7

Commutative Ideal Theory:
General Theory and Noetherian Rings

The ideal theory of commutative rings was initiated in Dedekind’s successful “restoration” of unique factorization in the rings of algebraic integers of number fields by the introduction of ideals. Some of these, e.g., are not factorial, that is, do not have unique factorization of elements into irreducible elements (see BAI, pp. 141 – 142). However, Dedekind showed that in these rings unique factorization does hold for ideals as products of prime ideals (definition on p. 389). A second type of ideal theory, which was introduced at the beginning of this century by E. Lasker and F. S. Macaulay, is concerned with the study of ideals in rings of polynomials in several indeterminates. This has obvious relevance for algebraic geometry. A principal result in the Lasker-Macaulay theory is a decomposition theorem with comparatively weak uniqueness properties of ideals in polynomial rings as intersections of so-called primary ideals (definition on p. 434). In 1921 Emmy Noether gave an extremely simple derivation of these results for arbitrary commutative rings satisfying the ascending chain condition for ideals. This paper, which by its effective use of conceptual methods, gave a new direction to algebra, has been one of the most influential papers on algebra published during this century.

In this chapter we shall consider the ideal theory—once called additive ideal theory—which is an outgrowth of the Lasker-Macaulay-Noether theory. In recent years the emphasis has shifted somewhat away from the use of primary decompositions to other methods, notably, localization, the use of the prime spectrum of a ring, and the study of local rings. The main motivation has continued to come from applications to algebraic geometry. However, other developments, such as the study of algebras over commutative rings, have had their influence, and of course, the subject has moved along under its own power.

We shall consider the Dedekind ideal theory in Chapter 10 after we have developed the structure theory of fields and valuation theory, which properly precede the Dedekind theory.

Throughout this chapter (and the subsequent ones) all rings are commutative unless the contrary is explicitly stated. From time to time, mainly in the exercises, applications and extension to non-commutative rings will be indicated. The first nine sections are concerned with arbitrary commutative rings. The main topics considered here are localization, the method of reducing questions on arbitrary rings to local rings via localization with respect to the complements of prime ideals, the prime spectrum of a ring, rank of projective modules, and the projective class group. The ideal theory of noetherian rings and modules is developed in sections 7.10–7.18. Included here are the important examples of noetherian rings: polynomial rings and power series rings over noetherian rings. We give also an introduction to affine algebraic geometry including the Hilbert Nullstellensatz. Primary decompositions are treated in section 7.13. After these we consider some of the basic properties of noetherian rings, notably the Krull intersection theorem, the Hilbert function of a graded module, dimension theory, and the Krull principal ideal theorem. We conclude the chapter with a section on I-adic topologies and completions.

7.1 PRIME IDEALS. NIL RADICAL

We recall that an element p of a domain D is called a prime if p is not a unit and if p|ab in D implies p|a or p|b in D. This suggests the following

DEFINITION 7.1. An ideal P in a (commutative) ring R is called prime if P ≠ R and if ab ∈ P for a,b ∈ R implies either a ∈ P or b ∈ P.

In other words, an ideal P is prime if and only if the complementary set P′ = R – P is closed under multiplication and contains 1, that is, P′ is a submonoid of the multiplicative monoid of R. In congruence notation the second condition in Definition 7.1 is that if ab ≡ 0(mod P), then a ≡ 0(mod P) or b ≡ 0(mod P). The first condition is that the ring = R/P ≠ 0. Hence it is clear that an ideal P is prime if and only if R/P is a domain. Since an ideal M in a commutative ring R is maximal if and only if R/M is a field and since any field is a domain, it is clear that any maximal ideal of R is prime. It is clear also that an element p is prime in the usual sense if and only if the principal ideal (p)( = pR) is a prime ideal. Another thing worth noting is that if P is a prime ideal in R and A and B are ideals in R such that AB ⊂ P, then either A ⊂ P B ⊂ P. If not, then we have a ∈ A, P and b ∈ B, ∈ P. Then ab ∈ AB ⊂ P, contrary to the primeness of P. It is clear by induction that if P is a prime ideal and a₁ a₂ … a_n ∈ P, then some a_i ∈ P and if A₁ A₂^…A_n ⊂ P for ideals A_i, then some A_i ⊂ P.

We recall the elementary result in group theory that a group cannot be a union of two proper subgroups (exercise 14, p. 36 of BAI). This can be strengthened to the following statement: If G₁, G₂, and H are subgroups of a group G and H ⊂ (G₁ ∪ G₂), then either H ⊂ G₁ or H ⊂ G₂. The following result is a useful extension of this to prime ideals in a ring.

PROPOSITION 7.1. Let A, I₁, …, I_n be ideals in a ring such that at most two of the I_j are not prime. Suppose A ⊂ ⁿ₁ I_j. Then A ⊂ I_j for some j.

Proof.   We use induction on n. The result is clear if n = 1. Hence we assume n > 1. Then if we have A ⊂ I₁ ∪ ^… ∪ _k ∪ ^… ∪ I_n for some k, the result will follow by the induction hypothesis. We therefore assume for k = 1, 2,…,n and we shall complete the proof by showing that this leads to a contradiction. Since , there exists an . Since A ⊂ I_j, a_k ∈ I_k. If n = 2, it is readily seen that (as in the group theory argument) a₁ + a₂ ∈ A but a₁ + a₂ I₁ ∪ I₂, contrary to hypothesis. If n > 2, then at least one of the I_j is prime. We may assume it is I₁. Then it is readily seen that

is in A but is not in I_j. Again we have a contradiction, which proves the result.

The foregoing result is usually stated with the stronger hypothesis that every I_j is prime. The stronger form that we have proved is due to N. McCoy, who strengthened the result still further by replacing the hypothesis that A is an ideal by the condition that A is a subrng (BAI, p. 155) of the ring R. It is clear that the foregoing proof is valid in this case also. In the sequel we shall refer to Proposition 7.1 as the “prime avoidance lemma.” The terminology is justified since the contrapositive form of the proposition is that if A and I₁, …, I_n are ideals such that A I_j for any j and at most two of the I_j are not prime, then there exists an a ∈ A such that a I _j.

There is an important way of obtaining prime ideals from submonoids of the multiplicative monoid of R. This is based on

PROPOSITION 7.2. Let S be a submonoid of the multiplicative monoid of R and let P be an ideal in R such that (1) P ∩ S = Ø. (2) P is maximal with respect to property (1) in the sense that if P′ is any ideal such that P′ P, then P′ ∩ S ≠ Ø. Then P is prime.

Proof.   Let a and b be elements of R such that a P and b P. Then the ideals (a) + P and (b) + P properly contain P and so meet S. Hence we have elements p₁, p₂ ∈ P, x₁, x₂ ∈ R, s₁, s₂ ∈ S such that s₁ = x₁a + p₁, s₂ = x₂b + p₂. Then s₁ s₂ ∈ S and

Hence if ab ∈ P, then s₁s₂ ∈ P, contrary to P ∩ S = Ø. Thus ab P and we have shown that a P, b P implies ab P, so P is prime.

If S is a submonoid not containing 0, then the ideal 0 satisfies 0 ∩ S = Ø. Now let A be any ideal such that A ∩ S = Ø and let be the set of ideals B of R such that B ⊃ A and B ∩ S = Ø. It is an immediate consequence of Zorn’s lemma that contains maximal elements. Such an element is an ideal P ⊃ A satisfying the hypotheses of Proposition 7.2. Hence the following result follows from this proposition.

PROPOSITION 7.3. Let S be a submonoid of the multiplicative monoid of R and A an ideal in R such that S ∩ A = Ø. Then A can be imbedded in a prime ideal P such that S ∩ P = Ø.

Let N denote the set of nilpotent elements of R. Evidently if z ∈ N so zⁿ = 0 for some n and if a is any element of R, then (az)ⁿ = aⁿzⁿ = 0. Hence az ∈ N. If z_i ∈ N, i = 1, 2, and n_i is an integer such that = 0, then by the binomial theorem for n = n₁ + n₂ – 1 we have (z₁ + z₂)ⁿ = . This is 0, since if i < n₁ then n – i ≥ n₂, so and if i ≥ n₁, then clearly . Thus N is an ideal. We call this ideal the nil radical of the ring R and we shall denote it as nilrad R. If = z + N is in the nil radical of = R/N, N = nilrad R, then zⁿ ∈ N for some n and so z^mn = 0 for some integer mn. Then z ∈ N and = 0. Thus nilrad = 0. We recall that any nil ideal of a ring is contained in the (Jacobson) radical rad R (p. 192). We recall also that rad R is the intersection of the maximal left ideals of R (p. 193), so for a commutative ring rad R is the intersection of the maximal ideals of the ring. The analogous result for the nil radical is the following

THEOREM 7.1 (Krull). The nil radical of R is the intersection of the prime ideals of R.

