do geometric fibers determine scheme-theoretic image? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T04:58:14Z http://mathoverflow.net/feeds/question/73959 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/73959/do-geometric-fibers-determine-scheme-theoretic-image do geometric fibers determine scheme-theoretic image? IMeasy 2011-08-29T10:48:10Z 2011-08-29T18:36:24Z <p>Let $X,Y,Z$ be reduced algebraic varieties, and let $Y$ and $Z$ be normal. Let $f:X \to Y$ and $g:X \to Z$ two surjective projective morphisms of algebraic varieties such that the geometric fibers of $f$ and $g$ coincide. Is there an isomorphism $h:Y\to Z$ such that $g=h \circ f$?</p> http://mathoverflow.net/questions/73959/do-geometric-fibers-determine-scheme-theoretic-image/73995#73995 Answer by Laurent Moret-Bailly for do geometric fibers determine scheme-theoretic image? Laurent Moret-Bailly 2011-08-29T18:36:24Z 2011-08-29T18:36:24Z <p>In positive characteristic, you get a counterexample by taking $X=Y=Z=$ the affine line (say), $f$ the identity and $g$ the Frobenius map.</p> <p>Assume now that the ground field is algebraically closed of characteristic zero. Consider the map $(f,g):X\to Y\times Z$. Its image $\Gamma$ is a closed subvariety of $Y\times Z$. The assumption on the fibers exactly means that both projections from $\Gamma$ to $Y$ and $Z$ are bijective (on closed points). Since they are proper they must be (in char. zero) finite birational, hence (by normality) isomorphisms. So, $\Gamma$ is the graph of the isomorphism we are looking for.</p>