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.

  1. Spec(k) consists of a single point if k is a field.
  2. Spec(Z) = {0, 2, 3, 5, 7, }.
  3. If k is an algebraically closed field, then Spec(k[t]) = {*} k.

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.

Remark The previous definition allows us to think of an element f R as a "function" on X=Spec(R), in the sense that we have "values" f(P) defined for each point P X . However, these functions have the somewhat disturbing property that the space where their values live varies depending on the point where we're doing the evaluating. The only reason we weren't bothered earlier---in the case where R is a finitely generated algebra without nilpotents over an algebraically closed field k---is that the value of a regular function f A(X) at a point x corresponding to a maximal ideals M lives in the quotient ring A(X)/M, and all of these quotients can be canonically identified with the field k.

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 }

Remark As before, one can check that the collection consisting of all zero sets of functions satisfies the requisite properties to be the closed sets in a topology on X; the resulting topology on the spectrum of a ring is called the Zariski topology.

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} = PY P.

Lemma For any ideal J in a commutative ring R, the radical of J is

{ f R : N>0, fN J} = PJ 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 = { fN, 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 fi (for i=1, 2) lives in R but not in Q, and that f1 f2 Q. By maximality, Q + <fi> meets S; let's say it contains an element qi + ei fi with qi Q. Since S is a multiplicative set, it contains the product

(q1 + e1 f1)(q2 + e2 f2).

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)
  1. If J is a proper ideal in R, then Z(J) is nonempty.
  2. For any ideal J R, one has I(Z(J)) equal to the radical of J.
  3. 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.

Remark The first item is the analog of the weak Nullstellensatz; the second item is the analog of Hilbert's Nullstellensatz.

Remark The Zariski topology is almost never Hausdorff. For example, suppose R is an integral domain. Then the zero ideal <0> is prime in R. So, it can be identified with a point η X, usually called the generic point of X. Observe first that I(η) = 0, this being the intersection of the prime ideals that contain 0. Next, the associated zero set Z(0) is the whole space, since every prime ideal contains the zero ideal. So, the point η is not only non-closed, but actually dense in the entire spectrum of R. In general, the only thing you can say is that the spectrum of a ring is T0; i.e.., given a pair of points, there exists a neighborhood of one not containing the other.

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.

Remark Quasicompactness isn't terribly useful without the Hausdorff axiom to accompany it. Later, we'll consider categorical notions (separated, proper) that provide good substitutes for some standard topological notions (Hausdorff, compact).

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(fi), for i I. Since the union of the D(fi) covers X, the intersection of the complements Z(fi) is empty. Thus, Z(< fi, 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 = < fi, i I >

or to the existence of a representation

1 = ei fi

for some ei R. But any such representation must actually be a finite sum; we only need to retain the open sets in the cover for which fi 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
  1. D(f) D(g) = D(fg); and
  2. D(fN) = 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(fi) for i ranging through some index set I. Taking the complement of the covering relation, we find that

Z(f) i Z(fi) = Z(< fi, 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(< fi, i I>)) = Rad(< fi, i I>).

So, the covering condition reduces to the existence of a nonnegative integer N and a relation

fN = ei fi

for some ei R. Once again, any such relation must be finite; we just keep the sets in the open cover where the coefficients are nonzero.

Remark The kinds of relations that we've seen twice in the proof of quasicompactness are the algebraic analogs of partitions of unity. They can (and will) be used in similar ways.

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