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A scheme $X$ defines a (covariant) functor from commutative rings to sets by

$A\mapsto X(A)=Mor(SpecA,X)$

Does this functor commutes with inverse limits? It does when $X$ is affine, but I can't see the answer in the general case.

Edit:

As Zhen nicely explained, Spec does not preserve limits in general, but if the inverse system is the the specific one:

$\dots\to\mathbb{C}[t]/t^{n+1}\to\mathbb{C}[t]/t^n\to\dots\to\mathbb{C}[t]/t^2\to\mathbb{C}$

Whose inverse limit is $\mathbb{C}[[t]]$, is it true then? Namely, is it true that $X(\mathbb{C}[[t]])$ is the inverse limit of $X(\mathbb{C}[t]/t^n)?$ If so, what would be the proof?

Sorry for not asking the correct question from the beginning.

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  • $\begingroup$ As indicated in my answer, if $\operatorname{Spec} \mathbb{C}[[t]]$ is the filtered colimit of $\cdots \to \operatorname{Spec} \mathbb{C}[t]/(t^n) \to \cdots$ in $\textbf{Sch}$ (and I have not checked this, but it appears to be true) then $X(\mathbb{C}[[t]])$ will indeed be the inverse limit of $\cdots \to X(\mathbb{C}[t]/(t^n)) \to \cdots$, by the first paragraph of my answer. $\endgroup$
    – Zhen Lin
    Commented Nov 21, 2012 at 11:41

1 Answer 1

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The functor you define is a composite of two functors: the scheme-representable functor $\textbf{Sch}(-, X) : \textbf{Sch}^\textrm{op} \to \textbf{Set}$ and the functor $\textrm{Spec} : \textbf{CRing} \to \textbf{Sch}^\textrm{op}$. Now, it is a general fact about representable functors that they map all colimits to the corresponding limits, so for example $\textbf{Sch}(-, X)$ will map disjoint unions of schemes to products of sets, etc.

The problem is in the $\textrm{Spec}$ functor. Unfortunately it is not true that $\textrm{Spec}$ sends limits in $\textbf{CRing}$ to colimits in $\textbf{Sch}$: in fact this is not even true for products. For example, if $k$ is a field and $A$ is the product of infinitely many copies of $k$, then $\operatorname{Spec} A$ is not going to be the disjoint union of infinitely many copies of $\operatorname{Spec} k$. (Assuming the axiom of choice anyway...) However, it is well-known that $\operatorname{Spec} 0 = \emptyset$ and $\operatorname{Spec} A \times B = \operatorname{Spec} A \amalg \operatorname{Spec} B$, so $\textrm{Spec}$ at least sends finite products to finite coproducts.

Also, essentially by design $\operatorname{Spec}$ fails to preserve equalisers. Indeed, if $\operatorname{Spec}$ preserved equalisers then there would be no chance of constructing any projective schemes. (Consider the construction of the projective line by gluing together two copies of the affine line.)

In summary, the functor of points does not preserve all limits in general; but it does at least preserve finite products.

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  • $\begingroup$ Thank you for the answer. I would still be happy to see a specific counter example with a simple filtered inverse limit like $\dots\to\mathbb{C}[t]/t^{n+1}\to\mathbb{C}[t]/t^{n}\to\dots$. $\endgroup$
    – KotelKanim
    Commented Nov 21, 2012 at 6:28
  • $\begingroup$ As far as I can tell, $\textrm{Spec}$ transforms that into a filtered colimit, so that's not a counterexample. $\endgroup$
    – Zhen Lin
    Commented Nov 21, 2012 at 7:45

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