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Let $f: X\to Y$ be a flat morphism between smooth complex affine varieties. Let $Z$ be the closed set of most singular points of $f$ (in the sense: $p$ is a most singular point of $f$ if the tangent map $Tf_p$ at $p$ has the lowest rank among the closed points in $X$). Denote $r:=\text{rank} Tf_p$ for $p\in Z$. Is it true that $\text{dim}f(Z)=r$ (generic smooth theorem tells us it is $\leq r$). If not in general, is there a quick example for $\text{dim}f(Z)<r$?

Newly updated: Is there a counterexample for proper flat morphism $f: X\to Y$ between smooth complex manifolds $X$ and $Y$?

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    $\begingroup$ Am I missing something? Take the standard blowup $f:\mathbb{C}^2\rightarrow \mathbb{C^2}$, $f(x,y)=(x,xy)$. Then $Z$ is the line $x=0$, $\operatorname{rk} T_p(f)=1$ for $p\in Z$ (and 2 otherwise), but $f(Z)=(0,0)$. $\endgroup$
    – abx
    Aug 10 '20 at 19:31
  • $\begingroup$ @abx Sorry I overlooked the trivial examples. I just added the necessary flatness condition. $\endgroup$
    – Feng Hao
    Aug 10 '20 at 19:52
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    $\begingroup$ This answer is cribbed from Johan de Jong's blog (if he would like to post this as an answer, I will delete this comment). Consider the morphism $f:\mathbb{A}^3\to \mathbb{A}^2$ given by $f(x,y,z)=(x,xz+y^2)$. This is flat of relative dimension $1$. The singular locus in $X$ equals $\text{Zero}(x,y)$, and the rank of the derivative map equals $1$ on this locus. Yet the image in $\mathbb{A}^2$ of the critical locus is the origin, which has dimension $0$. $\endgroup$ Aug 11 '20 at 9:28
  • $\begingroup$ @JasonStarr Thank you for pointing out the interesting example on de Jong's blog! I just voted for your comment as a right example. $\endgroup$
    – Feng Hao
    Aug 11 '20 at 9:55

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