Proof.     Let N = nilrad R and let N′ = P where the intersection is taken over all of the prime ideals P of R. If z ∈ N, we have zⁿ = 0 for some integer n. Then zⁿ ∈ P for any prime ideal P and hence z ∈ P. Hence N ⊂ P for every prime ideal P, so N ⊂ N′. Now let s N, so s is not nilpotent and S = {sⁿ|n = 0, 1, 2, …} is a submonoid of the multiplicative monoid of R satisfying S ∩ {0} = Ø. Then by Proposition 7.3 applied to A = 0 there exists a prime ideal P such that P ∩ S = Ø. Then s P and s N′. This implies that N′ ⊂ N and so N = N′ = P.

If A is an ideal in R, we define the nil radical of A, nilrad A (sometimes denoted as ), to be the set of elements of R that are nilpotent modulo A in the sense that there exists an integer n such that zⁿ ∈ A. This is just the set of elements z such that = z + A is in the nil radical of = R/A. Thus

where v is the canonical homomorphism of R onto . It follows from this (or it can be seen directly) that nilrad A is an ideal of R containing A. It is clear also that iteration of the process of forming the nil radical of an ideal gives nothing new : nilrad (nilrad A) = nilrad A.

We have the bijective map of the set of ideals of R containing A onto the set of ideals of = R/A. Moreover, (BAI, p. 107). Hence B is prime in R if and only if is prime in . Thus the set of prime ideals of is the set of ideals = P/A where P is a prime ideal of R containing A. Since (P/A) = (P)/A = (BAI, p. 67, exercise 2), and taken over the prime ideals of is the nil radical of , we have

Hence nilrad A = P taken over the prime ideals P of R containing A. We state this as

THEOREM 7.2. The nil radical of an ideal A is the intersection of the prime ideals of R containing A.

EXERCISES

        1. Let A_i, 1 ≤ i ≤ n, be ideals, P a prime ideal. Show that if P ⊃ ⁿ₁ A_i then P ⊃ A_i for some i and if P = ⁿ₁ A_i, then P = A_i for some i.

        2. Show that if P is a prime ideal, then S = R – P is a submonoid of the multiplicative monoid of R, which is saturated in the sense that it contains the divisors of every s ∈ S. Show more generally that if {P_x} is a set of prime ideals, then S = R – P_x is a saturated submonoid of the multiplicative monoid of R. Show that conversely any saturated submonoid of the multiplicative monoid of R has the form R – P_x, {P_x} a set of prime ideals of R.

        3. Show that the set of zero divisors of R is a union of prime ideals.

        4. (McCoy.) Show that the units of the polynomial ring R[x], x an indeterminate, are the polynomials a₀ + a₁ x + ^… + a_nxⁿ where a₀ is a unit in R and every a_i, i > 0, is nilpotent. (Hint: Consider the case in which R is a domain first. Deduce the general case from this by using Krull’s theorem.)

The next two exercises are designed to prove an important result on the radical of a polynomial ring due to S. Amitsur. In these exercises R need not be commutative.

        5. Let f(x) ∈ R[x] have 0 constant term and suppose f(x) is quasi-regular with quasi-inverse g(x) (p. 194). Show that the coefficients of g(x) are contained in the subring generated by the coefficients of f(x).

        6. (Amitsur.) Show that R[x] is semi-primitive if R has no nil ideals ≠ 0. (Hint: Assume that rad R[x] ≠ 0 and choose an element ≠ 0 in this ideal with the minimum number of non-zero coefficients. Show that these coefficients are contained in a commutative subring B of R. Apply exercises 4 and 5.)

7.2 LOCALIZATION OF RINGS

The tool of localization that we shall now introduce is one of the most effective ones in commutative algebra. It amounts to a generalization of the familiar construction of the field of fractions of a domain (BAI, pp. 115–119). We can view this generalization from the point of view of a universal construction that is a solution of the following problem:

Given a (commutative) ring R and a subset S of R, to construct a ring R_s and a homomorphism λ_s of R into R_s such that every λ_s(s), s ∈ S, is invertible in R_s, and the pair (R_s, λ_s) is universal for such pairs in the sense that if η is any homomorphism of R into a ring R′ such that every η(s) is invertible, then there exists a unique homomorphism : R_s → R′ such that the diagram

is commutative.

Before proceeding to the solution of the problem (which is easy), we shall make some remarks about the problem.

1. Since the product of two elements of a ring is invertible if and only if the elements are invertible, there is no loss in generality in assuming that S is a submonoid of the multiplicative monoid of R. The solution of the problem for an arbitrary set S can be reduced to the case in which S is a monoid by replacing the set S by the submonoid generated by S. For example, if S is a singleton {s}, then we can replace it by = {sⁿ|n = 0, 1, 2, …}.

2. The special case of the field of fractions is that in which R is a domain and S = R*, the submonoid of non-zero elements of R. In this case nothing is changed if we restrict the rings R′ in the statement of the problem to be fields. This is clear since the image under a homomorphism of a field either is the trivial ring consisting of one element or is a field.

3. Since a zero divisor of a ring ( ≠ {0}) is not invertible, we cannot expect λ_s to be injective if S contains zero divisors.

4. If the elements of S are invertible in R, then there is nothing to do: We can simply take R_S = R and λ_S = 1_R, the identity map on R. Then it is clear that = η satisfies the condition in the problem.

5. If a solution exists, it is unique in the strong sense that if (R_S⁽¹⁾, λ_S⁽¹⁾) and (R_S⁽²⁾, λ_S⁽²⁾) satisfy the condition for the ring R and subset S, then there exists a unique isomorphism ζ : R_S⁽¹⁾ → R_S⁽²⁾ such that

is commutative (see p. 44).

We now proceed to a construction of a pair (R_s, λ_s) for any ring R and submonoid S of its multiplicative monoid. As in the special case of a domain R and its monoid S = R* of non-zero elements, we commence with the product set R × S of pairs (a, s), a ∈ R, s ∈ S. Since the monoid S may contain zero divisors, it is necessary to modify somewhat the definition of equivalence among the pairs (a, s) that we used in the construction of the field of fractions of a domain. We define a binary relation ~ on R × S by declaring that (a₁, s₁) ~ (a₂, s₂) if there exists an s ∈ S such that

(or ss₂a₁ = ss₁a₂). If R is a domain and S does not contain 0, then (4) can be replaced by the simpler condition s₂a₁ = s₁a₂. In the general case it is readily verified that ~ is an equivalence relation. We denote the quotient set defined by this equivalence relation as R_S and we denote the equivalence class of (a, s) by a/s.

We define addition and multiplication in the set R_S by

and

It is a bit tedious but straightforward to check that these compositions are well-defined and that if we put 0 = 0/1 ( = 0/s for any s ∈ S) and 1 = 1/1 ( = s/s, s ∈ S), then (R_S, + , ·, 0, 1) is a ring. We leave the verifications to the reader.

We now define a map λ_S (or λ_S^R) of R into R_S by

It is clear from (5) and (6) and the definition of 1 in R_S that λ_S is a homomorphism of R into R_S. Moreover, if s ∈ S, then λ_S(S) = s/1 has the inverse 1/s since (s/1) (1/s) = s/s = 1. Now let η be a homomorphism of R into R′ such that η(s) is invertible for every s ∈ S. It is readily verified that

is well-defined and this is a homomorphism of R_S into R′. Moreover, λ_S(a) = (a/l = η(a). Thus we have the commutativity of (3). The uniqueness of is clear also since a/s = (a/1) (1/s) = (a/1) (s/l)^{– 1} and hence any satisfying the commutativity condition satisfies (a/s) = (λ_s(a)λ_s(s)^–1) = η(a)η(s) ^{– 1}. Thus is the map we defined in (8) and (R_s, λ_s) is a solution of the problem we formulated at the outset. We shall call (R_s, λ_s) (or simply R_s) the localization of R at S.

We shall now study relations between R_s and R_s′ for S′, a submonoid of the monoid S. Note first that λ_s(s′) is invertible in R_s, s′ ∈ S′. Hence we have a unique homomorphism ζ_{S′, S} : R_s′ → R_s such that λ_s = ζ_{s′, s} λ_s′. Now suppose that λ_s′(s) is invertible in R_s′ , s ∈ S. Then we also have a unique homomorphism ζ_s, s′ : R_s → R_s′ such that λ_s′. = ζ_{s, s′} λ_s. It follows that ζ_{S, S′} ζ_{S′, S} = 1_Rs, and ζ_S′,Sζ_{S, S′} = l_Rs. Hence ζ_{S′, S} and ζ_{S, S′} are isomorphisms. We shall now show that in general if S′ ⊂ S, then R_s is, in fact, a localization of R_s′. We state the result in a somewhat imprecise manner as follows.

PROPOSITION 7.4. Let S′ be a submonoid of S and let S/S′ = {s/s′|s ∈ S, s′ ∈ S′}. Then S/S′ is a submonoid of the multiplicative monoid of R_s′ and we have canonical isomorphisms of R_s and (R_S′)_λS′(S) and of (R_S′)_λs′(S) and (R_s′)_s/s′.

