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I am sorry for asking two questions in one but they are strongly related.

  • What is an example of (affine?) schemes $X=(|X|,\mathcal{O}_X)$ and $Y=(|Y|,\mathcal{O}_Y)$ and a map of topological spaces $|f|\colon|X|\to |Y|$ that cannot be promoted into a map $f\colon X\to Y$ of schemes?

I guess something like $exp:\mathbb{R}\to\mathbb{R}$ is an example but I cannot prove that it is an example.

  • What is an example of (affine?) schemes $X=(|X|,\mathcal{O}_X)$ and $Y=(|Y|,\mathcal{O}_Y)$ and a map of topological spaces $|f|\colon|X|\to |Y|$ that can be promoted into a map $f_1\colon X\to Y$ of schemes and into a map $f_2\colon X\to Y$ a map of schemes with $f_1\neq f_2$?
user8463524
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  • The exponential map $\mathbb R \to \mathbb R$ cannot be a polynomial function, because it grows too fast. In attempting to make this an answer for the first question, you need to define a continuous map $\text{Spec}\mathbb R[x] \to \text{Spec}\mathbb R[x]$ of sets. You've already said what you want the ideal $(x-c)$ to be carried to, and you might as well send the generic point ${0}$ to itself. That leaves the remaining maximal ideals that are not of the form $(x-c)$, which exist because $\mathbb R$ is not algebraically closed. Can you work out what the continuity condition means here? – Dustan Levenstein Aug 19 '14 at 14:55
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    I like your idea of promotion from topology to algebraic geometry: hierarchy should be preserved, at least if you are at the upper echelon :-) – Georges Elencwajg Aug 19 '14 at 15:05

3 Answers3

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Just take $X=\mathrm{Spec}(K)$ and $Y=\mathrm{Spec}(L)$ for two fields $K,L$.

There is a unique map $|X| \to |Y|$. The morphisms $X \to Y$ correspond to field homomorphisms $L \to K$. There may be no such homomorphisms, but there may be also many of them. (For example, consider $\mathbb{Q}(\sqrt{2}) \to \mathbb{Q}$ or $\mathbb{Q}(\sqrt{2}) \to \mathbb{Q}(\sqrt{2})$).

Martin Brandenburg
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1) Take $X=Y=\mathbb A^1_\mathbb C$, so that $|X|=\mathbb C\sqcup \{\eta\}$.
Any permutation of $|X|$ fixing the generic point $\eta$ induces a homeomorphism $f:|X|\to |X|$ .
If the induced permutation on the subset of closed points $\mathbb C\subset |X|$ is not continuous in the classical topology, then the map $f$ cannot come from a scheme morphism $X\to X$.
As an example you can take the permutation exchanging $0$ and $1$ and fixing everything else.

2) Given a field $k$ and a non-trivial automorphism $\phi:k\to k$, the induced scheme morphisms $\phi^*, Id^*:\text {Spec}(k)\to \text {Spec}(k)$ are different but induce the same homeomorphism (of one point spaces!) $|\text {Spec}(k)|\to |\text {Spec}(k)|$

Georges Elencwajg
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An interesting example is afforded by a discrete valuation ring $R$ and its quotient field $k$. $spec \ R=\{(0), \mathfrak{p}\}$, with $\mathfrak{p}$ closed and $(0)$ open. And $spec \ k=\{(0)\}$. Now the inclusion map $R\rightarrow k$ induces $spec \ k\rightarrow spec \ R$ where $(0)\mapsto (0)$ the map on sheaves is $k\rightarrow k$ which is local. However $(0) \mapsto \mathfrak{p}$ is also a continous map. However the induced map in sheaves is $R\rightarrow k$, the inclusion map and this is not a local map since the inverse image of $(0)$ is not the maximal ideal of $R$.

Rene Schipperus
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