2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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) $\endgroup$ – user87543 Aug 12 '13 at 14:34
  • $\begingroup$ 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 $Gal(\Bbb Q(\sqrt2i)/\Bbb Q)$. $\endgroup$ – walcher Aug 12 '13 at 14:43
  • $\begingroup$ 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 $\endgroup$ – user87543 Aug 12 '13 at 15:19
  • 1
    $\begingroup$ 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. $\endgroup$ – walcher Aug 12 '13 at 15:28
  • 1
    $\begingroup$ 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). $\endgroup$ – walcher Aug 14 '13 at 15:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy