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.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Question is to solve for Galois Group of $x^4-2x^2-2$ over $\mathbb{Q}$.

I know the roots of this polynomial are $\sqrt{1+\sqrt{3}},-\sqrt{1+\sqrt{3}},\sqrt{1-\sqrt{3}},-\sqrt{1-\sqrt{3}}$.

But, $\sqrt{2}i=\sqrt{1+\sqrt{3}}.\sqrt{1-\sqrt{3}}$. So, I concluded that splitting field would be $\mathbb{Q}(\sqrt{1+\sqrt{3}},\sqrt{2}i)$ (for some reason i am not very sure about, i have not written splitting field to be $\mathbb{Q}(\sqrt{1-\sqrt{3}},\sqrt{2}i)$).

So, i have splitting field as $\mathbb{Q}(\sqrt{1+\sqrt{3}},\sqrt{2}i)$.

What i do usually after i know that extension is of the form $F(a,b)$ with $F(a)\cap F(b)=F$, I calculate Galois group of $F(a,b)/F(b)$ and Galois group of $F(a,b)/F(a)$ and write their product as in $G=Gal(F(a,b)/F)$

I do the same here for $Gal(\mathbb{Q}(\sqrt{1+\sqrt{3}},\sqrt{2}i)/\mathbb{Q}(\sqrt{1+\sqrt{3}}))$, this is of order 2, sending $\sqrt{2}i\rightarrow -\sqrt{2}i$ (fixing $\sqrt{1+\sqrt{3}}$) So, I have $Gal(\sqrt{1+\sqrt{3}},\sqrt{2}i)/\mathbb{Q}(\sqrt{1+\sqrt{3}}))\cong \mathbb{Z}_2$.

The problem is with $Gal(\mathbb{Q}(\sqrt{1+\sqrt{3}},\sqrt{2}i)/\mathbb{Q}(\sqrt{2}i))$, I Have to prove that Galois group of $x^4-2x^2-2$ is dihedral group of order 8.I already have an element of order 2, I have to get an element of order 4 from $Gal(\mathbb{Q}(\sqrt{1+\sqrt{3}},\sqrt{2}i)/\mathbb{Q}(\sqrt{2}i))$ which is becoming more cumbersome.

Any help would be appreciated. Thank You.

share|cite|improve this question

Since $\Bbb Q(\sqrt 2i) \cap \Bbb Q(\sqrt{1+ \sqrt3}) = \Bbb Q$, they are linearly disjoint. Thus $$[\Bbb Q(\sqrt{1+ \sqrt3}, \sqrt2i): \Bbb Q(\sqrt 2i)]=[\Bbb Q(\sqrt{1+ \sqrt3}): \Bbb Q]=4,$$ because $x^4-2x^2-2$ is irreducible by Eisenstein.

share|cite|improve this answer
yes, I understand this, I am looking for a generator of $Gal(\mathbb{Q}(\sqrt{1+\sqrt{3}},\sqrt{2}i)/\mathbb{Q}(\sqrt{2}i))$ (assuming the extension is cyclic) – Praphulla Koushik Aug 12 '13 at 14:34
This would probably be done by brute force (ugly calculations), but you don't need it. The linear disjointness of $\Bbb Q(\sqrt{1+ \sqrt3})$ and $\Bbb Q(\sqrt2i)$ implies that $Gal(\Bbb Q(\sqrt{1+ \sqrt3}, \sqrt2i)/ \Bbb Q)$ is the semidirect product of $Gal(\Bbb Q(\sqrt{1+ \sqrt3})/\Bbb Q)$ and $Gal(\Bbb Q(\sqrt2i)/\Bbb Q)$. – walcher Aug 12 '13 at 14:43
I did not think before writing previous statement, $\Bbb Q(\sqrt{1+ \sqrt3})/\Bbb Q$ is not even galois, where is the question of finding galois group :O – Praphulla Koushik Aug 12 '13 at 15:19
We know that the Galois group of an irreducible degree $4$ polynomial is a transitive subgroup of $S_4$. As I have explained above, for this particular polynomial, the Galois group is a semidirect product of a group of order $2$ and a group of order $4$, so it has order $8$. The only order $8$ transitive subgroup of $S_4$ is $D_8$ (in fact it is the only subgroup of order $8$), so it must be the group you are looking for. – walcher Aug 12 '13 at 15:28
As I said, you don't need the transitive part, because $D_8$ is the only subgroup of $S_4$ with 8 elements (so even without the 'transitive' requirement it is completely determined by its order). – walcher Aug 14 '13 at 15:36

Your Answer


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

Not the answer you're looking for? Browse other questions tagged or ask your own question.