Subject: Hovey's other question Date: Tue, 7 Oct 2003 21:46:55 -0400 From: Tom Goodwillie >2. This is really an algebraic geometry question, but since I was asking > >the other one...Suppose you have a map X --> Y of schemes such that the >induced map of S-valued points Sch(Spec S, X) --> Sch(Spec S, Y) is an >isomorphism for every connected ring S (in particular, every local ring >S). Must the map of schemes be an isomorphism? If every Spec A was the > >coproduct of connected affine schemes Spec S this would be true, but I >guess I >doubt that every Spec A is such a coproduct. > > Mark Hovey No. Here is an example. Let Y be Spec R, where R=(Z/2)x(Z/2)x... is the product of infinitely many copies of Z/2. Every map from R to a connected ring S must factor R->Z/2->S, since every element of R satisfies x^2=x while in S only 1 and 0 satisfy that equation. So every map from R to S factors through the local ring R_P for some prime ideal P of R. (By the way, R_P is always Z/2.) Let X be the disjoint union of Spec R_P over all P. The obvious map X->Y is not iso, since each point in X is open while the only open points in Y are the ones associated with the projections of the product R on its factors. Tom Goodwillie _________________________________________________________________ Subject: Answer to my own question Date: 08 Oct 2003 06:59:17 -0400 From: Mark Hovey In case anyone was interested in my second question, here is an example that I think shows that two schemes can agree on all connected rings but still be different. Let k be a field and consider Spec k x Z_p, where Z_p denotes the p-adics in the discrete topology. This is the coproduct in the category of schemes of uncountably many copies of Spec k. As a topological space, it is an uncountable discrete set, so it can't be the underlying space of an affine scheme, because those are all quasi-compact (which is the French word for compact). An S-valued point of Spec k x Z_p is a map k --> S together with a continuous map Spec S --> Z_p, where, again, the p-adics have the discrete topology. Now consider the inverse limit of the schemes Spec k x Z/p^n . In general, I don't really know how to take the inverse limit of schemes, but if I take the inverse limit in the category of functors from rings to sets, and the answer happens to be a scheme, that answer has to also be the inverse limit in the category of schemes. In this case that is what happens. Spec k x Z/p^n is Spec F(Z/p^n, k), and the inverse limit is therefore Spec (colim F(Z/p^n, k)), which is the same thing as Spec C(Z_p, k), where C denotes the continuous maps from the p-adics with their profinite topology and k with the discrete topology. An S-valued point of this scheme is a map k --> S together with a continuous map Spec S --> Z_p, where the p-adics are now given the profinite topology. There is therefore a map Spec S x k --> invlim (Spec S x Z/p^n). If S is connected, since the p-adics are totally disconnected, a continuous map Spec S --> Z_p will have to be constant. So these two schemes have the same S-valued points for all connected rings S, but they are still different. Mark Hovey