Proof.   We shall obtain the first isomorphism by showing that (R_S′ )_λS′(S) has the universal property of R_s. First, we have the composite homomorphism λ_λs,_(s) λ_s′ of R into (R_S′)_λs′(S) obtained from the sequence of homomorphism

Now let η be a homomorphism of R into R′ such that η(s) is invertible, s ∈ S. Since S′ ⊂ S, we have a unique homomorphism of R_s′ into R′ such that λ_s′ = η. If s ∈ S, then ′(λ_S′(s)) = η(s) is invertible in R′. Accordingly, by the universal property of (R_S′)λ_s′(S) we have a unique homomorphism of (R_s′)_λs′(s) into R′ such that λ_λs′(S) = ′. Then . Moreover, is the only homomorphism of (R_S′)_λs′(s) into R′ satisfying .For, this condition implies that , which implies that by the universality of . Then we obtain the uniqueness of by the universality of . Thus has the universal property of (R_S, λ_S ) and so we have the required isomorphism.

The isomorphism of (R_S′)_λS′(S) and (R_s′)_s/s′ can be seen by observing that λ_s′(S) is a submonoid of S/S′ . Hence we have the canonical homomorphism of (R_S′)_λs′(s) into (R_S′)_S/S′. If s ∈ S, s′ ∈ S′, then s/s′ = (s/1)(s′/l)^{– 1} in R_s′ and and are invertible in (R_S′)λ_s′(s). Hence is invertible. It follows from the result we obtained above that the canonical homomorphism of (R_s′)_λs′(S) into (R_s′)_s/s′ is an isomorphism.

EXERCISES

        1. Let S and T be submonoids of the multiplicative monoid of R. Note that ST = {st|s ∈ S, t ∈ T} is the submonoid generated by S and T. Show that R_ST ≅ (R_s)λ_s′(T).

        2. Let a, b ∈ R. Show that .

        3. Let S be a submonoid of the multiplicative monoid of R. If a, b ∈ S, define if a|b in S. In this case there is a unique homomorphism of such that . Show that (with respect to the ζ ’s).

7.3 LOCALIZATION OF MODULES

It is important to extend the concept of localization to R-modules. Let M be a module over R, S a submonoid of the multiplicative monoid of R. We shall construct an R_s-module M_s in a manner similar to the construction of R_s. We consider M × S the product set of pairs (x, s), x∈ M, s ∈ S, and we introduce a relation ~ in this set by (x₁, s₁) ~ (x₂, s₂) if there exists an s ∈ S such that

The same calculations as in the ring case show that ~ is an equivalence. Let M_s denote the quotient set and let x/s be the equivalence class of (x, s). We can make M_s into an R_s-module by defining addition by

and the action of R_s on M_s by

We can verify as in the ring case that (10) and (11) are well-defined, that + and 0 = 0/s constitute an abelian group, and that (11) defines a module action of R_s on M_s. We shall call the R_s-module M_s the localization of M at S or the S-localization of M.

Although we are generally interested in M_s as R_s-module, we can also regard M_s as R-module by defining the action of R by a(x/s) = (a/1) (x/s) = ax/s. Since a a/1 is a ring homomorphism, it is clear that this is a module action. We have a map λ_s (or λ_s^M if it is necessary to indicate M) of M into M_s defined by x x/1. This is an R-module homomorphism. The kernel of λ_s is the set of x ∈ M for which there exists an s ∈ S such that sx = 0, that is, if ann x denotes the annihilator ideal in R of x, then ann x ∩ S ≠ Ø. It is clear that λ_s need not be injective. For example, if Mis a -module and S = – {0}, then the torsion submodule of M is mapped into 0 by λ_s. It is clear that if S includes 0, then M_s = 0 and ker λ_s = M. Another useful remark is

PROPOSITION 7.5. If M is a finitely generated module, then M_s = 0 if and only if there exists an s ∈ S such that sM = 0.

Proof.     If sM = 0, every x/t = 0 = 0/1 since s(lx) = s(t0) = 0. Conversely, suppose M_s = 0 and let {x_l, …, x_n} be a set of generators for M. Then x_i/1 = 0 implies there exists an s_i ∈ S such that s_ix_i = 0. Then sx_i = 0 for s = ∏ⁿ₁ s_i ; hence sx = 0 for any ∑ r_ix_i, r_i ∈ R Thus sM = 0.

Let f : M → N be a homomorphism of .R-modules of M into N. Then we have a corresponding R_s-homomorphism f_s of M_s into N_s defined by

Again we leave the verifications to the reader. The maps M M_s, f f_s define a functor, the S-localization functor, from the category of R-modules to the category of R_s-modules. We shall now show that this functor is naturally isomorphic to the functor R_{s R} from R-mod to R_s-mod. Thus we have

PROPOSITION 7.6. For every R-module M we can define an R_s-isomorphism η_M of M_s onto R_sR M that is natural in M.

Proof.   We show first that there is a map η_M of M_s into R_sR M such that

Suppose x₁/s₁ = x₂/s₂, which means that we have an s ∈ S such that ss₂x₁ = ss₁x₂. Then

Hence (13) is well-defined. Direct verification shows that η_M is a group homomorphism. We note next that we have a well-defined map of the product set R_s × M into M_s such that

To check this we have to show that if a₁/s₁ = a₂/s₂ in R_s, then a₁x/s₁ = a₂x/s₂. Now if a₁/s₁ = a₂/s₁, then we have an s ∈ S such that sa₁s₂ = sa₂s₁. Then sa₁s₂x = sa₂s₁x, which implies the required equality a₁x/s₁ = a₂x/s₂. Direct verification shows that (14) satisfies the condition for a balanced product of the .R-modules R_s and M. Hence we have a group homomorphism η′ _M of R_sRM into M_s sending (a/s)x into ax/s. Following this with η_M we obtain (1/s)ax = (a/s)x. On the other hand, if we apply η_M to x/s, we obtain (l/s)x and the application of η′_M to this gives x/s. It follows η_Mη′_M is the identity map on R_sRM and η′_Mη_M is the identity on M_s. Hence η_M is a group isomorphism and so is η′_M = η^{– 1}_M. Direct verification using the definitions shows that η_M and hence is an R_s-map, hence an R_s-isomorphism. It remains to show the naturality, that is, the commutativity of

for a given R-homomorphism f : M → N. This follows from the calculation

which completes the proof.

This result gives rise to a useful interplay between tensor products and localization. We recall first that the functor N_R is right exact, that is, exactness of implies that of N . Applying this with N = R_s in conjunction with Proposition 7.6 shows that is exact. Next we prove directly that if is exact, then is exact. Suppose f_s(x′ /s) = 0, so f(x′)/s = 0. Then we have a t ∈ S such that tf(x′) = 0. Then f(tx′) = 0 and since f is injective, tx′ = 0 and hence x′ /s = 0. Hence ker f_s = 0 and so is exact. We can now apply Proposition 7.6 to conclude that is exact. We recall that this is the definition of flatness for R_s as R-module (p. 153). Hence we have the important

PROPOSITION 7.7. R_s is a flat R-module.

We remark also that we have shown that if 0 → M′ → M → M″ → 0 is an exact sequence of R-modules, then 0 → M′_s → M_s → M″_s → 0 is an exact sequence of R_s-modules. Thus the S-localization functor from R-mod to R_s-mod is exact.

Now let N be a submodule of M. Then the exact sequence where l is the injection gives the exact sequence and the definition shows that l_s is the injection of N_s. We have the exact sequence where v is the canonical homomorphism. Hence we have the exact sequence , which shows that M_s/N_s ≅ (M/N)_s. More precisely, the foregoing exactness shows that the map

is an isomorphism of M_s/N_s with (M/N)_s.

If M and N are R-modules and f is a homomorphism f : M → N, then we have the exact sequence

where coker f = N/f (M). Hence we have the exact sequence

This implies that (ker f)_S is ker f_S. Also (coker f)_S = (N/f(M))_S ≅ N_S/f(M)_S = N_S/f_S(M) = coker f_S.

We recall next that we have an R_S-isomorphism of onto sending (exercise 13, p. 148). Hence, by Proposition 7.6, we have an R_s-isomorphism of (M_RN)_s onto M_{s Rs}N_s such that

Now let M be an R_s-module. Then M becomes an R-module by defining ax for a ∈ R to be (a/l)x. It is clear that if f is a homomorphism of M as R_s-module, then f is an R-homomorphism. Now consider M_s where M is regarded as an R-module. Since s/1 is invertible in R_s for s ∈ S, sx = (s/l)x = 0 for x ∈ M implies x = 0. Hence the homomorphism λ_s : x x/1 of M into M_s is a monomorphism. Since x/s = (l/s)(x/l), λ_s is surjective. Thus we can identify M as R_s-module with the S-localization M_s of M as R-module.

