# Is $\mathbb Q(\sqrt 2) \times \mathbb Q(\sqrt 3)=\mathbb Q(\sqrt 2,\sqrt 3)$ if I prove $\sqrt 2,\sqrt 3$ are L.I. over $\mathbb Q$? [duplicate]

I proved that $$\{1,\sqrt 2\}$$ and $$\{1,\sqrt 3\}$$ are respective bases of $$\mathbb Q(\sqrt 2)$$ and $$\mathbb Q(\sqrt 3)$$ over $$\mathbb Q$$. I want to show in some sense that since $$\sqrt 2,\sqrt 3$$ are Linearly Independent over $$\mathbb Q$$, that the product of the bases $$\{1,\sqrt 2\}$$ and $$\{1,\sqrt 3\} =\{1,\sqrt 2,\sqrt 3, \sqrt 6\}$$ is exactly a base $$\mathbb Q(\sqrt 2,\sqrt 3)$$

Can this make sense?

• What have you tried? – jgon Feb 23 '19 at 21:02
• Yes, it works, see the duplicate (or here). We also have $[KL:\mathbb{Q}]=\frac{[K:\mathbb{Q}][L:\mathbb{Q}]}{[K\cap L:\mathbb{Q}]}$ for $K=\Bbb Q(\sqrt{2})$ and $L=\Bbb Q({\sqrt{3}})$. – Dietrich Burde Feb 23 '19 at 21:05
• In fact the indepedence result generalizes to any number of radicals as long as the radicands are multiplicatively independent - see here. – Bill Dubuque Feb 23 '19 at 21:18