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Similar questions were asked here before but I still can't find exactly what I want.

I want to find the degree and a basis for $\mathbb{Q}( \sqrt2, \sqrt3)$ over $ \mathbb{Q}( \sqrt2 +\sqrt 3)$

I've read that $\mathbb{Q}( \sqrt2, \sqrt3)$ is equivalent to $\mathbb{Q}( \sqrt2 +\sqrt 3)$ but we haven't seen it in the lectures so I don't know if I can use it. I would not even know how t use it actually. But that means finding the degree of $\mathbb{Q}( \sqrt2, \sqrt3)$ over $ \mathbb{Q}( \sqrt2,\sqrt 3)$ but there is no irreducible polynomial in $ \mathbb{Q}( \sqrt2,\sqrt 3)$ that is not in $ \mathbb{Q}( \sqrt2,\sqrt 3)$ if that makes sense.

Using the degree formula I get :

$[\mathbb{Q}( \sqrt2, \sqrt3)$ : $ \mathbb{Q}( \sqrt2 +\sqrt 3)$] = $ [\mathbb{Q}( \sqrt2, \sqrt 3 )$:$ \mathbb{Q}( \sqrt2 +\sqrt 3)(\sqrt2)$]=$[\mathbb{Q}( \sqrt2, \sqrt 3 )(\sqrt2)$:$ \mathbb{Q}( \sqrt2 +\sqrt 3)$]

But $\sqrt2 \in \mathbb{Q}( \sqrt2 +\sqrt 3)$ so I don't think that works.

Any ideas ? Thanks !

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If you want to show that the two field are equal, it is the same to show that they are included in each other.

But by definition of $\Bbb Q(\sqrt2,\sqrt3)$,
$\Bbb Q(\sqrt2,\sqrt3) \subset \Bbb Q(\sqrt2 + \sqrt3) \iff \sqrt 2 \in \Bbb Q(\sqrt2 + \sqrt3) $ and $\sqrt 3 \in \Bbb Q(\sqrt2 + \sqrt3) $

And by definition of $\Bbb Q(\sqrt2 + \sqrt3)$,
$\Bbb Q(\sqrt2 + \sqrt3) \subset \Bbb Q(\sqrt2, \sqrt3) \iff \sqrt 2 + \sqrt 3 \in \Bbb Q(\sqrt2, \sqrt3) $

So you only need to find explicit expressions showing that

  • $\sqrt 2 \in \Bbb Q(\sqrt2 + \sqrt3) $ (apparently you know of this one)
  • $\sqrt 3 \in \Bbb Q(\sqrt2 + \sqrt3) $ (it shouldn't be more difficult than the above one)
  • $\sqrt 2 + \sqrt 3 \in \Bbb Q(\sqrt2, \sqrt3)$ (well this one is pretty easy anyway)

Once you know that they are equal, then the degree is $1$ and a basis is $\{1\}$.

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  • $\begingroup$ Thanks, the exercice is then more about proving the equality than looking for the degree and basis ! This question was helpful as well : math.stackexchange.com/questions/93463/… $\endgroup$ – ALM Dec 9 '13 at 16:36

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