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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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    It is pretty hard to come up with examples where it does *hold*, im fact :)2011-03-29

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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...2011-03-29