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?

  • $\begingroup$ What have you tried? $\endgroup$ – jgon Feb 23 '19 at 21:02
  • $\begingroup$ 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 $L=\Bbb Q({\sqrt{3}})$. $\endgroup$ – Dietrich Burde Feb 23 '19 at 21:05
  • $\begingroup$ In fact the indepedence result generalizes to any number of radicals as long as the radicands are multiplicatively independent - see here. $\endgroup$ – Bill Dubuque Feb 23 '19 at 21:18