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All varieties here is defined over complex number. Let $X$ be a normal projective irreducible variety and let $Y$ be a projective irreducible variety. Suppose that there is a bijective morphism $f : X \to Y$. This does not imply that $Y$ is normal. One can easily construct a counterexample for curve case.

My question is:

1) Is it true if $Y$ is regular in codimension 1?

2) Is there any extra condition guarantee normality of $Y$?

I searched some related questions and answers such as

Nonsingular/Normal Schemes or

Checking whether a variety is normal ,

but I cannot find or prove a method working on my situation. In my problem, finding local defining equations is almost impossible.

Edit: One of motivation is a kind of inverse problem of Zariski's main theorem. Suppose that $g : X \to Y$ is a morphism between projective varieties with connected fiber from normal variety $X$. By taking Stein factorization, we can reduce to the situation of question.

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1 Answer

To answer your questions.

1) No, for example, $k[x^2, x^3, xy, x^2y, y^2, xy^2, y^3] \subseteq k[x,y]$ induces a bijection on points. The former is also regular in codimension 1. It is worth noting that you can make these projective without much work.

If you mean honest points and not just geometric ones, then $$\mathbb{R}[x, y, ix, iy] \subseteq \mathbb{C}[x,y]$$ also induces a bijection on points and is an isomorphism away from the origin.

2) The condition you are probably looking for is called seminormality. If $Y$ is seminormal and you are working in characteristic zero, then what you are looking for holds.

EDIT: However, BIJECTIVE is not enough in the non-projective case (so it doesn't really matter to you), in general one should also assume that the map $X \to Y$ is finite/proper (see the Erratum to the Leahy-Vitulli paper I list below).

Here's a definition of seminormal.

Definition (Traverso, Greco-Traverso): An excellent scheme $Y$ is seminormal if every finite morphism $X \to Y$ which

  • is a bijection on points, and
  • induces isomorphisms at all residue fields

is an isomorphism.

When working over algebraically closed fields of characteristic zero, only the first condition matters.

Things like nodes and more generally normal crossings singulities are always seminormal. Cusps are not seminormal (they are the canonical example of non-semi-normality).

In fact, semi-normality also behaves well with respect to Zariski's main theorem and Stein factorization constructions because if $X$ is semi-normal, then $\Gamma(X, \mathcal{O}_X)$ is seminormal as a ring (it's $\text{Spec}$ is seminormal). A very algebraic definition of a seminormal ring $R$, which makes this trivial to see, is as follows.

Definition (Swan): A reduced Noetherian ring $R$ is said to be seminormal if whenever there are elements $a, b \in R$ such that $a^2 = b^3$, then there exists an element $c \in R$ such that $c^3 = a$ and $c^2 = b$.

The main sources for learning about seminormality are probably

  1. Greco-Traverso, ``On seminormal schemes''.
  2. Swan, ``On seminormality''.
  3. Leahy-Vitulli, ``Seminormal rings and weakly normal varieties''.
  4. Vitulli, ``Weak Normality and Seminormality'' (a survey paper, but quite algebraic).
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Dear Karl, don't you mean that $Y$ should be semi-normal ? –  Georges Elencwajg Apr 5 '11 at 14:51
    
Yes, I was mapping $Y$ to $X$ in my head. I'll fix it now, thanks! –  Karl Schwede Apr 5 '11 at 14:53
    
Dear Karl, thank you for great answer. I have another question. To show the normality of given variety, is it the best way checking R1 and S2? In my situation, given variety $X$ is an incident variety of points and curves in projective space. By dimension estimate, R1 is not too difficult, but I can't figure out how can I prove S2 (or CM) condition. –  Moon Apr 7 '11 at 2:16
    
Moon. If your variety is (locally) a complete intersection, then it is automatically S2 (and CM). Outside of that case, it can be hard, so it depends on exactly what you know. Of course, the fact that you have an incidence variety might be helpful, those sort of things often be S2 I would suspect. For trying to prove CM, one approach is to show that $H^i(X, L^{-n}) = 0$ for $L$ ample, $n \gg 0$ and $i < \dim X$. Grothendieck duality and Serre vanishing then forces $X$ to be Cohen-Macaulay. –  Karl Schwede Apr 7 '11 at 2:27
    
Thank you. I'll try it. –  Moon Apr 8 '11 at 6:00
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