MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $L=\mathbb{P}^l\subset\mathbb{P}^N$ be a fixed linear space, $l\geq0$, and let $M=\mathbb{P}^{N-l-1}$ be a linear space skew to $L$, i.e. $L\cap M=\emptyset$ and $\langle L, M\rangle=\mathbb{P}^N$. Let $X\subseteq\mathbb{P}^N$ be a closed irreducible variety not contained in $L$ and let $$ \pi_L:X\dashrightarrow\mathbb{P}^{N-l-1}=M $$ be the linear projection, i.e. the rational map defined on $X\setminus(L\cap X)$ by $$ \pi_L(x)=\langle L,x\rangle\cap M. $$

I say that (denoting by $x$ the general point of $X$):

  • $\pi_L$ is generically quasi-finite if $\pi_L^{-1}(\pi_L(x))$ is a finite set;
  • $\pi_L$ is generically unramified if $\pi_L^{-1}(\pi_L(x))$ coincide, as a scheme, with the point $x$ in a neighbourhood of $x$.

Is it true that if $\pi_L$ is generically quasi-finite then it's generically unramified ?

share|cite|improve this question
If I understood correctly what you are after, I think that the strange curves of Hartshorne's Definition on page 311 would give you problems. In particular, if $C$ is a smooth plane conic in characteristic 2 there is a point $p$ in the plane contained in all tangent lines to $C$. In this case, choosing $L=p$, all fibers are generically quasi-finite, but they are all non-reduced, so that the morphism is not generically unramified. It is true though that smooth conics in characteristic 2 are (essentially) the only strange curves... – M P Sep 6 '11 at 8:12
up vote 3 down vote accepted

The answer is yes in characteristic $0$.

In fact, take the non-empty Zariski open set $X^0$ where $\pi \colon X \to M$ has finite fibres, and let $M^0 \subset M$ be the image of $X^0$. Then $M^0$ is a Zariski open set of $M$ and the restriction $\pi^0 \colon X^0 \to M^0$ is a finite map. Passing to function fields, this means that $$(\pi^{0})^*(K(M^0)) \subset K(X^0)$$ is a finite field extension, and since we are working in characteristic $0$ it is also separable.

Now we can apply the following result, see [Shafarevich, Basic Algebraic Geometry I, Theorem 4 of Chapter II, page 144]:

Theorem. The set of points where a finite map $f \colon X \to Y$ is unramified is open, and it is nonempty if $f^*(K(Y)) \subset K(X)$ is a separable field extension.

It follows that $\pi^0 \colon X^0 \to M^0$ is generically unramified, so a fortiori $\pi \colon X \to M$ is generically unramified.

If the characteristic of the base field is $p >0$ the result is no longer true, as it is shown by MP's comment about strange curves.

share|cite|improve this answer
Thank you very much. Bye. – gio Sep 7 '11 at 7:21

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.