If I let $ \alpha = \sqrt{\sqrt5 + 2}$ then the minimal polynomial of $\alpha$ is:

$$f(x) = x^4 - 4x^2 - 1$$

It's also easy to see that this has two non-real roots: $\beta = \sqrt{2-\sqrt5}$ and the negative.

$\Bbb Q(\alpha)/\Bbb Q$ is a degree 4 extension and because of there are imaginary parts getting added $\Bbb Q(\alpha, \beta)/\Bbb Q (\alpha) \geq 2$.

I think it's equal to $2$ because:

$$\alpha^2\beta^2 = (2+\sqrt5)(2-\sqrt5) = -1$$

$$\beta^2 = -\frac1{\alpha^2}$$

So we're actually just adding the square root of an element we already have.

Here is where I get stuck: how do I determine the Galois group of $\Bbb Q(\alpha, \beta)/\Bbb Q$? It has order 8.

I think it's the order 8 dihedral group because it makes intuitive sense, but this might be wrong.

It has a non-normal subfield $\Bbb Q(\alpha)/\Bbb Q$, and subfields correspond to subgroups (fundamental theorem of Galois theory) so maybe it's something to do with the normal subgroups? I don't know what else I can use, there are quite a few groups of order 8.

Also, what actually are the automorphisms in the Galois group? I know complex conjugation always works which gives me one, but what are the others? If it really is just the dihedral group, then I guess I can just put the roots on a square and see what the automorphisms are by rotating and flipping the square, but how can I see those are really field automorphisms?

  • $\begingroup$ Actually there is only one group of order 8 which acts transitively on 4 letters. $\endgroup$ – sharding4 May 20 '18 at 22:57
  • $\begingroup$ Oh right of course, since $f$ is irreducible, its group has to act transitively on the roots so it has to be $D_8$. Thanks. $\endgroup$ – pizzaroll May 21 '18 at 16:23

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