Subject: Hovey's other question
Date: Tue, 7 Oct 2003 21:46:55 -0400
From: Tom Goodwillie <tomg@math.brown.edu>

>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 <hovey@picard.math.wesleyan.edu>

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