + (a_3 a_1 - 4 a_0) y + (-a_3^2 a_0 - a_1^2 + 4 a_2 a_0).$$
Then $R(y)$ is reducible iff it has a root in $\Bbb Q$, which in turn happens iff one of the combinations $r_a r_b + r_c r_d$ is rational, and again this occurs iff the Galois group $\operatorname{Gal}(p)$ is contained inside a copy of $D_4$.
This implies that, for an irreducible quartic, checking (1) whether the discriminant is square and (2) whether the resolvent cubic is reducible is enough to distinguish when the Galois group is isomorphic to $S_4, A_4, V_4$. More subtle conditions are sometimes required to distinguish between $D_4$ and $\Bbb Z_4$.
