Is $\mathbb{Z}[\sqrt{2},\sqrt{3}]$ flat over $\mathbb{Z}[\sqrt{2}]$?

Question

Is $\mathbb{Z}[\sqrt{2},\sqrt{3}]$ flat over $\mathbb{Z}[\sqrt{2}]$? The definitions doesn't seem to help. An idea of how to look at such problems would be helpful.

Answer 1

$\mathbb Z[\sqrt{2},\sqrt{3}]$ is freely generated as a $\mathbb Z[\sqrt{2}]$-module (exercise). Free modules are flat. QED

Answer 2

I have a way of deciding this, although I don't like it very much.

The ring $\mathbb Z[\sqrt2]$ is a Dedekind domain — it's the ring of integers of $\mathbb Q(\sqrt2)$. A module over a Dedekind domain is flat if and only if it is torsion-free. Why? Well, flatness can be checked at each prime, each localization of a Dedekind domain at a prime is a PID, and the result is true for PIDs.