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The title says it all:

Let $A$ be a commutative ring.

Are there any interesting known conditions on $A$ (other then being noetherian of course...) to ensure existence of a (non-zero) commutative noetherian ring $B$, and a flat ring map $A\to B$?

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2 Answers

up vote 3 down vote accepted

If $A$ is integral, you can take $B$ to be the fraction field $K(A)$ of $A$. The morphism $A\to K(A)$ is flat. In general, you (edit: can't) probably take $B$ to be the total field of fractions.

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Are you claiming that the generalized fraction field will always be noetherian? –  anonymous Nov 13 '12 at 22:12
Actually, let me take that back. That can't be correct. What about taking $B$ to be the localization of $A$ at some minimal prime ideal? In this case $B$ is a local zero-dimensional ring. Aren't these always noetherian? –  Harry Nov 13 '12 at 22:19
I proved a theorem which is correct for any ring $A$ which satisfies the condition of the above question. I know that my theorem is not true for any ring, so assuming my proof is correct, the answer to the above question cannot be - for any ring. –  anonymous Nov 13 '12 at 22:20
Dear @Harry: Dimension zero local rings do not have to be Noetherian. See the link in the comments to the answer below. –  Rankeya Nov 14 '12 at 0:29
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Inspired by Harry's observation in the discussion, here is another way to get examples of what you seek. If you take any ring $A$ having a minimal prime ideal $p$ that is finitely generated (but $A$ is not necessarily Noetherian), then $A_p$ is a Noetherian ring (this follows from the result that if all the prime ideals of a ring are finitely generated, then the ring is Noetherian). You will then get a flat ring map $A \rightarrow A_p$, but $A$ is not Noetherian.

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Thanks, so am I correct that not all rings have minimal primes which are finitely generated? –  anonymous Nov 13 '12 at 22:32
Yes you are. See the discussion here: math.stackexchange.com/questions/39458/… –  Rankeya Nov 13 '12 at 22:38
Thanks for both of you. That's funny, because now I actually proved that there is a ring which does not satisfy the above question. Any flat extension of it is not noetherian :) –  anonymous Nov 13 '12 at 22:57
well, any domain will do the case... –  anonymous Nov 13 '12 at 23:40
You are correct. I will edit my answer. –  Rankeya Nov 14 '12 at 0:13
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