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This question already has an answer here:

Prove, that $ \Bbb Q [ \sqrt{2} + \sqrt{3} ] = \Bbb Q [ \sqrt{2} , \sqrt{3} ] $

I don't know the definition of $\Bbb Q [ \sqrt{2} , \sqrt{3} ]$, can anyone help me with this?

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marked as duplicate by Najib Idrissi, egreg, user26857, user1337, Joe Johnson 126 Dec 8 '13 at 17:14

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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First we have : $\mathbb Q(\sqrt 2,\sqrt 3)=\mathbb Q[\sqrt 2,\sqrt 3]$ because $\sqrt 2,\sqrt 3$ are algebraic over $\mathbb Q$

$\mathbb Q(\sqrt 2,\sqrt 3)$ this is the smallest field containing $\mathbb Q,$ $\sqrt 2$ and $\sqrt 3$

Since $\sqrt 2+\sqrt 3\in \mathbb Q(\sqrt 2,\sqrt 3) $ we have $\mathbb Q(\sqrt 2+\sqrt 3)\subset\mathbb Q(\sqrt 2,\sqrt 3)$ then it suffice to prove that $[\mathbb Q(\sqrt 2,\sqrt 3):\mathbb Q]=[\mathbb Q(\sqrt 2+\sqrt 3):\mathbb Q],$ as $[\mathbb Q(\sqrt 2,\sqrt 3):\mathbb Q]=[\mathbb Q(\sqrt 2,\sqrt 3):\mathbb Q(\sqrt 2)][\mathbb Q(\sqrt 2):\mathbb Q]$ we have $[\mathbb Q(\sqrt 2,\sqrt 3):\mathbb Q]=4$ (because Irr($\sqrt 3$,$\mathbb Q(\sqrt 2))=X^2-3.$) you can easily check by calculating that : deg(Irr($\sqrt 3+\sqrt 2$,$\mathbb Q)$=4

so $\mathbb Q(\sqrt 2,\sqrt 3)=\mathbb Q(\sqrt 2+\sqrt 3)$

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    $\begingroup$ The question was about $\mathbb{Q}[\sqrt{2},\sqrt{3}]$ - which is not a field directly from the definitions. It's nontrivial that $\mathbb{Q}[\alpha] = \mathbb{Q}(\alpha)$. $\endgroup$ – user98602 Dec 8 '13 at 16:20
  • $\begingroup$ @Mike because $\sqrt 2,\sqrt 3$ algebraic over $\mathbb Q$ $\endgroup$ – Med Dec 8 '13 at 16:22
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    $\begingroup$ I know why it's true. But to a student who's just seen these for the first time, they're different, and the question he's looking to answer likely wants him to explicitly show that they're the same (by showing they have the same elements). $\endgroup$ – user98602 Dec 8 '13 at 16:23

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