# Does the functor of points commutes with inverse limits?

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.

• 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. Nov 21, 2012 at 11:41

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