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If I have an extension $L/K$ of number fields, then I can take the inclusion $\mathcal{O}_K \hookrightarrow \mathcal{O}_L$ and get a morphism of "curves" $\operatorname{Spec} \mathcal{O}_L \to \operatorname{Spec} \mathcal{O}_K$.

Can I do something like this for an arbitrary field extension? Or at least for a tower $F \subseteq K \subseteq L$ with $\operatorname{trdeg}_F K = \operatorname{trdeg}_F L$?

Daniel McLaury
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  • Why wouldn't you? – Mariano Suárez-Álvarez Apr 07 '14 at 19:21
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    @MarianoSuárez-Alvarez, I wasn't sure what the appropriate analogue of $\mathcal{O}_L$ would be for an arbitrary field. My guess was to take the prime subring and adjoin an independent set of transcendentals, then take the integral closure, but I wasn't quite sure how to go forward: for instance, it wasn't entirely clear to me how to show that didn't depend on the choice of transcendentals, for instance. – Daniel McLaury Apr 07 '14 at 20:27
  • Of course, the answer to your titular question is "yes": $\text{Spec}(L)\to\text{Spec}(K)$. Now, it is possible to phrase your question like this: does every map of fields occur as the map between generic points of a map of positive dimensional schemes. Of course, this is generically possible (except for finite fields!), as is implicit in Georges's answer. But, as Georges also points out, there is not canonical choice. – Alex Youcis Apr 08 '14 at 05:03

1 Answers1

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a) No, I don't think you can do that: if $F=K=\mathbb R\subset L=\mathbb C$, what on earth would be $\mathcal O_\mathbb R$ ?

b) Here is a comment on this rhetorical question.
Strangely, $\mathbb R$ is the fraction field of some non-trivial subring $A\subsetneq \mathbb R$:
Take a transcendence basis $(r_i)_{i\in I}$ of $\mathbb R$ over $\mathbb Q$, then consider the ring $P=\mathbb Q[r_i|i\in I]$ and take for $A$ the integral closure of $P$ in $\mathbb R$.
Then $A$ is not a field because $P$ isn't (Atiyah-Macdonald, Proposition 5.7) but $\operatorname {Frac(A)}=\mathbb R$ (Atiyah-Macdonald, Proposition 5.12) .

c) So, why not take $\mathcal O_\mathbb R=A ?$
Because $A$ is incredibly not canonical and I don't even think that such an $A$ could be proved to exist without the axiom of choice.

Georges Elencwajg
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    I have just asked on MathOverflow whether indeed the existence of such an $A$ with $A\subsetneq \mathbb R=\operatorname{Frac}(A)$ necessitates the axiom of choice. – Georges Elencwajg Apr 07 '14 at 23:24
  • Well, in the case that $F = \mathbb{R}$, both $K=\mathbb{R}$ and $L=\mathbb{C}$ have transcendence degree zero over $F$, so it seems like the right thing there would be the map $\operatorname{Spec} \mathbb{C} \to \operatorname{Spec} \mathbb{R}$ induced by the inclusion $\mathbb{R} \hookrightarrow \mathbb{C}$. – Daniel McLaury Apr 10 '14 at 01:09