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Let $f:X \rightarrow Y$ and $g:Y \rightarrow Z$ be morphisms of irreducible varieties such that $f$ is surjective on scheme points (equivalently geometric points) and $gf$ is flat. Is it necessarily true that $g$ is flat?

The fibres of $g$ at least are equidimensional, so for many cases it is automatic by miracle flatness that this map is flat. I haven't seen this written anywhere which makes me nervous.

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    $\begingroup$ Here's a counterexample: Let $Z$ be the nodal curve, $X$ its nontrivial double etale cover. In other words, $Z$ is $P^1$ glued at 0 and $\infty$, and $X$ is two copies of $P^1$ glued at two points. Let $Y$ be $P^1$ so that $g$ is the normalization. Then $f$ is surjective, $gf$ is etale hence flat, but $g$ is not flat $\endgroup$ Aug 2, 2017 at 2:07
  • $\begingroup$ Fine, but your $X$ is reducible. $\endgroup$
    – abx
    Aug 2, 2017 at 8:58
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    $\begingroup$ Suppose $f,g$ are finite, and $Z$ is regular. Then $X$ (or $Y$) is flat over $Z$ if and only if it is Cohen-Macaulay. So just take for $Y$ a surface with a singular point which is not Cohen-Macaulay (e.g. the cone over a rational quartic curve in $\mathbb{P}^3$) and for $X$ its normalization (which is Cohen-Macaulay). $\endgroup$
    – abx
    Aug 2, 2017 at 9:11
  • $\begingroup$ @abx Oops..Thanks for reminding me, I didn't see the irreducibility assumption. $\endgroup$ Aug 2, 2017 at 14:01

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