Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
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) –  DonAntonio Dec 11 '12 at 21:11

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.