6

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?

Kuba
  • 1,099

1 Answers1

2

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

Med
  • 1,598
  • 1
    The question was about $\mathbb{Q}[\sqrt{2},\sqrt{3}]$ - which is not a field directly from the definitions. It's nontrivial that $\mathbb{Q}[\alpha] = \mathbb{Q}(\alpha)$. –  Dec 08 '13 at 16:20
  • @Mike because $\sqrt 2,\sqrt 3$ algebraic over $\mathbb Q$ – Med Dec 08 '13 at 16:22
  • 2
    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). –  Dec 08 '13 at 16:23