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Suppose $A$, $B$ and $C$ are commutative noetherian rings. Suppose we are given maps $f:A\to B$ and $g:B \to C$.

Suppose further that $f$ is flat and that $g \circ f$ is also flat. Does it follows that $g$ is flat?

Thank you!

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I am not thrilled about this question. As Akhil's answer shows, counterexamples are almost immediate. The material involved is at the graduate level and a graduate student should be practicing and improving her skills at exploring implications of definitions, searching for theorems and counterexamples and so forth. It is okay not to be able to come up with the answer to any given question (it had better be...), but the way the question is written it looks like the OP hasn't thought about it or tried anything. So what will he learn from it? Almost nothing. –  Pete L. Clark Mar 29 '11 at 16:37
    
It is pretty hard to come up with examples where it does hold, im fact :) –  Mariano Suárez-Alvarez Mar 29 '11 at 16:53

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No; consider $A = k, B = k[x], C = k$ (considered as a $B$-module by, say, having $x$ act by zero).

In general, in algebraic geometry, there is a general property that you can cancel many properties of schemes (if $g \circ f$ satisfies something, so does $f$; this works for separatedness, among other things), but to do this, you need the diagonal map to satisfy this property. (E.g. the diagonal map is always separated, as an immersion.) But generally the diagonal map is not flat, so this kind of cancellation doesn't work. (N.B. If a composition $X \to Y \to Z$ is \'etale and $Y \to Z$ is unramified, then $ X \to Y$ is, nonetheless, etale; this is because the diagonal of an unramified morphism is an open immersion.)

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thank you for your answer! what I really have is that f and $g\circ f$ are formally etale and I want to know if $g$ is, so I guess trying to refer to the different ingredients of formal etaleness will not work here... –  the L Mar 29 '11 at 22:08

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