Skip to main content
MathOverflow

Return to Question

Post Closed as "Duplicate" by Igor Rivin, abx ag.algebraic-geometry
added 326 characters in body
Source Link
Julia
Julia
  • 397
  • 3
  • 6

Let $k$ be an algebraically closed field of characteristic $0$ and let $f:X \rightarrow Y$ be a map between smooth varieties over $k$ that is bijective on closed points. Musr $f$ be an isomorphism?

The restriction to fields of characteristic $0$ is necessary to rule out the Frobenius map, and the restriction to smooth varieties is necessary to rule out resolutions of singularities of curves. My guess is that there will be counterexamples with $X$ and $Y$ algebraic surfaces, but I do not know a rich enough store of examples of surfaces to find one.

Edit: I assert nothing about the maps on tangent spaces (unlike the question that someone has claimed is a duplicate of this one), and would be happy for a counterexample where that is the issue (but as I said, I insist that my varieties be smooth, which rules out the counterexamples in the purported duplicate question).

Let $k$ be an algebraically closed field of characteristic $0$ and let $f:X \rightarrow Y$ be a map between smooth varieties over $k$ that is bijective on closed points. Musr $f$ be an isomorphism?

The restriction to fields of characteristic $0$ is necessary to rule out the Frobenius map, and the restriction to smooth varieties is necessary to rule out resolutions of singularities of curves. My guess is that there will be counterexamples with $X$ and $Y$ algebraic surfaces, but I do not know a rich enough store of examples of surfaces to find one.

Let $k$ be an algebraically closed field of characteristic $0$ and let $f:X \rightarrow Y$ be a map between smooth varieties over $k$ that is bijective on closed points. Musr $f$ be an isomorphism?

The restriction to fields of characteristic $0$ is necessary to rule out the Frobenius map, and the restriction to smooth varieties is necessary to rule out resolutions of singularities of curves. My guess is that there will be counterexamples with $X$ and $Y$ algebraic surfaces, but I do not know a rich enough store of examples of surfaces to find one.

Edit: I assert nothing about the maps on tangent spaces (unlike the question that someone has claimed is a duplicate of this one), and would be happy for a counterexample where that is the issue (but as I said, I insist that my varieties be smooth, which rules out the counterexamples in the purported duplicate question).

Source Link
Julia
Julia
  • 397
  • 3
  • 6

Bijective map of smooth varieties that is not an isomorphism

Let $k$ be an algebraically closed field of characteristic $0$ and let $f:X \rightarrow Y$ be a map between smooth varieties over $k$ that is bijective on closed points. Musr $f$ be an isomorphism?

The restriction to fields of characteristic $0$ is necessary to rule out the Frobenius map, and the restriction to smooth varieties is necessary to rule out resolutions of singularities of curves. My guess is that there will be counterexamples with $X$ and $Y$ algebraic surfaces, but I do not know a rich enough store of examples of surfaces to find one.