# Dimension of highest discriminants of a morphism

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)?

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

• 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 $f(Z)=(0,0)$.
– abx
Aug 10 '20 at 19:31
• @abx Sorry I overlooked the trivial examples. I just added the necessary flatness condition. Aug 10 '20 at 19:52
• 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 $0$. Aug 11 '20 at 9:28
• @JasonStarr Thank you for pointing out the interesting example on de Jong's blog! I just voted for your comment as a right example. Aug 11 '20 at 9:55