Finding Galois group $Gal(x^4 + 4 / Q)$ I have a problem with group $Gal(x^4 + 4 / Q)$. I think that splitting field is $Q(\sqrt{2i})$. I have 4 roots: $\sqrt{2i}$, $\sqrt{-2i}$, $-\sqrt{2i}$ and $-\sqrt{-2i}$. Knowing that roots must be mapped to roots, I have at most 4 Q-automorphism, and according to theorem I obtaint that at least Q-automorphism, so $|Gal(x^4 + 4 / Q)| = 4$. (If it isn't correct, please correct me.)
And I have to determine if $Gal(x^4 + 4 / Q)$ is isomorphic to $Z_2×Z_2$ or to $Z_4$. I know if $Gal(x^4 + 4 / Q)$ contains an element of order 4, then $Gal(x^4 + 4 / Q)$ is isomorphic to $Z_4$, but I don't have a clue how to prove or disprove it.
Thank you.
 A: Well with quartics of certain forms you can complete the square rather easily to find a factorisation into quadratics - sometimes in an extension field, but here over $\mathbb Q$:
$x^4+4=x^4+4x^2+4-4x^2=(x^2+2)^2-4x^2=(x^2+2x+2)(x^2-2x+2)$
For an example which comes up regularly $x^4+1=(x^2+1)^2-2x^2=(x^2+\sqrt 2 x+1)(x^2-\sqrt 2 x +1)$. And you can do similar things in other cases when all the powers are even.
In your case the polynomial you are given can be factored over $\mathbb Q$. The negatives of the roots of the first factor are the roots of the second factor. The first factor splits in a quadratic extension, and the second factor splits too.

Note that if you have a polynomial $p(x^2)$ you know, in fact, that if $\alpha$ is a root, then so is $-\alpha$, so the roots pair up, and you get various possibilities of factorising $p(x^2)=q(x)q(-x)$ corresponding to different ways of splitting the roots - allocating $\alpha$ to one factor and $-\alpha$ to the other. That means that you only have to split $q(x)$ to split $p(x^2)$. If there are $n$ distinct pairs of roots we can fix $\alpha$ and find that there are $2^{n-1}$ ways of choosing the signs for the others. In the case in the question there are four roots in two pairs, and two factorisations of the kind under consideration.
As an illustration we have $x^4+4=(x^2+2ix-2)(x^2-2ix-2)$ from the alternative pairing of roots in the case in your question. This clearly belongs in $\mathbb Q(i)$ and solving the quadratics shows that they split in $\mathbb Q(i)$.

We always expect three ways of splitting a quartic into quadratics. The third way here is to pair each root with its negative obtaining $x^4+4=(x^2+2i)(x^2-2i)$
