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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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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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    Dear @Harry: Dimension zero local rings do not have to be Noetherian. See the link in the comments to the answer below.2012-11-14
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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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    There was an unfortunate mistake in my proof, so I don't know it anymore :(2012-11-14