Affine Schemes
We're on the verge of being able to define an affine scheme.
It's more than the set Spec(R); it's even more than
the Zariski topology on this set; it's all that plus a structure sheaf.
Definition
Let
X=Spec(R) be the spectrum of a commutative
ring. We define the
structure sheaf on
X to be the sheaf
OX
whose ring of sections
OX(U) on an open
U ⊂ X consists of those functions
s : U → ∐P∈U RP
that satisfy the following properties:
- for all P ∈ U, one has s(P)
∈ RP; and
- for all P ∈ U, there exists an open
neighborhood P ∈ V ⊂ U and elements
f, g ∈ R such that for all Q ∈
V, one has f ∉Q and s(Q)
= g/f ∈ RQ.
Lemma
Let X=Spec(R) and let f ∈
R. There is a natural ring homomorphism αf
: Rf → OX(D(f)).
Proof:
An element of Rf can always be represented as a
fraction g/fN with g ∈ R
and N a nonnegative integer. If Q ∈
D(f) = D(fN), then fN ∉ Q, and
so the fraction g/fN determines a well-defined
element in each of the local rings RQ. We define
αf(g/fN) to be the function that assigns
to a prime ideal Q the element g/fN ∈
RQ.
Lemma
The maps αf are injective.
Proof:
Suppose
s=g/fN ∈
Rf and
αf(s) = 0. That means,
for every prime ideal
Q ∈ D(f), the
fraction
g/fN represents zero when considered as an
element of the local ring
RQ. So, there exist
elements
hQ (depending on
Q, of
course), such that
hQ g = 0 ∈ R. From this
relation, we can conclude that the fraction
g/fN
actually represents
0 as soon as both
f and
hQ have been inverted;
i.e., in
RfhQ, and hence as functions on
D(f) ∩ D(hQ) = D(fhQ). Without
loss of generality, we can replace
hQ by
fhQ. Now we have such a relation for every point
Q ∈ D(f). So, the collection of standard open
sets of the form
D(hQ) is an open cover of
D(f). By the proof of quasicompactness, this means
that we have a (finite partition of unity) relation of the form
fM = ∑ eQ hQ.
Multiplying this relation through by
g, we get
fM g = ∑ eQ hQ g = ∑ 0 = 0.
But that equation tells us that
g/fN already
represents the zero element when viewed in the ring
Rf, which was exactly what we needed to show.
Proposition
The maps αf are isomorphisms.
Proof:
Since we've already proved that they are injective, we just need to
verify surjectivity. So, take a section
s ∈
OX(D(f)). By definition, near each point of
D(f) there is an open neighborhood on which
s can be represented as a quotient of elements of
R. Using the facts that the standard open sets form
a basis for the topology and that
D(f) is
quasicompact, we reduce to considering the following situation.
We are given a finite collection of elements
fi, gi ∈
R and nonnegative integers
mi
such that:
- The sets D(fi) cover
D(f);
- The fractions gi/fimi ∈ Rfi
define sections in OX(D(fi)) that agree when
restricted to the intersections D(fi) ∩
D(fj) for different i and
j.
The first simplification is to observe that we can replace the
various exponents
mi appearing in the denominators
with a single nonnegative integer
M. (Just multiply
numerator and denominator by the right power of
fi, which we can do by finiteness.) Using the fact
that
D(f) = D(fM), we can then make things even
simpler by assuming that
M=1. Now the restrictions
in question live in
OX(D(fi fj)). The
elements being restricted all live in the image of
αfi fj. Since that map is injective,
the second condition reduces to the statement that
[gi / fi] =
[gj / fj]
∈ Rfi fj.
Translating this into a condition in
R, we learn
that there exist nonegative integers
N (depending
on
i and
j, but we can ignore
that because of finiteness) such that
(fi fj)N (gi fj - gj fi) = 0 ∈ R.
Rewriting this expression a tad, we get
[gi fiN] [fjN+1] - [gj fjN] [fiN+1] = 0.
Replacing our original pairs of data
(fi, gi) by
the new pairs
(fiN+1, gi fiN), we can then
assume that
N=0 as well. After these
simplifications, we can turn our attention to the first
condition. Because we have a cover, the proof of quasicompactness
gives us a partition of unity relation of the form
fT = ∑ ei fi.
Define
g = ∑ ei gi.
Multiply by
fj and compute:
g fj = ∑ ei gi fj = ∑ ei fi gj = fT gj ∈ R.
In other words, there is an equality of fractions
g/fT =
gj/fj in all of the rings
Rfj
showing that
αf(g/fT) must hit our
original section
s.
