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.

  • 1
    $\begingroup$ 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) $\endgroup$
    – DonAntonio
    Commented Dec 11, 2012 at 21:11

1 Answer 1


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 $D_8$: enter image description here 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 $\mathbb{Q}(\sqrt{2})$.


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