# $\Bbb Q [ \sqrt{2} + \sqrt{3} ] = \Bbb Q [ \sqrt{2} , \sqrt{3} ]$ [duplicate]

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?

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

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 $\mathbb Q(\sqrt 2,\sqrt 3)=\mathbb Q(\sqrt 2+\sqrt 3)$
• The question was about $\mathbb{Q}[\sqrt{2},\sqrt{3}]$ - which is not a field directly from the definitions. It's nontrivial that $\mathbb{Q}[\alpha] = \mathbb{Q}(\alpha)$. – user98602 Dec 8 '13 at 16:20
• @Mike because $\sqrt 2,\sqrt 3$ algebraic over $\mathbb Q$ – Med Dec 8 '13 at 16:22