Before stating some of the important corollaries of this
result, I need to introduce yet another piece of notation.
Definition
Let F be a sheaf of abelian groups on a topological
space X. For each open U ⊂
X, define Γ(U, F) = F(U).
Corollary
Γ(D(f), OX) = Rf.
Proof:
This is a simple restatement of the proposition.
Corollary
Γ(X, OX) = R.
Proof:
This is a special case of the previous corollary, since
X=D(1).
Corollary
Let x ∈ X = Spec(R) correspond to the prime
ideal P ⊂ R. Then the stalk
OX,x is isomorphic to the local ring
RP.
Proof:
Because the standard open sets form a basis for the topology, we can
compute the direct limit that defines the stalk by looking at the
sections of OX over those
D(f) that contain P. However,
P ∈ D(f) if and only if f ∉
P. So, the stalk is the limit of OX(D(f)) =
Rf = R[1/f], where we end up inverting precisely the
elements of R \ P.
Now let's look at a pair of rings and a homomorphism
φ : R → S between them. Since we're
supposed to think of elements of the rings as functions on the
corresponding spectra, we can at least hope that
φ is related in some sensible way to a map on
spectra that goes in the other direction (f : Spec(S) →
Spec(R)). After a moment's reflection, we can see what
this map should be on the level of sets: Given a prime ideal
P in S, its inverse image
φ-1(P) is a prime ideal in
R.
Lemma
Given a homomorphism of rings, φ : R →
S, it induces a function f: Spec(S) →
Spec(R) on spectra, given by f : P |→
φ-1(P). Moreover, f is a
continuous map of topological spaces.
Proof:
Under the map f, the inverse image
f-1(Z(J)) of the closed set defined by an ideal
J ⊂R consists of the set of prime ideals
P ⊂ S such that f(P) ∈
Z(J). Equivalently, φ-1(P) ⊃
J or P ⊃ φ(J). In other
words, the inverse image of Z(J) is just the closed
set Z(φ(J)).
So far, we've defined a map on spectra, coming from a
homomorphism, that reflects the topological part of the
structure. What can we expect on the sheaf-theoretic part? Well, let's
write X=Spec(R) and Y=Spec(S),
and keep the rest of our notation the same, so that φ :
R → S is a ring homomorphism that defines a
continuous function f : Y → X. Take an open
subset U ⊂ X and a section s
∈ OX(U). Write, for the moment,
V=f-1(U). By composing the set-theoretic
functions that underly the whole structure, we get a function
s°f : V → ∐P∈U RP.
Let's now think about a point Q ∈ V. This
point represents a prime ideal of S such that
P := f(Q) = φ-1(Q). The ring homomorphism
φ : R → S can be composed with the
localization map S → SQ to give a natural
homomorphism R → SQ which just happens to
take every element of R \ P to an
invertible element of SQ. By the universal
property of localizations, we get a naturally induced homomorphism
RP → SQ. Composing yet again, we get
"φ"
°s°f : V → ∐Q∈V SQ.
It is (relatively) easy to check that this composite represents a
section of OY(f-1(U)).
There appear to be some technical difficulties with using this
construction to understand how the structure sheaves on the two
spectra are related. First, we haven't explained what to do with
points P ∈ U that are not in the image of
f. Second, we've only described what's going on
with sections on open sets of Y of the form
V = f-1(U), and not on general open sets.
Definition
Let
f : Y → X be a continuous function
between two topological spaces, and let
F be a
sheaf of abelian groups defined on
Y. Define the
direct image of
F along
f to be the sheaf
f*F on
X whose sections on an open set
U
⊂ X are
f*F(U) = F(f-1(U))
Now we can correctly interpret the construction we started
describing earlier. Given a ring homomorphism φ : R
→ S, it not only determines a continuous function on
spectra f : Y = Spec(S) → X = Spec(R), but
it also defines a morphism of sheaves
f# : OX → f*OY
by composing any section of the structure sheaf of
X with f on one side and with
the localizations of φ on the other.
Lemma
With this notation, f#X : OX(X) →
OY(Y=f-1(X)) is nothing other than the original
ring homomorphism φ.
Proof:
By our earlier computation, we can at least see that
f#X is a homomorphism from
R to S. Now an element
r∈R is the section that assigns to each prime
P⊂R the element
r∈RP. Given any prime
Q⊂S, the composition defining
f#(r) assigns to Q
the element φ(r)∈SQ. But this is
exactly the same as the section corresponding to
φ(r)∈S.
Comments on this web site should be addressed to the
author:
Kevin R. Coombes
Department of Biomedical Informatics
The Ohio State University
Columbus, Ohio 43210