7.4 LOCALIZATION AT THE COMPLEMENT OF A PRIME IDEAL.
LOCAL-GLOBAL RELATIONS

We recall that an ideal P in R is prime if and only if the complement, R – P of P in R, is a submonoid of the multiplicative monoid of R. Of particular importance are localizations with respect to such monoids. We shall usually write M_P for M_R–P, f_P for f_{R – P}, etc. and call M_P the localization of M at the prime ideal P.

We consider first a correspondence between the ideals of R and R_P and we shall begin by considering more generally the ideals in R and in R_s for any submonoid S. Let A′ be an ideal in R_s. Then

is an ideal in R. Clearly (j(A′))_s = A′. Again, if we begin with an ideal A in R, then it is easily seen that A_s = R_s if and only if A contains an element of S. For this reason it is natural to confine our attention to the ideals A of R that do not meet S.

We observe next that if P is a prime ideal of R such that P ∩ S = Ø, then j(P_S) = P. For, let a ∈ j(P_S). Then we have an element p ∈ P and elements s, t ∈ S such that a/s = p/t. Hence we have a u ∈ S such that uat = ups ∈ P. Since ut∈ S and S ∩ P = Ø, this implies that a ∈ P. Hence j(P_S) ⊂ P. Since the reverse inclusion j(A_s) ⊃ A holds for any ideal A, we have j(P_S) = P. It is straightforward to verify two further facts: If P is a prime ideal of R such that P ∩ S = Ø, then P_s is a prime ideal in R_s and if P′ is a prime ideal in R_s, then j(P′) is prime in R that does not meet S. We leave this to the reader. Putting together these results we obtain

PROPOSITION 7.8. The map P P_s is bijective and order-preserving from the set of prime ideals of R, which do not meet S with the set of prime ideals of R_s. The inverse map is P′ j(P′) (defined by (18)).

We now consider the important case of localization at a prime ideal P. Since Q ∩ (R – P) = Ø means Q ⊂ P, the foregoing result specializes to

PROPOSITION 7.9. The map Q Q_P is a bijective order-preserving map of the set of prime ideals Q contained in P with the set of prime ideals of R_P. The inverse map is P′ j(P′).

It is clear that P_p contains every prime ideal of R_P. It is clear also that the elements not in P_p are units in R_P and since R_P ≠ P_p, no element of P_p is a unit. Thus P_p is the set of non-units of R_P and hence R_P is a local ring with rad R_P = P_p as its only maximal ideal. We repeat the statement of this result as

PROPOSITION 7.10. R_P is a local ring with rad R_P = P_p as its only maximal ideal.

This fact accounts for the central importance of localization : It often permits a reduction of questions on commutative rings and modules over such rings to the case of local rings, since in many important instances a result will be valid for R if it holds for every R_P, P a prime ideal in R. The following result gives some basic properties of modules that hold if and only if they hold at all the localizations at prime ideals.

PROPOSITION 7.11. (1) Let M be an R-module. If M = 0, then M_s = 0 for every localization and if M_P = 0 for every maximal P, then M = 0. (2) If M and N are R-modules and f is a homomorphism of M into N, then f injective (surjective) implies that f_s is injective (surjective) for the localizations at every S. On the other hand, if f_P is injective (surjective) for every maximal ideal P, then f is injective (surjective). (3) If M is flat, then so is every M_s and if M_P is flat for every maximal ideal P, then M is flat.

Proof.     (1) Evidently M = 0 implies M_S = 0. Now assume M ≠ 0 and let x be a non-zero element of M. Then ann x ≠ R, so this ideal can be imbedded in a maximal ideal P of R. We claim that x/1 ≠ 0 in M_P, so M_P ≠ 0. This is clear since x/l is the image of x under the canonical homomorphism of M into M_P and if x/l = 0, then ann x ∩ (R – P) ≠ Ø. Since ann x ⊂ P, this is ruled out.

(2) Let f : M → N. Then we have seen that ker f_s = (ker f)_s and coker f_s ≅ (coker f)_s for any submonoid S of the multiplicative monoid of R. Since a homomorphism is injective (surjective) if and only if its kernel (cokernel) is 0, (2) is an immediate consequence of (1).

(3) Suppose that M is a flat R-module and is an exact sequence of R_S-modules. Regarding N′ and N as R-modules, we see that is exact, and by (16), is exact. We have seen that N′_s and N_s can be identified with N′ and N respectively. Hence is exact and so M_S is R_S-flat. Now suppose that M is an R-module such that M_P is R_p-flat for every maximal ideal of P of R. Let be an exact sequence of R-modules and consider Since M_P is flat and is exact, is exact. Then is exact. Since this holds for every maximal is exact by (2). Hence M is flat.

EXERCISES

        1. Show that the nil radical of R_S is (nilrad R)_S.

        2. Show that if P is a prime ideal of R, then R_P/P_P is isomorphic to the field of fractions of the domain R/P.

        3. Show that if R is a factorial domain (BAI, p. 141) and S is a submonoid of the multiplicative monoid of R not containing 0, then R_S is a factorial domain.

        4. Let {P_i| 1 ≤ i ≤ n} be a set of prime ideals in R and let S = ⁿ₁(R –P_i). Show that any prime ideal of R_S has the form P_S where P is a prime ideal contained in one of the P_i.

7.5 PRIME SPECTRUM OF A COMMUTATIVE RING

Let R be a commutative ring and let X = X(R) denote the set of prime ideals of R. There is a natural way of introducing a topology on the set X that permits the introduction of geometric ideas in the study of the ring R. We proceed to define this topology.

If A is any subset of R, we let V(A) be the subset of X consisting of the prime ideals P containing A. Evidently V(A) = V(I(A)) where I(A) is the ideal generated by A and since the nilradical of an ideal I is the intersection of the prime ideals containing I, it is clear that V(I) = V(nilrad I). Also if P is a prime ideal containing I_l I₂ for I₁, I₂ ideals, then either P ⊃ I₁ or P ⊃ I₂. Hence V(I₁ I₂) = V(I₁) ∪ V(I₂). We can now verify that the sets V(A), A a subset of R, satisfy the axioms for closed sets in a topological space:

        (1) Ø and X are closed sets, since Ø = V({1}) and X = V({0}).

        (2) The intersection of any set of closed sets is closed, since if {A_α} is a set of subsets of R, then

        (3) The union of two closed sets is closed, since we can take these to be V(I₁) and V(I₂), I_j ideals, and then V(I₁) ∪ V(I₂) = V(I₁ I₂).

We shall call X equipped with this topology the prime spectrum of R and denote it as Spec R. The subset X_max of X consisting of the maximal ideals of R with the induced topology is called the maximum spectrum. This will be denoted as Maxspec R. Such topologies were first introduced by M. H. Stone for Boolean rings and were considered by the present author for the primitive ideals of an arbitrary ring. In the case of commutative rings the topology is called the Zariski topology of X.

The open sets in X = Spec R are the complements X – V(A) = X – _{a ∈ A} V({a}) = _{a ∈ A}(X – V({a})). We denote the set X – V({a}) as X_a for a ∈ R. This is just the set of prime ideals P not containing a, that is, the P such that = a + P ≠ 0 in R/P. Since any open set is a union of sets X_a, these open subsets of X constitute a base for the open sets in Spec R. It is worthwhile to list the following properties of the map a X_a of R into the set of open subsets of Spec R:

        (1) X_ab = X_a ∩ X_b.

        (2) X_a = Ø if and only if a is nilpotent (since nilrad R is the intersection of the prime ideals of R).

        (3) X_a = X if and only if a is a unit in R.

If Y is a subset of X, put Δ_Y = _{P ∈ Y} P. This is an ideal in R and V(Δ_Y) is a closed set containing Y. On the other hand, if V(A) ⊃ Y, then P ⊃ A for every P ∈ Y so Δ_Y ⊃ A and V(A) ⊃ V(Δ_y). Thus v(Δ_Y) is the closure of Y, that is, the smallest closed set in X containing Y. In particular, we see that the closure of a point P is the set of prime ideals containing P. If R is a domain, 0 is a prime ideal in R and hence the closure of 0 is the whole space X.

EXAMPLES

1. R = . As we have noted, the closure of the prime ideal 0 is the whole space X(). Hence Spec is not a T₁-space (a space in which points are closed sets). Now consider Maxspec . The maximal ideals of are the prime ideals (p) ≠ 0. Hence the closure of (p) ≠ 0 is (p) and so Maxspec is a T₁-space. Let Y be an infinite set of primes (p) ≠ 0 in . Evidently _(p)∈y(p) = 0, so V(Δ_Y) = X. Thus the closure of any infinite subset of Maxspec is the whole space. Hence the closed sets of Maxspace are the finite subsets (including Ø) and the whole space. Evidently the Hausdorff separation axiom fails in Maxspec .

2. R = F[x₁, …, x_r], F a field, x_i indeterminates. For r = 1, the discussion of Spec and Maxspec is similar to that of the ring . For arbitrary r we remark that F[x₁, …, x_r]/(x₁) F[x₂, …, x_r], so the prime ideals of F[x₁, …, x_r] containing (x₁) are in 1–1 correspondence with the prime ideals of F[x₂, …, x_r]. Hence the closure of (x₁) in Spec F[x₁, …, x_r] is in 1–1 correspondence with Spec F[x₂, …, x_r].

