# On quasi-finite and unramified linear projections.

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 ?

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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

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.

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Thank you very much. Bye. – gio Sep 7 '11 at 7:21