Chapter 3

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia

Last update: 3 June 2013

The Spectrum of a commutative ring

Let A be a commutative ring with 1. Let X=Spec(A) denote the set of all prime ideals of A. (p is a prime ideal A/p is an integral domain; thus A itself is not a prime ideal.) If xX it is sometimes convenient to write jx for the ideal x. For each subset E of A, let V(E)={xX:jxE}. If E consists of a single element f, we write V(f) in place of V({f}).

Lemma (3.1).

(i) V(0)=X; V(1)=.
(ii) If EE, then V(E)V(E).
(iii) V(λEλ)= λV(Eλ).
(iv) V(EE)= V(E) V(E).


Only (iv) is not entirely trivial. Clearly V(EE) V(E) V(E). Conversely, if xV(E)V(E) then there exist fE and fE such that fjx and fjx; since jx is prime, we have ffjx, hence xV(EE).

It follows from (3. 1) that the sets V(E) satisfy the axioms for closed setsJn a topology on X. This topology is called the Zariski topology or spectral topology on X, and it is the only one we shall use.

If a is an ideal in A, the radical r(a) of a is the set of all fA such that some power of f lies in a; it is also the intersection of all the prime ideals of A which contain a. In particular, the radical r(0) of the zero ideal is the set N of all nilpotent elements of A; this ideal is called the nilradical of A.

If E is a subset of A and if a is the ideal generated by E, then V(E)=V(a)=V(r(a)).

We need some more notation:

Ax=Ajx= local ring of A with respect to the prime ideal jx;

mx=jxAx= maximal ideal of Ax:

k(x)=Ax/mx= residue field of Ax field of fractions of A/jx.

If fA, f(x) denotes the class of f mod. jx in A/jxk(x). Thus f(x)=0 if and only if fjx.

D(f)=X-V(f) ={xX:f(x)0} = 'support' of fA; it is an open set.

Finally, if YX, j(Y) denotes yYjy. Thus j({x})=jx. Then we have the following formulas:

Lemma (3.2).

(i) j()=A, j(X)=N (the nilradical of A).
(ii) If YY, then j(Y)j(Y).
(iii) j(λYλ)= λj(Yλ).
(iv) j(V(E))= radical of the ideal generated by E.
(v) V(j(Y))=Y.

It follows from (iv) and (v) that aV(a), Yj(Y) gives an order-reversing one-one correspondence between closed subsets of X and ideals a in A such that a=r(a). Hence, if the ring A is Noetherian, X=Spec(A) is a Noetherian space. (The converse of this is false: X can be Noetherian and A not Noetherian. For example, let B be a polynomial ring k[x1,x2,] over a field in a countable infinity of indeterminuates, let b be the ideal generated by x1,x22,,xnn,, and let A=B/b. Then A is not Noetherian but has exactly one prime ideal.)

If x,yX then y{x} (i.e., y is a specialization of x) if and only if jxjy. Hence {x} is a closed set (by abuse of language, x is a closed point of X) if and only if jx is a maximal ideal of A. Thus X is a T1 space (every point is closed) if and only if every ideal of A is maximal, i.e., dimA=0. However, X is always a T0-space (this means that, given any two distinct points x,y in X, then either there is a neighbourhood of y which does not contain x, or else a neighbourhood of x which does not contain y).

Next, let us look at the open sets D(f), fA. First, from (3.1) (iv) we have

D(fg)=D(f) D(g) (f,gA).

Proposition (3.3).

(i) The open sets D(f) form a base of open sets for the topology of X.
(ii) Each D(f) is quasi-compact. In particular X=D(1) is quasi-compact.


(i) If U is an open set in X, then U=X-V(E) for some EA; we have V(E)= fE ???y (3.1) (iii), hence U=fED(f).

(ii) By virtue of (i) it is enough to show that every covering of a set D(f) by open sets D(fλ) has a finite subcovering. Suppose then that D(f)λLD(fλ); let a be the ideal of A generated by the fλ, then V(f)V(fλ)=V(a), hence V(r(f))V(r(a)) and therefore r(f)r(a), so that fr(a) and therefore fna for some n>0. Say fn=λJaλfλ, where J is some finite subset of L. Then fnb, where b is the ideal generated by the fλ, λJ; hence V(f)=V(fn)V(b)=λJV(fλ). Taking complements, we have D(f)λJD(fλ), as required.

The open sets D(f) (fA) will be called basic open sets.

Let a be an ideal of A. Then the ideals of A/a correspond one-to-one to the ideals of A which contain a, and therefore Spec(A/a) is canonically homeomorphic to the closed subspace V(a) of Spec(A). In particular, Spec(A) and Spec(A/N) are canonically homeomorphic (N= nilradical of A).

Proposition (3.4). X=Spec(A) is irreducible A/N is an integral domain.


From what has just been said, we may as well take N=0. Suppose X is reducible; then there exist proper closed subsets Y1,Y2 in X such that Y1Y2=X, and therefore j(Y1)j(Y2)=j(X)=N=0 (by (3.2)). But j(Y1) and j(Y2) are 0, hence there exist fij(Yi) such that fi0, and f1f2j(Y1)j(Y2)=0. Hence A is not an integral domain.

Conversely, if A is not an integral domain we have f,g in A such that f0, g0 and fg=0. Hence V(f)X, V(g)X (since N=0); but X=V(fg)=V(f)V(g). Consequently X is reducible.

