Suppose $X\rightarrow {\rm Spec}(A)$ is a morphism between a scheme and an affine scheme that

1) is a bijection topologically and

2) each scheme theoretic fiber is a single reduced point

3) both $X$ and $A$ are reduced

Then is $X$ affine?

  • $\begingroup$ Do you have a counterexample when only 1) holds? $\endgroup$ – MooS Mar 18 '17 at 20:29
  • 1
    $\begingroup$ If $f$ is an isomorphism of underlying topological spaces, $f_*$ is exact, in particular $H^i(X,F)=H^i(\operatorname{Spec} A,f_*F)=0$ for any quasi-coherent sheaf $F$ on $X$. Hence $X$ is affine. What am I missing? $\endgroup$ – MooS Mar 18 '17 at 20:32
  • $\begingroup$ I guess maybe the topology of $X$ is finer than that of ${\rm Spec}(A)$. But that's a good idea…I didn't think of that. $\endgroup$ – DCT Mar 18 '17 at 20:35
  • $\begingroup$ Ah ok, so maybe $f$ is only a bijection, but not a homeomorphism. That might be a problem. $\endgroup$ – MooS Mar 18 '17 at 20:38

Here is one example where $X$ is not affine. A similar idea would work if $A$ is reduced with infinitely many points in $\operatorname{Spec}A$, and you took the same definition for $X$ below using residue fields of points in $\operatorname{Spec}A$. A more interesting question might be if your question is true assuming $X$ is connected.

Let $k$ be a field, and consider the morphism $$X := \coprod_{x \in \mathbf{A}^1_k} \operatorname{Spec} \kappa(x) \longrightarrow \mathbf{A}^1_k$$ where $\kappa(x)$ denotes the residue field at a point $x \in \mathbf{A}^1_k$, and the map is defined by mapping the unique point in $\operatorname{Spec} \kappa(x)$ to $x$. This is a bijection on topological spaces by construction, each scheme theoretic fiber is a spectrum of a field, hence is a single reduced point, and $X$ is reduced. On the other hand, $X$ is not quasicompact, hence $X$ cannot be affine.

  • $\begingroup$ Maybe one should change the wording of 1) to a set-theoretic bijection then. $\endgroup$ – MooS Mar 19 '17 at 8:50
  • $\begingroup$ @MooS Right, I guess I understood OP's replies to your comments as saying the map on underlying sets need not be a homeomorphism. $\endgroup$ – Takumi Murayama Mar 19 '17 at 17:01

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