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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.

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Dear rola, It is pretty obviously free, and free modules are flat. Regards, – Matt E Jul 24 '12 at 5:31
@MattE That's a lot better — would you mind posting that as an answer? I can incorporate it into what I wrote if you don't have the time, but it wouldn't be as good. I somehow convinced myself that the module structure would be weird, but of course it isn't. Cheers, – Dylan Moreland Jul 24 '12 at 5:38
Dear Dylan, Done. Cheers, – Matt E Jul 24 '12 at 5:40
@rola Note that there is still something to do here: take what you believe is a basis for $\mathbb Z[\sqrt2, \sqrt3]$ over $\mathbb Z[\sqrt2]$ and prove that it is one. – Dylan Moreland Jul 24 '12 at 5:46
up vote 6 down vote accepted

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

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Thanks a lot Matt for taking time to answer. – rola Jul 24 '12 at 6:09

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.

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Unlike you, I like this answer: +1 ! – Georges Elencwajg Jul 24 '12 at 8:36

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