Equality of rational maps

2012-10-14T20:03:28Z

Hi.

Let $f,g:X\dashrightarrow\mathbb{P}^N$ be two rational maps from a complex smooth irreducible projective variety $X$ to a projective space.

Suppose that for every general point $x\in X$ we have $\overline{f^{-1}(f(x))}=\overline{g^{-1}(g(x))}$.

Is true that $\overline{f(X)}$ and $\overline{g(X)}$ are isomorphic (resp. projectively equivalent) ?

If not, is true that $\overline{f(X)}$ is smooth if and only if $\overline{g(X)}$ is smooth?

Thanks.

Answer by Piotr Achinger for Equality of rational maps

2012-10-14T20:13:54Z

The answer all your questions is no. Let $X = \mathbb{P}^1$ and $N=2$. Let $f:X\to \mathbb{P}^2$ be $f(x:y) = (x:y:0)$ and $g:X\to \mathbb{P}^2$ be $g(x:y) = (x^2y:x^3:y^3)$. Then $f$ and $g$ are both bijective, but the image of $g$ is the singular curve $x_0^3 = x_1^2 x_2$.

EDIT. My guess is that when $f$ and $g$ are regular and both images are normal then the answer that $f(X)$ and $g(X)$ are isomorphic could be yes thanks to Zariski's Main Theorem, but I don't see an argument that would show this.

EDIT. Suppose that $f$ and $g$ are both regular. Let $Y_0 = f(X)$ and $Y_1 = g(X)$. Denote the image of $X$ in $Y_0\times Y_1$ under $(f, g)$ by $Z$. Then by your assumption $Z$ projects bijectively onto both $Y_0$ and $Y_1$. If $Y_0$ and $Y_1$ are normal, then by ZMT we have $Y_0 = Z = Y_1$. In any case (even if $f$ and $g$ are rational) we get that $Y_0$ and $Y_1$ are birational.

Equality of rational maps - MathOverflow

Equality of rational maps

Answer by Piotr Achinger for Equality of rational maps