We shall now derive some of the basic properties of the prime spectrum. We prove first

PROPOSITION 7.12. Spec R is quasi-compact.

Proof.     This means that if we have a set of open subsets O_α such that O_α = X, then there exists a finite subset O_α1, …, O_αn of the O_α such that O_α1 = X. (We are following current usage that reserves “compact” for “quasi-compact Hausdorff.”) Since the sets X_a form a base, it suffices to show that if a∈A X_a = X, then X_a1 = X for some finite subset {a_i} of A. The condition a∈A X_a = X gives X – V(A) = X and V(A) = Ø. Then V(I(A)) = Ø for the ideal I(A) generated by A and so I(A) = R. Hence there exist a_i ∈ A, x_i ∈ R such that ∑^r₁a_ix_i = 1. Retracing the steps, we see that V({a_i}) = Ø and X_al = X.

Let N = nilrad R, = R/N, and let v be the canonical homomorphism a → a + N of R onto . Any prime ideal P of R contains N and v(P) is a prime ideal of . Moreover, every prime ideal of has the form v(P), P a prime ideal of R. We have

PROPOSITION 7.13. The map P v(P) is a homeomorphism of Spec R onto Spec .

Proof.     The map is injective, since P ⊃ N for every prime ideal P and we have seen that the map is surjective. Now if a ∈ R, v(a) ∈ v(P) if and only if a ∈ P. This implies that if A is a subset of R, then the image of the closed set V(A) in Spec R is V(v(A)) in Spec and if is a subset of , then the inverse image of the closed set V() is V(v^{– 1}()). Hence P v(P) is a homeomorphism.

We recall that a space X is disconnected if it contains an open and closed subset ≠ Ø, ≠ X. We shall show that Spec R is disconnected if and only if R contains an idempotent ≠ 0, 1. This will follow from a considerably stronger result, which gives a bijection of the set of idempotents of R and the set of open and closed subsets of Spec R. To obtain this we shall need the following result on lifting of idempotents, which is of independent interest.

PROPOSITION 7.14. Let R be a ring that is not necessarily commutative, N a nil ideal in R, and = u + N an idempotent element of = R/N. Then there exists an idempotent e in R such that = . Moreover, e is unique if R is commutative.

Proof.   We have u² – u = z where z is nilpotent, say, zⁿ = 0 and z ∈ N. Then (u(l – u))ⁿ = uⁿvⁿ = 0 where v = 1 – u. From u + v = 1 we obtain

where e is the sum of the terms uⁱv^{2n – 1 – i} in which n ≤ i ≤ 2n– 1 and f is the sum of the terms uⁱv^{2n – 1 – i} in which 0 ≤ i ≤ n – 1. Since uⁿvⁿ = 0, any term in e annihilates any term in f. Hence ef = 0 = fe. Since e + f = 1, this gives e² = e, f² = f. Every term in e except u^{2n – 1} contains the factor uv = – z. Hence u^{2n – 1} ≡ e (mod N). Since u ≡ u² ≡ u³ ≡ ^… ≡ u^{2n– 1} (mod N), we have e ≡ u (mod N). This proves the first assertion.

Now assume that R is commutative. The uniqueness of e will follow if we can show that if e is an idempotent, then the only idempotent of the form e + z, z nilpotent, is e. The condition (e + z)² = e + z gives (1 – 2e)z = z². Then z³ = (1 – 2e)z² = (1 — 2e)²z and by induction we have (1 – 2e)ⁿz = zⁿ⁺¹. Since (1 – 2e)² = 1 – 4e + 4e = 1, this implies that z = 0 and hence e + z = e.

If e and f are idempotents in R, then so are e′ = 1 – e, ef, and e o f = l – (l – e)(l – f) = e + f – ef. It is readily verified that the set E of idempotents of R is a Boolean algebra with the compositions e ∧ f = ef and e ∨ f = e o f (exercise 1, p. 479 of BAI). We remark next that the open and closed subsets of a topological space X constitute a subalgebra of the Boolean algebra of subsets of X. We can now prove

THEOREM 7.3. If e is an idempotent in R, then X_e is an open and closed subspace of Spec R and the map e X_e is an isomorphism of the Boolean algebra E onto the Boolean algebra of open and closed subsets of Spec R.

Proof.     Let e = e² ∈ R. Then e(1 – e) = 0 , so any prime ideal P of R contains one of the elements e, 1 – e but not both. Hence X_e ∪ X_{1 – e} = X and X_e ∩ X_{1 – e} = Ø, so X_e is open and closed. Now let Y be an open and closed subset of X, Y′ = X – Y. Let R = R/N, N = nilrad R, v the canonical homomorphism of R onto . We use the homeomorphism P v(P) of X = Spec R with = Spec to conclude that v(Y) is an open and closed subset of and v(Y′) is its complement. Consider Δ_v(Y) and Δ_v(y′) as defined on p. 404. These are ideals in and V(Δ_v(Y)) = v(Y), V(Δ_v(Y′) = v(Y′) since v(Y) and v(Y′) are open and closed in . If is a prime ideal of containing Δ_v(y) + Δ_v(y′), then ⊃ Δ_v(y) and ⊃ Δ_v(Y′), so ∈ v(Y) ∩ v(Y′). Since v(Y) ∩ v(Y′) = Ø, there are no such and so Δ_v(Y) + Δ_v(Y′) = . If is any prime in , either ⊃ Δ_v(Y) or ⊃ Δ_v(Y′), so ⊃ Δ_v(Y) ∩ Δ_v(Y′). Since this holds for all and has no nilpotent elements ≠ 0, Δ_v(Y) ∩ Δ_v(y′) = 0. Hence . Let be the unit of Δ_v(Y′) and let e be the idempotent in R such that v(e) = . Now if P is an ideal in R, the condition P e is equivalent to v(P) , which in turn is equivalent to v(P) ⊃ Δ_v(Y′) and to v(P) ∈ v(Y′) and P ∈ Y′ . Hence X_e = Y, which shows that the map e X_e is surjective. To see that the map is injective, let e be a given idempotent in R, = v(e). Then we have and and ( – ) are ideals. Now v(X_e) is the set of prime ideals of containing – , hence (– ), and v(X_{1 – e}) is the set of prime ideals containing . We have shown that , we have . Since 1 – e is the only idempotent in the coset — , by Proposition 7.14, this implies the injectivity of e X_e. If e and f are idempotents, then X_ef = X_e ∩ X_f. Also X_{1 – e} = X – X_e, X₁ = X, X₀ = Ø. These relations imply that e X_e is an isomorphism of Boolean algebras.

Evidently we have the following consequence of the theorem.

COROLLARY.   Spec R is connected if and only if R contains no idempotent ≠ 0, 1.

Because of this result, a commutative ring R is called connected if the only idempotents in R are 0 and 1.

Let f be a homomorphism of R into a second ring R′. If P′ is a prime ideal in R′, then P = f ^{– 1}(P′ ) is a prime ideal in R since if ab ∈ P, then f(a)f(b) = f(ab) ∈ P′. Thus either f(a) or f(b) ∈ P′ and hence either a or b ∈ P. This permits us to define a map

of X′ = Spec R′ into X = Spec R. If a ∈ R, then a prime ideal P′ of R′ does not contain a′ = f (a) if and only if P = f*(P′) does not contain a. Hence

This implies that the inverse image of any open subset of Spec R under f* is open in Spec R′, so f* is a continuous map of Spec R′ into Spec R. It is clear that if R′ = R and f = 1_R, then f* = l_{spec R} and if g is homomorphism of R′ into R″ then (fg)* = g*f*. Thus the pair of maps R Spec R, f f* define a contravariant functor from the category of commutative rings (homomorphisms as morphisms) into the category of topological spaces (continuous maps as morphisms).

EXERCISES

        1. Let f be a homomorphism of R into R′, f* the corresponding continuous map of Spec R′ into Spec R. Show that if f is surjective, then f*(Spec R′) is the closed subset V(ker f) of Spec R. Show also that f* is a homeomorphism of Spec R′ with the closed set V(ker f).

        2. Same notations as exercise 1. Show that f*(Spec R′) is dense in Spec R if and only if nilrad R ⊃ ker f. Note that this holds if f is injective.

        3. Give an example of a homomorrphism f of R into R′ such that for some maximal ideal M′ of R′, f ^{– 1} (M′) is not maximal in R.

        4. (D. Lazard.) Show that if P is a prime ideal in R and I is the ideal in R generated by the idempotents of R contained in P, then R/I contains no idempotents ≠ 0, 1. Show that the set of prime ideals P′ ⊃ I is the connected component of Spec R containing P (the largest connected subset containing P). Hint: Let be an idempotent of = R/I where = u + I. Then u(l – u) ∈ I ⊂ P and we may assume u ∈ P. Show that there exists an f ∈ I such that f ² = f and f(u – u²) = u – u². Then (1 – f)u = g is an idempotent contained in P, so g ∈ I. Then u = g + fu ∈ I and = 0.)

