The following question occurred to me while doing exercises in Hartshorne. If $A \to B$ is a homomorphism of (commutative, unital) rings and $f : \text{Spec } B \to \text{Spec } A$ is the corresponding morphism on spectra, does $f$ bijective imply that $f$ is a homeomorphism? If not, can anyone provide a counterexample? The reason this seems reasonable to me is because I think that the inverse set map should preserve inclusions of prime ideals, which is the meaning of continuity in the Zariski topology, but I can't make this rigorous.
bijective morphism of affine schemes
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algebraic-geometry
schemes
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0One observation is that you can reduce to the case where $A\to B$ is an embedding: the result clearly holds if $A\to B$ is onto, because then $P\subseteq Q$ implies $\varphi(P)\subseteq \varphi(Q)$. The map $\mathrm{Spec}(B)\to\mathrm{Spec}(A)$ factors through $\mathrm{Spec}(\varphi(A))$, and the conditions ensure that the contraction map to $\varphi(A)$ is also a bijection, while the map $\mathrm{Spec}(\varphi(A))\to\mathrm{Spec}(A)$ is also a bijection and thus a homeomorphism. So you want to know if the bijection of primes means if $A\subseteq B$, $P\cap A\subseteq Q\cap A$ then $P\leq Q$. – 2011-06-17
1 Answers
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No. Let $A$ be a DVR. Let $k$ be the residue field, $K$ the quotient field. There is a map $\mathrm{Spec} k \sqcup \mathrm{Spec} K \to \mathrm{Spec} A$ which is bijective, but not a homeomorphism (one side is discrete and the other is not). Note that $\mathrm{Spec} k \sqcup \mathrm{Spec}K = \mathrm{Spec} k \times K$, so this is an affine scheme.
As Matt E observes below in the comments, one can construct more geometric examples of this phenomenon (e.g. the coproduct of a punctured line plus a point mapping to a line): the point is that things can go very wrong with the topology.
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2Dear Akhil, Of course my example doesn't exhibit the dimension-theoretic feature that yours does! (This goes hand-in-hand with its "less exotic" nature.) Best wishes, – 2011-06-18