Use Galois Theory to prove the existence of $A$ and $B$ such that $\mathbb{Q}(\sqrt{6+3\sqrt{3}})=\mathbb{Q}(\sqrt{A}, \sqrt{B})$

Question

Use Galois Theory to prove the existence of $A$ and $B$ such that $\mathbb{Q}(\sqrt{6+3\sqrt{3}})=\mathbb{Q}(\sqrt{A}, \sqrt{B})$

So $\mathbb{Q}(\sqrt{6+3\sqrt{3}})$ is the field of rational numbers with the extra element generated by $\sqrt{6+3\sqrt{3}}$. Is this correct? How could we proceed using Galois Theory?

Find these $A$ and $B$

From inspection I think, without loss of generality, $A=\sqrt{2}$ and $B=\sqrt{3}$.

How could I prove this?

Edit, linked question gives useful approach but further answers are welcome (and useful)

Possible duplicate of Degree of the extension $\mathbb{Q}(\sqrt{a+\sqrt{b}})$ over $\mathbb{Q}$; for $a=6$ and $b=27$. — Dietrich Burde, May 29 '16 at 18:47
Its a different question though. Unless the proof requires calculating the degree, is this the case? — amiz9, May 29 '16 at 18:48
The answer at the duplicate gives $m$ and $n$ such that the biquadratic extension is $\mathbb{Q}(\sqrt{n},\sqrt{m})$ - that's what you want. — Dietrich Burde, May 29 '16 at 18:49
ok thanks I will check it out. I still would like this question open since the answer below for example is helpful — amiz9, May 29 '16 at 18:51
Sure. And this answer will stay, even if the question were closed as a duplicate. — Dietrich Burde, May 29 '16 at 18:53
@amiz9, something worth noting is that $A$, $B$ are not unique in the slightest, (In fact, you could look up the primitive element theorem, which says you don't even need $B$ as $\mathbb{Q}$ is char $0$). So you shouldn't say "Find these $A$, $B$,". You should say "Find a pair $A$, $B$". — mdave16, Apr 20 '17 at 20:29

score 3 · Answer 1 · edited May 29 '16 at 20:54

3

Notice your element satisfies

$$(x^2-6)^2-27=0\iff x^4-12x^2+9=0$$

Then if you adjoin the square root of $3$ you get it to split as

$$x^2-6=3\sqrt3$$

i.e. the degree of the total extension is $4$, so the Galois group has order $4$. But there are only two groups of order $4$, and since you can exhibit more than one element of order $2$, you know the Galois group is the Klein $4$ group. But then all such extensions with that Galois group are of the form $\Bbb Q(\sqrt{A},\sqrt{B})$, completing the proof.

edited May 29 '16 at 20:54

Nicholas Stull

3,014

answered May 29 '16 at 18:48

Adam Hughes

36,777

thank you, makes sense. The only thing is how would I know that the Klein 4 Galois group has the form $\mathbb{Q}(\sqrt{A}, \sqrt{B})$? – amiz9 May 29 '16 at 18:50
@amiz9 The Klein 4 group has a normal subgroup of index 2, hence there is an intermediate field of degree 2, which must be of the form $\Bbb Q(\sqrt A)$. As there are three such normal subgroups, we also get $\Bbb Q(\sqrt B)$ and $\Bbb Q(\sqrt C)$. Can you see why the big field is $\Bbb Q(\sqrt A,\sqrt B,\sqrt C)$ and why one of the square roots can be dropped? – Hagen von Eitzen May 29 '16 at 18:57
@amiz9 the Galois theory shows that subfields correspond to subgroups. Since the only other finite group of order $4$ has only one proper, non-trivial subgroup, showing multiple ones reduces to the Klein $4$ group case, which is the case you're seeking. – Adam Hughes May 29 '16 at 19:10

Use Galois Theory to prove the existence of $A$ and $B$ such that $\mathbb{Q}(\sqrt{6+3\sqrt{3}})=\mathbb{Q}(\sqrt{A}, \sqrt{B})$

1 Answers1