# Galois Group of $\sqrt{1+\sqrt{2}}$

I'm studying algebra, and came upon the following problem.

Let $E = \mathbb{Q}(a)$, where $a = \sqrt{1 + \sqrt2}$. Find the irreducible polynomial of $a$, and determine the degree of $E$ over $\mathbb{Q}$. Identity the Galois group of $E/ \mathbb{Q}$, and find how many subfields of $E$ there are.

I can see that the irreducible polynomial is $$(x-\sqrt{1+\sqrt2})(x-\sqrt{1-\sqrt2})(x+\sqrt{1+\sqrt2})(x+\sqrt{1-\sqrt2}) = x^4-2x^2-1,$$ and thus $E$ is of degree $4$ over $\mathbb{Q}$.

I think that there's a problem with the statement, because $E$ is not Galois - it's not the splitting field of a separable polynomial (unless this is a use of terminology unfamiliar to me?). However, I'm wondering how to describe the Galois group of $x^4-2x^2-1$.

I can see that there are eight automorphisms in the group. If I call the roots $\{\pm \alpha, \pm \beta\}$, then we could map $\pm \alpha$ to $\pm \alpha$ or to $\pm \beta$. In the former case, $\pm \beta$ can map to $\pm \beta$, and in the later $\pm \beta$ can map to $\pm \alpha$. This makes eight. They're all valid isomorphisms because $$\left[\mathbb{Q}\left(\sqrt{1+\sqrt2}, \sqrt{1-\sqrt2}\right): \mathbb{Q}\right] = \left[\mathbb{Q}\left(\sqrt{1-\sqrt2},\sqrt{1+\sqrt2}\right) : \mathbb{Q}\left(\sqrt{1+\sqrt2}\right)\right]\left[\mathbb{Q}\left(\sqrt{1+\sqrt2}\right): \mathbb{Q}\right] = 2 \cdot 4 = 8.$$

I'm wondering how to move from here to identifying the Galois group subgroups and associated subfields. Is there any method that makes this less computationally heavy? Thanks.

• Very much recommended: jmilne.org/math/CourseNotes/ft.htmlDownload the above and read in particular chapter 4 and more in particular, page 87-88 (the regular pdf, not the one for ereaders) Dec 11, 2012 at 21:11

Galois group is sometimes used to mean automorphism group. The Galois closure here is $$E^{gal} = \mathbb{Q}(i, \sqrt{ 1 + \sqrt{2} })$$, so that $$E$$ is indeed not Galois. One can calculate that the Galois group of $$E^{gal}$$ is $$D_8$$, and indeed you found that it has order $$8$$. This results from a computation with the Galois group as a subgroup of $$S_8$$, where we view $$S_8$$ as a group of permutations on the roots of the minimal polynomial of $$a$$. Also it is good to know how to write the various small groups in terms of generators and relations.
You found the minimal polynomial has degree $$4$$, so $$E / \mathbb{Q}$$ has degree $$4$$. By the fundamental theorem of Galois theory, subfields of $$E^{gal}$$ contained in $$E$$ correspond to subgroups of $$\text{Aut}(E^{gal})$$ containing $$\text{Aut}_E (E^{gal})$$. $$E$$ corresponds to a non-normal subgroup of $$D_8$$ of order $$2$$ in $$D_8$$. It's not hard to calculate the subgroup diagram of $$D_8$$: There are 4 non-normal subgroups, and each of these contains only two subgroups of $$D_8$$- $$D_8$$, and one of $$\{ e, ax, a^2, a^3x \}$$ and $$\{ e , x ,a^2, a^2 x \}$$. Hence $$E$$ contains $$\mathbb{Q}$$ and one other subfield. That subfield must be $$\mathbb{Q}(\sqrt{2})$$.