In the correspondence between closed subsets of X and ideals of A which are equal to their radicals, the irreducible closed subsets correspond to the prime ideals. In particular the irreducible components of X correspond to the minimal prime ideals of A. Furthermore, x{x} gives a one-to-one correspondence between the points of X and the irreducible closed subsets of X, i.e., every irreducible closed subset of X has exactly one generic point. For if xX, then {x} is irreducible by (2.1) (iv). If {x}={y}, then each of x and y is a specialization of the other, so that jx=jy, i.e. x=y. Conversely, if Y is an irreducible subset of X,Y corresponds to a prime ideal jx of X, i.e. Y=V(jx)={x}.

Comparison with Affine algebraic varieties

Let k be a field, K an algebraically closed extension of k, and let V be a (k,K)-affine variety as in Chapter 1; let A be the coordinate ring of V (a k-algebra, finitely generated with no nilpotent elements), and let X=Spec(A). What is the relationship between V and X? Let us assume that K is a universal domain in the sense of Weil, i.e. that K has infinite transcendence degree over k; this is just to give us plenty of elbow room. Let xV, then x determines a homomorphism AK, whose kernel is a prime ideal of A, i.e. an element x of X. Conversely, if p is any prime ideal of A, we can embed A/p in K (for the field of fractions of A/p is a finitely generated field extension of k, hence is an algebraic extension of a pure transcendental extension of k) and thus we have a homomorphism AK with kernel p. Hence xx is a map of V onto X, and X is obtained from V by identifying 'equivalent' points in V, i.e. points which are generic specializations of each other.

At the other extreme, if k=K, then V may be identified with the set of maximal ideals of A, i.e. with the set of closed points of X: so in this case the map VX described above is injective (and not in general surjective).

Functorial properties

Let A,A be two rings and let φ:AA be a ring homomorphism (which is always assumed to map identity element to identity element). If xX=Spec(A), then φ-1(jx) is a prime ideal in A, hence a point of X=Spec(A). Thus we have a mapping

Spec(φ)=aφ: XX,

said to be associated with φ. Let φx denote the embedding of A/φ-1(jx) in A/jx induced by φ; then φx extends to a field monomorphism

φx:k (aφ(x)) k(x).

Lemma (3.5).

(i) aφ-1 (V(E)) =V(φ(E)), for any subset E of A. In particular:
(ii) aφ-1(D(f)) =D(φ(f)) (fA).
(iii) aφ(V(a))= V(φ-1(a)) (aany ideal ofA).


(i) is straightforward and (ii) follows from (i). To prove (iii) we may assume that a=r(a), since V(r(a))=V(a) and r(φ-1(a))=φ-1(r(a)). Put Y=V(a), and let a=j(aφ(Y)); then V(a)=aφ(Y) by (3.2) (v). Also:

fa f(x)=0 for allx aφ(Y) fφ-1 (jx)for all xY φ(f)j(Y) =j(V(a))=a fφ-1(a) .

Hence aφ(V(a))= aφ(Y)= V(a)= V(φ-1(a)).

From (i) or (ii) above it follows that aφ is continuous. Clearly, if A is another ring, φ:AA another ring homomorphism, then a^(φφ)= a^φaφ; so that Spec is a contravariant functor from the, category of rings and ring homomorphisms to the category of topological spaces and continuous maps.


(1) If a is an ideal in A and φ:AA/a the projection, then aφ:Spec(A/a)Spec(A) is a homeomorphism of Spec(A/a) onto V(a).

(2) Let S be a multiplicatively closed subset of A (i.e. S is closed under finite products, so that in particular 1S (take the empty product!)). Then we can form the ring of fractions S-1A, and we have a canonical mapping φ:AS-1A, hence aφ:Spec(S-1A)Spec(A). It is a well-known and not difficult fact of commutative algebra that the prime ideals of S-1A are in one-one correspondence (under aφ) with the prime ideals of A which don't meet S, and consequently aφ is a homeomorphism of Spec(S-1A) onto the set of all xX such that jxS=. (In general this subset of X is neither open nor closed, nor even locally closed.)

(3) In particular, Spec(Ax) may be canonically identified with the subspace of X consisting of all generizations of x, i.e. all y such that x{y}.

(4) As another example, let fA and let S be the set of all fn (n0). In this case S-1A is usually denoted by Af. Then Spec(Af) is identified with the set of all xX such that jx contains no power of f, i.e. such that fjx. Hence

Proposition (3.6). If φ:AAf is the canonical homomorphism (fA), then aφ is a homeomorphism of Spec(Af) onto the open set D(f).

(5) The 'characteristic morphism'. Since A has an identity element, there is a canonical mapping φ:A, where is the ring of integers; hence aφ:XSpec(). Now the points of Spec() are (0) and the prime ideals (p) (p a positive prime number), and aφ(x) is just the ideal generated by the characteristic of the residue field k(x) of x.

Proposition (3.7). Let φ:AA be a ring homomorphism, aφ:XX the associated map.

(i) If φ is surjective, aφ is a closed embedding (i.e. a homeomorphism of X onto a closed subset of X).
(ii) If φ is injective, aφ is dominant (i.e. aφ(X) is dense in X).


(i) is just Example 1 above.

(ii) follows from (3.5) (iii): aφ(X)= aφ(V(0))= V(φ-1(0))= V(0) (since φ is injective) =X.

Notes and References

This is a typed excerpt of the book "Algebraic Geometry: Introduction to Schemes - I.G. Macdonald".

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