Equality of rational maps - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T02:41:40Z http://mathoverflow.net/feeds/question/109649 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/109649/equality-of-rational-maps Equality of rational maps gio 2012-10-14T20:03:28Z 2012-10-14T21:26:19Z <p>Hi.</p> <p>Let $f,g:X\dashrightarrow\mathbb{P}^N$ be two rational maps from a complex smooth irreducible projective variety $X$ to a projective space.</p> <p>Suppose that for every general point $x\in X$ we have $\overline{f^{-1}(f(x))}=\overline{g^{-1}(g(x))}$.</p> <p>Is true that $\overline{f(X)}$ and $\overline{g(X)}$ are isomorphic (resp. projectively equivalent) ?</p> <p>If not, is true that $\overline{f(X)}$ is smooth if and only if $\overline{g(X)}$ is smooth?</p> <p>Thanks.</p> http://mathoverflow.net/questions/109649/equality-of-rational-maps/109650#109650 Answer by Piotr Achinger for Equality of rational maps Piotr Achinger 2012-10-14T20:13:54Z 2012-10-14T21:26:19Z <p>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 $x_0^3 = x_1^2 x_2$.</p> <p><strong>EDIT</strong>. My guess is that when $f$ and $g$ are regular and both images are <em>normal</em> 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.</p> <p><strong>EDIT</strong>. 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.</p>