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(In following we are working in "classical" complex setting: i.e. all involved schemes are considered to be varieties over $k=\mathbb{C}$)
Let $X \subset \mathbb{P}^n$ be irreducible surface and $L $ some general $(n-3)$-plane disjoint from $X$. We consider now the projection map $f = \pi_L: X \to \mathbb{P}^2$ from $L$.

Let $Z \subset X$ be the locus of critical points, i.e. points $p \in X$ that are singular or such that the tangent map $df_p: T_pX \to T_{f(p)}\mathbb{P}^2$ has a non trivial kernel $\text{Ker}(df_p) \neq 0 $. It's a fact that this is a proper closed subset of $X$.
Let $B = f(Z)$ be it's image in $\mathbb{P}^2$ which by properness of $f$ has to be closed too. $B$ is from naive point of view given as union of a plane curve and a finite collection of some points.

There is an interesting remark in Harris' book Algebraic Geometry, on p 290 in proof of Prop. 18.10 that one knows moreover that $B$ has no $0$-dimensional components. Unfortunately this fact is not needed for the rest of the proof, so the author not gave a justification for this.

Question: How to check this last statement? Ot looks rather counterintuitive, since thinking of contractions of curves one might expect $B$ might have $0$-dimensional components.

Thoughts: It seems not to be true for all generically finite projective morphisms $f: X \to \mathbb{P}^2$ of surfaces, since for example the blowup at center $0 \in \mathbb{P}^2$ gives $B=0$. So if the statement is true then it must be based on special structure of projections from linear subspaces, but I not find an argument. I also not know how "deep" this result is, ie which "tools" are required to prove it. Even if the quoted book is for undergrades, Harris often quotes there some facts going far beyond the scope of the book, and so I'm not sure if this property of $B$ is a result of advanced research or can it be seen with rather "elmentary" tools.

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    $\begingroup$ This type of result is called "purity of the branch locus." Here's a discussion with some references in the stacks project: stacks.math.columbia.edu/tag/0BMB $\endgroup$
    – Samir Canning
    Commented Feb 24, 2023 at 15:21
  • $\begingroup$ @SamirCanning: Thank you, this looks rather promising. So it seems that we can extract from this purity result the following: if $f: X \to Y$ is a finite map of varieties over $\mathbb{C}$ with $X$ normal and $Y$ smooth (regular suffice), then the branch locus has pure codimension one. $\endgroup$
    – user267839
    Commented Feb 24, 2023 at 16:24
  • $\begingroup$ By the way: is the term "branch locus" in modern setting compatible with the old or "classical" notation of the "image of critical points" $Z$ as defined in the question? $\endgroup$
    – user267839
    Commented Feb 24, 2023 at 16:24
  • $\begingroup$ I assume in following that branch locus in the question is the same as $B=f(Z)$ in the question, please correct me if I'm wrong. Let come back to our generically finite morphism $\pi_L: X \to \mathbb{P}^2$. There is an open subset $U \subset \mathbb{P}^2$ over which $\pi_L$ the fibers are finite and over the closed complement $C:=\mathbb{P}^2-U$ all the fibers have positive dimension, ie "non finite locus". $\endgroup$
    – user267839
    Commented Feb 24, 2023 at 16:25
  • $\begingroup$ If the notations of branch locus and $B=f(Z)$ in the question coinside, then the branch locus equals to the union of the closed $C$ and the branch locus of the restriction of $\pi_L$ to preimage of $U$, which is finite. So we can split the problem to analyze $C$ and the finite restriction $(\pi_L)_{\vert V}$, where $V:= \pi_L^{-1}(U)$. $\endgroup$
    – user267839
    Commented Feb 24, 2023 at 16:26

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The morphism $f$ is not just generically finite but finite. I think this resolves all the difficulties.

To see this, since $f$ is proper it just suffices to check it is quasi-finite, and thus to check the fiber over a point can't contain a curve. But the fiber over a point is a $(n-2)$-plane containing $L$, and any curve in an $(n-2)$-plane intersects every $(n-3)$-plane, so if such a curve were contained in the fiber it would contradict the assumption that $X$ is disjoint from $L$.

Alternately, we can argue that $\mathbb P^n \setminus L \to \mathbb P^2$ is affine and thus $f$ is affine, and, since affine and proper, is finite.

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  • $\begingroup$ Thank you, I see. But one problem stays still unsolved (compare with comments above): to apply purity here we have to ensure $X$ is normal, but Harris not assumed this. Can this problem be somehow fixed? Eg passing to normailzation of $X$ or restrict the map to smooth locus? But both naive approach are seemingly flawed, as in comments explained. $\endgroup$
    – user267839
    Commented Feb 24, 2023 at 16:52
  • $\begingroup$ @user7391733 No, I think normality is necessary for the conclusion to hold. If $X$ is obtained by gluing together two smooth points in a smooth surface then I think a generic projection will have the image of the glued point as an isolated point of the branch locus. $\endgroup$
    – Will Sawin
    Commented Feb 24, 2023 at 16:56
  • $\begingroup$ I see, presumingly Harris assumed $X$ indeed to be even smooth, this reduction make sense since the goal was to show that for $X$ irreducible, the intersection $Y:= X \cap H$ with general hyperplane $H$ is irreducible and a general hyperplane intersects $X$ in smooth locus, so we can assume $X$ to be even smooth. $\endgroup$
    – user267839
    Commented Feb 24, 2023 at 17:14
  • $\begingroup$ One question about your counterexample showing that "normality" cannot be dropped: Is this example with gluing two points on a smooth surface of "algebraic" or only "analytic" nature? Clearly we can construct such pathologies topologically and analytic/holomorphic functions are "flexible" enough to allow to implement such gluing construction to analytic geometry. Algebraic geometry is in that sense more "rigid", so I'm not sure why such identification of two points on a smooth surface (or even weirder identifications) works here in algebraic realm? $\endgroup$
    – user267839
    Commented Feb 24, 2023 at 17:14
  • $\begingroup$ @user7391733 Some gluings are hard to construct in the category of schemes but gluing pairs of points are relatively easy (one can consider first the case of an affine variety and do it by algebra). In general, gluings may not exist in the category of schemes but will in the category of algebraic spaces - the category of algebraic spaces is about as powerful as the category of analytic spaces for this. $\endgroup$
    – Will Sawin
    Commented Feb 24, 2023 at 18:15

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