In exercises 5 and 6, R need not be commutative.

        5. Let N be a nil ideal in R and let _l, …, _n be orthogonal idempotent elements of . Show that there exist orthogonal idempotents e_i in R such that _i = _l, 1 ≤ i ≤ n. Show also that if ∑ⁿ_1i = 1, then necessarily ∑e_i = 1

        6. Let R, N, be as in exercise 5 and let _ij , 1 ≤ i,j ≤ n, be elements of such that . Show that there exist e_ij ∈ R such that .

7.6   INTEGRAL DEPENDENCE

In BAI, pp. 278–281, we introduced the concept of R-integrality of an element of a field E for a subring R of E. The concept and elementary results derived in BAI can be extended to the general case in which E is an arbitrary commutative ring. We have the following

DEFINITION 7.2.   If E is a commutative ring and R is a subring, then an element u ∈ E is called R-integral if there exists a monic polynomial f (x) ∈ R[x], x an indeterminate, such that f (u) = 0.

If , then we have the relation uⁿ = a₀ + a₁ u + ^… + a_{n – 1} u^{n – 1}, from which we deduce that if M = R1 + Ru + ^… + Ru^{n – 1}, then uM ⊂ M. Evidently M is a finitely generated R-submodule of E containing 1. Since uM ⊂ M, M is an R[u]-submodule of E and since 1 ∈ M, M is faithful as R[u]-module. Hence we have the implication 1 2 in the following.

LEMMA.   The following conditions on an element u ∈ E are equivalent: 1. u is R-integral. 2. There exists a faithful R[u]-submodule of E that is finitely generated as R-module.

Proof.     Now assume 2. Then we have u_i ∈ M such that M = Ru₁ + Ru₂ + ^… + Ru_n. Then we have the relations uu_i = ∑ⁿ_{j = 1} a_iju_j, 1 ≤ i ≤ n, where the a_ij ∈ R.

Hence we have the relations

If we multiply the ith of these equations by the cofactor of the (i, j)-entry of the matrix u1 – (a_ij) and add the resulting equations, we obtain the equation f(u)u_j = 0, 1 ≤ j ≤ n, where f(x) is the characteristic polynomial of the matrix (a_ij). Since the u_j generate M and M is faithful as R[u]-module, we have f(u) = 0, so u is a root of the monic polynomial f (x). Hence 2 1.

We remark that the argument just used is slightly different from the one given in the field case in BAI, p. 279. If M and N are R-submodules of E that are finitely generated over R then so is MN = {∑m_in_i |m_i ∈ M, n_i ∈ N}. Moreover, if 1 ∈ M and 1 ∈ N then l ∈ M N and if either uM ⊂ M or uN ⊂ N then uMN ⊂ MN. These observations and the foregoing lemma can be used in exactly the same way as in BAI, pp. 279–280, to prove

THEOREM 7.4.   If E is a commutative ring and R is a subring, the subset R′ of elements of E that are R-integral is a subring containing R. Moreover, any element of E that is R′-integral is R-integral and hence is contained in R′.

The subring R′ of E is called the integral closure of R in E. If R′ = E, that is, every element of E is integral over R, then we say that E is integral over R (or E is an integral extension of R).

If A is an ideal of a ring E, then evidently R ∩ A is an ideal in the subring R of E. We call this the contraction of A to R and denote it as A^c. We also have the subring (R + A)/A of E/A and the canonical isomorphism of this ring with the ring R/A^c = R/(R ∩ A). One usually identifies (R + A)/A with R/A^c by means of the canonical isomorphism and so regards E/A as an extension of R/A^c. In this sense we have

PROPOSITION 7.15.   If E is integral over R and A is an ideal in E, then = E/A is integral over = R/A^c.

Proof.     Let = u + A be an element of . Then we have a_i ∈ R, 0 ≤ i ≤ n – 1, such that uⁿ = a₀ + a₁u + ^… + a_n–1u^{n – 1}. Then where _i = a_i + A. Hence is integral over (R + A)/A, hence over = R/A^c.

Next suppose that S is a submonoid of the multiplicative monoid of R. Then the localization E_s contains the localization R_s as a subring. Moreover, we have

PROPOSITION 7.16.   Let R be a subring of E, R′ the integral closure of R in E, S a submonoid of the multiplicative monoid of R. Then R′_S is the integral closure of R_s in E_s.

Proof.     Any element of R′_s has the form u/s, u ∈ R′, s ∈ S. We have a relation uⁿ = a₀ + a₁ u + ^… + a_{n – 1}, u^{n – 1}, a_i ∈ R. Hence . Hence u/s is integral over R_s. Conversely, suppose that u/s is integral over R_s where u ∈ E, s ∈ S. To show that u/s = v/t where v ∈ R′, t ∈ S, it suffices to show that there exists an s′ ∈ S such that us′ ∈ R′. For, u/s = us′/ss′ has the required form. Now u/S integral over R_s implies that u/1 is integral over R_s, since s/1 ∈ R_s and u/1 = (u/s)(s/1). Thus we have a relation of the form with a_i ∈ R, s_i ∈ S. Multiplication by where the a′ ∈ R. Hence . Then there exists a t₂ ∈ S such that . Multiplication of this relation by tⁿ₂^{– 1} shows that s′ u ∈ R′ for s′ = t₁ t₂.

We prove next

PROPOSITION 7.17.   If E is a domain that is integral over the subdomain R, then E is a field if and only if R is afield.

Proof.     Assume first that R is a field and let u ≠ 0 be an element of E. We have a relation uⁿ + a₁u^{n – 1} + ^… + a_n = 0 with a_i ∈ R and since u is not a zero divisor in E, we may assume a_n ≠ 0. Then a^{– 1}_n exists in R and we have . Hence u is invertible and E is a field. Conversely, suppose that E is a field and let a ≠ 0 be in R. Then a^{– 1} exists in E and we have a relation . Multiplication by a^{n – 1} gives . Hence a is invertible in R and R is a field.

An immediate consequence of this result and of Proposition 7.17 is the

COROLLARY 1.   Let E be a commutative ring, R a subring such that E is integral over R, and let P be a prime ideal in E. Then P^c = P ∩ R is maximal in R if and only if P is maximal in E.

Proof.    = E/P is a domain and by Proposition 7.15, is integral over = R/P^c. By Proposition 7.17, is a field if and only if is a field. Hence P is maximal in E if and only if P^c is maximal in R.

We also have the following

COROLLARY 2.   Let E and R be as in Corollary 1 and suppose P₁ and P₂ are ideals in R such that P₁ P₂. Then P^c₁ P₂.

Proof.     We have P^c₁ ⊃ P^c₂. Now suppose p = P^c₁ = P^c₂. Consider the localization E_s for S = R – p. By Proposition 7.16, E_S is integral over R_S. Since P_i ∩ R = p, P_i ∩ S = Ø. Then, by Proposition 7.8, P_1S P_2S. On the other hand P_iS ⊃ p_S which is the maximal ideal of the local ring R_S. Since E_S is integral over R_S it follows from Corollary 1 that P_iS is maximal in E_S. This contradicts P_lS P_2S. Thus P^c₁ P^c₂.

If P is a prime ideal in E, then it is clear that P^c is a prime ideal in R. Hence we have a map P P^c of Spec E into Spec R. This is surjective since we have the following

THEOREM 7.5 (“LYING-OVER” THEOREM).   Let E be a commutative ring, R a subring such that E is integral over R. Then any prime ideal p of R is the contraction P^c of a prime ideal P of E.

Proof.     We assume first that R is local and p is the maximal ideal of R. Let Pbe a maximal ideal in E. Then P^c is a maximal ideal in R, by the above corollary. Since p is the only maximal ideal in R, we have p = P^c.

Now let R be arbitrary and consider the localizations R_p and E_p. R_p is a local ring and E_p is integral over R_p (Proposition 7.16). Now there exists a prime ideal P′ in E_p whose contraction P′ ^c = P′ ∩ R_p = p_p, the maximal ideal of R_p. By Proposition 7.8 (p. 401), P′ = P_p for a prime ideal P of E such that P ∩ (R – p) = Ø or, equivalently, P ∩ R ⊂ p. Again, by Proposition 7.8, P = j(P_p) = {u ∈ E|u/s ∈ P_p} for some s ∈ R – p. Now p ⊂ j(p_p) ⊂ j(P_p) = P. Hence P ∩ R = p.

EXERCISES

        1. Let E be a commutative ring, R a subring such that (1) E is integral over R and (2) E is finitely generated as R-algebra. Show that E is finitely generated as R-module.

