Re: Basic Varieties, Sheaves, Prevarieties question

James wrote:

Dear all,

There are a few things confusing me. I state one
question below, and then
my second question towards the bottom. I am reading
Algebraic Geometry by
Daniel Bump, and he states the following on page 82
and 83 :

Let (X, O) be a prevariety. i.e. X is an
irreducible topological space, O
is a sheaf of rings on X, and X admits a finite
open cover of affine open
sets (i.e. open sets each isomorphic to some affine
variety).

Let U be an open subset of X. Then he says that
even though O(U) is just a
ring, we can think of elements of O(U) as being
functions U ---> k, where
k is the ground field. This is what I don't really
understand. Clearly if
X was just an affine variety, then O(U) are
literally the functions U
----> k that are regular at every point of U (i.e.
at x in U, it can be
written as g/h with h non-vanishing in a
neighborhood of x).

But if (X, O) is now a prevariety (i.e. a locally
ringed space with a
finite open cover of affines), then O(U) is merely
a ring.

He says that for an affine variety, if f is in
O(U), then we can think of
f(x) as the unique constant a in k such that when
you look at f - a in the
local ring O_x, then f - a is in the maximal ideal.
Well this is true
since if X was an affine variety, then f(x) is
certainly defined, and if
f(x) = a, then the function f - a vanishes at x,
and so the function f - a
certainly lies in the maximal ideal M_x.

For a general prevariety (X, O), let U be an open
subset of X. Let f be in
O(U). Why is f still a function from U to k when U
is an arbitrary open
subset and not an affine open subset? Maybe my
guess is that if x in U,
let V be an affine open subset containing x that is
contained in U. The
map O(U) ---> O(V) is injective, so consider the
image of f, call it f ',
in O(V). Now we have a notion of f ' (x) since V is
affine, and then
declare f(x) to be f ' (x)?

I'm far from being an expert in this area,
but I would have thought your idea was correct,
but more complicated than necessary.
Surely you only need to consider the local ring O(x),
and as you say this is determined by O|V, the
restriction of the sheaf
to an affine neighbourhood V of x.

However, isn't there an assumption in what you say
that
the ground field k is algebraically closed?
Otherwise I don't see why there should be an element
a of k
with the property you specify.

Also I don't see why O(U)->O(V) should be injective,
but on the other hand that doesn't seem to me to
matter anyway.

I also have a directly related question : Let (X,
O_X) and (Y, O_Y) be
prevarieties. A morphism of the ringed spaces
consists of a continuous map
f : X ----> Y (called the underlying map), and for
each subset U of Y, a
ring homomorphism f^* : O_Y(U) ---> O_X(f^(-1)(U))
that is compatible with
restriction.

The author's claim is that if (X, O_X) and (Y, O_Y)
are prevarieties, and
if f : X ---> Y is just a continuous map, then
there exists at most one
morphism of prevarieties with underlying map. He
says let U be an open
subset of Y. Then he says if we regard once again
the elements of O(U) as
functions on U taking values in k, then the map f^*
: O(U) --->
O(f^(-1)(U)) is just composition with f. Why is
this???

I take it X and Y are pre-varieties
over the same algebraically closed field k?

In that case, I suppose you have to show that if M is
a morphism
then the homomorphism M(x): O(fx)->O(x) of local
rings
is determined by the fact that if t is in O(U) and s
= f^*(t)
then s(x) = t(fx).

But as I said, I know almost nothing about the
subject,
so please ignore my rambling if someone else gives an
authoritative answer.


I think you're absolutely right on both accounts. Thanks.


--
Timothy Murphy
e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ie
tel: +353-86-2336090, +353-1-2842366
s-mail: School of Mathematics, Trinity College,
Dublin 2, Ireland
.