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.
Best Answer
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 $\mathbb{Q}(\sqrt{2})$.