        2. (“Going-up” theorem). Let E be a commutative ring, R a subring such that E is integral over R. Let p₁ and p₂ be prime ideals of R such that p₁ ⊃ p₂ and let P₂ be a prime ideal of E such that P^c₂ = p₂. Show that there exists a prime ideal P_l of E such that P₁ ⊃ P₂ and P^c₁ = p₁.

        3. Let R be a subring of a commutative ring E, R′ the integral closure of R in E, I an ideal in R, I^e = IR′ its extension to an ideal in R′. An element a ∈ E is integral over I if it satisfies an equation f(a) = 0 where f(λ) = λ^m + b₁λ^{m – l} + ^… + b_m, b_i ∈ I. The subset I′ of E of elements integral over I is called the integral closure of I in E.

   Show that I′ = nilrad I^e in R′. Sketch of proof: If a ∈ E and a^m + b₁ a^{m– 1} + ^… +b_m = 0 with the b_i ∈ I then a ∈ R′ and a^m ∈ I^e. Hence a ∈ nilrad I^e in R′. Conversely, let a∈ nilrad I^e in R′. Then a ∈ R′ and a^m = ∑ⁿ_ia_ib_i for some m > 0, a_i ∈ R′, b_i ∈ I. Let A be the R-subalgebra of E generated by the a_i. By exercise 1, A has a finite set of generators a₁, …, a_q, q ≥ n, as R-module. Then , where the b_jk ∈ I. As in the proof of the lemma on p. 408, this implies that a^m and hence a is integral over I.

   Remarks, (i) I′ is a subring of E (in the sense of BAI, that is, I′ is a subgroup of the additive group closed under multiplication), (ii) If R is integrally closed in E (R′ = R) then I′ = nilrad I in R.

        4. (Basic facts on contractions and extensions of ideals.) Let R be a subring of a commutative ring E. For any ideal I of E, I^c = I ∩ R is an ideal in R and for any ideal i of R, i^e = iE is an ideal in E. Note that

Hence conclude that

for any ideals i and I. Note that these imply that an ideal i of R is the contraction of an ideal of E i = i^ec and an ideal I of E is an extension of an ideal of R I = I^ce

7.7   INTEGRALLY CLOSED DOMAINS

An important property of domains that we shall encounter especially in the study of Dedekind domains (see Chapter 10) is given in the following

DEFINITION 7.3.   A domain D is called integrally closed if it is integrally closed in its field of fractions.

It is readily seen that is integrally closed. More generally any factorial domain is integrally closed (exercise 1, below). Our main objective in this brief section is to prove a “going-down” theorem (Theorem 7.6) for integrally closed domains that will be required later (in the proof of Theorem 8.37). For the proof of this theorem we require

PROPOSITION 7.18.   Let D be an integrally closed subdomain of a domain E, F the field offractions of D, I an ideal in D. Let a ∈ E be integral over I. Then a is algebraic over F and its minimum polynomial over F has its coefficients (– 1)^jb_j ∈ nilrad I, 1 ≤ j ≤ m.

Proof.     The first statement is clear. For the second let S be a splitting field over F of m(λ) and let m(λ) = (λ – a₁)^… (λ – a_m) in S[λ] where a₁ = a. For any i we have an automorphism of S/F such that a a_i. Since this stabilizes I it follows that every a_i is integral over I. Since b_j, 1 ≤ j ≤ m, is an elementary symmetric polynomial in the a_i it follows from Remark (i) following exercise 3 of section 7.6 that b_j is integral over I. Then, by Remark (ii), b_j ∈ nilrad I.

We shall need also the following criterion that an ideal be a contraction of a prime ideal.

PROPOSITION 7.19.   Let R be a subring of a commutative ring E, p a prime ideal of R. Then p = P^c ≡ P ∩ R for a prime ideal P of E p^ec = Ep ∩ R = p.

Proof.     If p = P^c then p^ec = p by exercise 4 of section 7.6. Now assume this holds for a prime ideal p of R. Consider the submonoid S = R – p of the multiplicative monoid of E. Since p^ec = p, Ep ∩ R = p and hence Ep ∩ S = Ø. Since Ep is an ideal of E and Ep ∩ S = Ø, the extension (Ep)_s is a proper ideal of the localization E_s of E relative to the monoid S. Then (Ep)_s is contained in a maximal ideal Q of E_s and P = j(Q) as defined by (18) is a prime ideal of R such that P ∩ (R – p) = P ∩ S = Ø. Then P ∩ R ⊂ p. Since P = _j(Q) ⊃ j((Ep)_s) ⊃ p we have P^c = P ∩ R = p.

We can now prove

THEOREM 7.6 (“Going-down” theorem).   Let D be an integrally closed subdomain of a domain E that is integral over D. Let p₁ and p₂ be prime ideals of D such that p₁ ⊃ p₂ and suppose P₁ is a prime ideal of E such that P^c₁ = p₁. Then there exists a prime ideal P₂ of E such that P^c₂ = p₂ and P₂ ⊂ P₁.

Proof.     Consider the localization E_pl ( = E_E – _P1). It suffices to show that p₂E_Pl ∩ D = p₂. For, if this holds, then by Proposition 7.19 there exists a prime ideal Q of E_P1 such that D ∩ _Q = p₂. Then P₂ = j(Q) is a prime ideal of E contained in P₁ and .

We now proceed to the proof that p₂E_Pl ∩ D = p₂. Let a ∈ p₂E_Pl. Then a = b/s where b ∈ Ep₂ and s ∈ E – P₁ By exercise 3 of section 7.6, b is integral over p₂ and hence by Proposition 7.18, the minimum polynomial of b over the field of fractions F of D has the form , 1 ≤ i ≤ m.

Suppose a = b/s ∈ p₂E_Pl ∩ D. Then s = b/a and the minimum polynomial of s over F is . Since s is integral over D, taking I = D in Proposition 7.18, we see that every b_i/aⁱ ∈ D. Then (b_i/aⁱ)aⁱ ∈ p₂ and b_i/aⁱ and aⁱ ∈ D. Now suppose a p₂. Then, since p₂ is prime in D, b_i/aⁱ ∈ p₂, 1 ≤ i ≤ m. Thus s is integral over p₂ and hence over p₁. Then again by exercise 3, s ∈ nilrad p₁ E ⊂ P₁ contrary to s ∈ E – P₁ This shows that a ∈ p₂ so p₂E_Pl ∩ D ⊂ p₂. Since the reverse inequality is clear we have p₂E_Pl ∩ D = p₂.

EXERCISES

        1. Show that any factorial domain is integrally closed.

        2. Let D be a domain, F its field of fractions. Show that if D is integrally closed then D_s is integrally closed for every submonoid S of the multiplicative monoid of R. On the other hand, show that if D_P is integrally closed for every maximal ideal P of D then D is integrally closed. (Hint: Use Proposition 7.11 on p. 402.)

7.8   RANK OF PROJECTIVE MODULES

We have shown in BAI, p. 171, that if R is a commutative ring and M is an R-module with a base of n elements, then any base has cardinality n. Hence the number n, called the rank of M, is an invariant. We repeat the argument in a slightly improved form. Let {e_i|l ≤ i ≤ n} be a base for M and let {f_j| l ≤ j ≤ m } be a set of generators. Then we have which gives . Assume n ≥ m and consider the n × n matrices

Then we have BA = l_n, the n × n unit matrix. Since R is commutative, this implies AB = 1_n (BAI, p. 97), which is impossible if n > m. Hence n = m. Thus we see that if M has a base of n elements, then any set of generators contains at least n elements. Hence any two bases have the same cardinality. The argument shows also that any set of n generators f_j = ∑a_ji e_i, 1≤ j ≤ n, is a base, since the argument shows that in this case the matrix A = (a_ij) is invertible in M_n(R), and this implies that the only relation of the form ∑c_if_i = 0 is the one with every c_i = 0. We summarize these results in

PROPOSITION 7.20   Let M be a free module over a commutative ring with base of n elements. Then (1) any base has cardinality n, (2) any set of generators contains at least n elements, and (3) any set of n generators is a base.

We shall now give a method, based on localization, for extending the concept of rank to finitely generated projective modules over commutative rings R. We shall prove first that such modules are free if R is local. For this we require an important lemma for arbitrary rings known as

NAKAYAMA’S LEMMA.   Let M be a module over a ring R (not necessarily commutative). Suppose that (1) M is finitely generated and (2) (rad R)_M = M. Then M = 0.

Proof.     Let m be the smallest integer such that M is generated by m elements x_l, x₂,…, x_m. If M ≠ 0, then m > 0. The condition (rad R)M = M implies that x_m = r₁x₁ + ^… + r_mx_m where the r_i ∈ rad R. Then we have

Since r_m ∈ rad R, 1 – r_m is invertible in R, and acting with (1 – r_m)^{– l} on the foregoing relation shows that x_m can be expressed in terms of x₁,…,x_{m – 1}. Then x₁,…, x_{m –1} generate M, contrary to the choice of m.

We recall that if B is an ideal in a ring R and M is an R-module, then BM is a submodule and the module M/BM is annihilated by B and so can be regarded in a natural way as R/B-module. In particular this holds for B = rad R. In this case we have the following consequence of Nakayama’s lemma.

