The Spectrum of a Ring
Definition
Let R be a commutative ring (as always, with
1). As a set, we define the
spectrum of R, written
Spec(R), to be the collection of prime ideals
P ⊂ R.
Definition
Let f ∈ R be an arbitrary element. Let
P ∈ X = Spec(R) be a point. Identify
P with a prime ideal in
R. Define f(P) to be the residue
class of f in the quotient ring
R/P.
Lemma
Let R be a commutative ring, f ∈
R an element, and P ⊂ R a prime
ideal. Then f(P)=0 if and only if f ∈
P.
Proof:
Obvious.
Definition
Let
X=Spec(R). Let
S ⊂
R be an arbitrary subset. We define the
zero
set of
S to be
Z(S) = {P ∈ X : f(P)=0 ∀ f ∈ S}
= {P ∈ X : P ⊃ S }
Definition
Let
X=Spec(R). Let
Y ⊂
X be an arbitrary subset. Define the ideal of functions
that vanish on
Y to be
I(Y) = { f ∈ R : f(P)=0 ∀ P ∈ Y}
= { f ∈ R : f ∈ P ∀ P ∈ Y}
= ∩_{P∈Y} P.
Lemma
For any ideal
J in a commutative ring
R, the radical of
J is
{ f ∈ R : ∃ N>0, f^{N}∈ J}
= ∩_{P⊃J} P.
Proof:
We have defined the radical to be the set on the left-hand side of the
equality. It is easy to see that the radical is contained in the set
on the right-hand side. Conversely, suppose that
f ∈
R is not contained in the radical. Consider the collection
T of all ideals of
R that
contain
J and are disjoint from the multiplicative
set
S = { f^{N}, N ≥ 1}. By
hypothesis,
J lies in the collection
T. By Zorn's Lemma,
T contains
a maximal element
Q. We claim that
Q is a prime ideal. To see this, suppose
f_{i} (for
i=1, 2) lives in
R but not in
Q, and that
f_{1} f_{2} ∈ Q. By maximality,
Q +
<f_{i}> meets
S; let's say it
contains an element
q_{i} + e_{i} f_{i} with
q_{i}
∈ Q. Since
S is a multiplicative
set, it contains the product
(q_{1} + e_{1} f_{1})(q_{2} + e_{2} f_{2}).
However, under the hypotheses, this product is also contained in
Q. This contradicts the fact that
Q does not meet
S, and thus
shows that
Q is a prime ideal. Therefore, no power
of
f can be contained on the right-hand side,
proving the result.
Lemma Let
X=Spec(R)
- If J is a proper ideal in
R, then Z(J) is nonempty.
- For any ideal J ⊂ R, one has
I(Z(J)) equal to the radical of
J.
- For any Y ⊂ X, one has
Z(I(Y)) equal to the closure of
Y in the Zariski topology.
Proof:
Item (i) follows from Zorn's Lemma; item (ii), from the previous lemma;
item (iii), directly from the definitions.
Definition
Let R be a commutative ring, and let f
∈ R. Define the standard open set
corresponding to f to be the complement of its zero
set; that is,
D(f) = Spec(R) \ Z(f).
Proposition
The standard open sets in Spec(R) form a basis for
the topology.
Proof:
This is the same as the proof given previously that the affine open
algebraic sets form a basis for the topology on an affine variety.
Definition
A topological space X is called
quasicompact if every open cover of
X contains a finite subcover.
Proposition
Let X=Spec(R). Then X is
quasicompact.
Proof:
Given any open cover of
X, we can refine it to a
cover in which every set is a standard open sets
D(f_{i}), for
i ∈ I. Since
the union of the
D(f_{i}) covers
X, the intersection of the complements
Z(f_{i}) is empty. Thus,
Z(< f_{i}, i ∈ I >) = ∅
. We've seen that the zero zet of a proper ideal can't be
empty, so the condition that a collection of standard open sets
covers
X reduces to
R = < f_{i}, i ∈ I >
or to the existence of a representation
1 = ∑ e_{i} f_{i}
for some
e_{i} ∈ R. But any such representation
must actually be a
finite sum; we only need to
retain the open sets in the cover for which
f_{i}
actually appears with a nonzero coefficient in this representation.
In a little while, we 're going to need a strengthened version
of this result: the standard open sets D(f) are
themselves quasicompact. Let's see how to spice up the current
result to get the one we really need.
Lemma
Let
f, g ∈ R, and let
N be
a nonnegative integer. Then
- D(f) ∩ D(g) = D(fg); and
- D(f^{N}) = D(f).
Proof:
D(f) is the set of points where
f is nonzero. In other words, it is the set of
prime ideals not containing f. Now the result
follows by translating the definition of prime ideal into the current
context.
Proposition
Every standard open set D(f) in
X=Spec(R) is quasicompact.
Proof:
Take an open cover. By refining the cover, we can assume all the open
sets in the cover are themselves standard open sets, of the form
D(f_{i}) for
i ranging through
some index set
I. Taking the complement of the
covering relation, we find that
Z(f) ⊃ ∩_{i} Z(f_{i}) = Z(< f_{i}, i ∈ I>).
Now we can reverse the containment by passing back to the ideals of
functions that vanish on these closed sets, obtaining
f ∈ I(Z(f)) ⊂ I(Z(< f_{i}, i ∈ I>))
= Rad(< f_{i}, i ∈ I>).
So, the covering condition reduces to the existence of a nonnegative
integer
N and a relation
f^{N} = ∑ e_{i} f_{i}
for some
e_{i} ∈ R. Once again, any such
relation must be
finite; we just keep the sets in
the open cover where the coefficients are nonzero.
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