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 $\mathbb Q$?

Asked Feb 23 '19 at 21:00

Active Feb 24 '19 at 02:37

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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?

edited Feb 24 '19 at 02:37

J. W. Tanner

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asked Feb 23 '19 at 21:00

Kam

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

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 $\mathbb Q$?

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