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 $\mathbb Q$?

Question

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?