COROLLARY.   Let M be a finitely generated R-module. Then x₁, …, x_m ∈ M generate M if and only if the cosets = x₁ + (rad R)M,…, _m = x_m + (rad R)M generate M/(rad R)M as R/rad R module.

Proof.   If x_l, … x_m generate M, then evidently _l, …, _m generate = M/(rad R)M as R-module and, equivalently, as = R/rad R-module. Conversely, suppose the _i generate as -module, hence as R-module. Then M = (rad R)M + N where N = ∑Rx_i. Then (rad R)(M/N) = M/N and M/N is finitely generated. Hence M/N = 0 by Nakayama’s lemma, so M = N = ∑^m₁Rx_i.

We can now prove

THEOREM 7.7.   If R is a (commutative) local ring, then any finitely generated projective module over R is free.

Proof.     We may assume that M is a direct summand of a free module F = R⁽ⁿ⁾ : F = M N where M and N are submodules. Then (rad R)F = (rad R)M + (rad R)N and (rad R)M ⊂ M, (rad R)N ⊂ _N. Hence = F/(rad R)F = where = (M + (rad R)F)/(rad R)F, = (N + (rad R)F)/(rad R)F. Now = R/rad R is a field and evidently is an n-dimensional vector space over , and and are subspaces. We can choose elements y₁, …, y_n so that y_l … y_r ∈ M, y_{r + 1}, …, y_n ∈ N and if _i = y_i + (rad R)F, then {_l, …, _n} is a base for over . Then, by the Corollary to Nakayama’s lemma, y_l, …, y_n generate F and, by Proposition 7.18, they form a base for F. Then {y₁ …, y_r} is a base for M so M is free.

We now consider finitely generated projective modules over an arbitrary ring K and we prove first

PROPOSITION 7.21.   If M is a free R-module of rank n then M_s is a free R_s-module of rank n for any submonoid S of the multiplicative monoid of R. If M is projective with n generators over R, then M_s is projective with n generators over R_s.

Proof.     To prove the first statement we recall that S-localization is a functor from R-mod to R_s-mod. Hence a relation M ≅ M₁ ^… M_s for R-modules implies M_s ≅ M_1S ^… M_sS for R_s-modules. Then M ≅ R ^… R (n copies) implies M_s ≅ R_s ^… R_S (n copies). Next suppose M is projective with n generators. This is equivalent to assuming that M is isomorphic to a direct summand of the free module R⁽ⁿ⁾. Then M_s is R_s-projective with n generators.

In particular, if M is projective with n generators, then for any prime ideal P, M_P is projective with n generators over the local ring R_P. Hence M_P is R_P-free of rank n_P ≤ n. We call n_P the P-rank of M and we shall say that M has a rank if n_P = n_Q for any two prime ideals P and Q of R. In this case the common value of the local ranks n_P is called the rank of the finitely generated projective module M. We shall now show that the map P n_P is a continuous map of Spec R into endowed with the discrete topology. This will imply that if Spec R is connected or, equivalently, R has no idempotents ≠ 0, 1, then M has a rank.

The continuity we wish to establish will be an easy consequence of

PROPOSITION 7.22.   Let M be a finitely generated projective R-module, P a prime ideal in R. Then there exists a P such that M is free as R-module.

Proof.     M_P is a free R_P-module of finite rank. If (x₁/s₁, …, x_n/s_n), x_i ∈ M, s_i ∈ R – P, is a base for M_P, and since the elements s_i/1 are units in R_P, (x_l/1, …, x_n/l) is also a base. Consider the R-homomorphism f of R⁽ⁿ⁾ into M such that (a_l, …,a_n) ∑a_ix_i. We have the exact sequence

Localizing at P, we obtain the exact sequence

Since M_P is free with base (x₁/l, …, x_n/l), f_P is an isomorphism and hence ker f_P = 0, coker f_p = 0. Since M is finitely generated, so is its homomorphic image coker f and since coker f_p, ≅ (coker f)_P, it follows from Proposition 7.5 that we have an element b P such that b(coker f) = 0. Then (coker f) = 0 and we have the exact sequence

Since M is projective, this splits, so ker f is isomorphic to a homomorphic image of R⁽ⁿ⁾ and hence this is finitely generated. Since is a submonoid of R – P, 0 = (ker f)_P ≅ ker f_p, is obtained by a localization of (ker f) (Proposition 7.4). Since (ker f) is finitely generated, this implies that there exists an element c/1, c P such that (c/1) (ker f) = 0. Then ker f = 0. Hence if we put a = bc, then a P and we have

so M ≅ R⁽ⁿ⁾ is free. This complates the proof.

We now have

THEOREM 7.8.   The map r_M: P n_P(M) of Spec R into (with discrete topology) is continuous.

Proof   Let P be any point in Spec R. We have to show that there exists an open set O containing P such that n_Q = n_P for all Q in O. Take O = X_a where a is as in Proposition 7.20. Then P ∈ X_a and if Q ∈ X_a, then M_Q is a localization of M, which is a free R-module of rank n. Then the rank of M_Q is n. Thus n_Q = n = n_P.

If Spec R is connected, the continuity of the map r_M implies that n_P = n_Q for all P, Q ∈ Spec R. Hence we have the

COROLLARY.   If R is connected, that is, R has no idempotents ≠ 0, 1, then the rank is defined for every finitely generated projective R-module.

Again let M be a finitely generated projective module and consider the continuous map r_M of Spec R into . Since has the discrete topology, any integer n is an open and closed subset of and r^–1_M(n) is an open and closed subset of Spec R. Moreover, Spec R is a disjoint union of these sets. Since Spec R is quasi-compact, the number of non-vacuous sets r^–1_M(n) is finite. Thus we have positive integers n_l, …, n_s such that r^–1_M(n_i) ≠ Ø and X = Spec R = r^–1_M(n_i). By Theorem 7.3, the open and closed subset r^–1_M(n_i) has the form X_el for an idempotent e_i ∈ R. Since , . Thus the e_i are orthogonal idempotents in R with sum 1 and hence R = R₁ ^… R_S where R_i = Re_i and M = M₁ ^… M_s where M_i = e_iM = R_iM. It is clear from the dual basis lemma (Proposition 3.11, p. 152) that M_i is a finitely generated projective R_i-module. We claim that M_i has a rank over R_i and this rank is n_i. Let P_i be a prime ideal in R_i. Then is a prime ideal in R not containing e_i, so P ∈ X_el and P contains every e_j, j ≠ i. We have and since e_i ∈ P and e_iR_j = 0 for j ≠ i, R_P = R_iP. Similarly M_P = M_iP. Hence the rank of M_iP over R_iP is n_i. We have the homomorphism r re_i of R into R_i in which the elements of R – P are mapped into elements of R_i–P_i. Following this with the canonical homomorphism of R_i into R_iPl gives a homomorphism of R into R_iPl in which the elements of R–P are mapped into invertible elements. Accordingly, we have a homomorphism of R_P into R_iPl and we can use this to regard R_iPl as R_P-module. Then M_iPl = R_iPlR_pM_iPl and since M_iP is a free R_iP module of rank n_i, M_iPl is a free R_iPl module of rank n_i. Since this holds for every prime ideal P_i of R_i, we see that the rank of M_i over R_i is n_i. A part of what we have proved can be stated as

THEOREM 7.9.   Let M be a finitely generated projective module over a commutative ring R. Then there are only a finite number of values for the ranks n_P(M) for the prime ideal P of R. If these values are n₁,…,n_s, then R = R ^… R_S where the R_i are ideals such that R_iM is finitely generated projective over R_i of rank n_i.

EXERCISES

        1. Let M be a finitely generated projective module over the commutative ring R and let P be a maximal ideal of R. Show that if M has a rank over R, then this is the dimensionality of M/PM regarded as vector space over the field R/P.

        2. Let M and R be as in exercise 1 and assume that R is an algebra over R′ such that R is finitely generated projective over R′ . Suppose that M has a rank over R and R has a rank over R′ . Show that M has a rank over R′ and rank M/R′ = (rank M/R) (rank R/R).

        3. Show that if D is a domain, then any finitely generated projective module M over D has a rank and this coincides with the dimensionality of F_DM over F where F is the field of fractions of D.

7.9   PROJECTIVE CLASS GROUP

We now make contact again with the Morita theory. We recall that if R′ and R are rings, an R′ -R-bimodule M is said to be invertible if there exists an R-R′-bimodule M′ such that M′ _R′ M ≅ R as R-R-bimodule and M_RM′ ≅ R′ as R′-R′ -bimodule. This is the case if and only if M is a progenerator of mod-R (R′ -mod). Then R′ ≅ End M_R as rings and M′ ≅ hom (M_R, R_R) as R-R′ -bimodule. If R′ = R, the isomorphism classes of invertible bimodules form a group Pic R in which the multiplication is given by tensor products. Now suppose R is commutative. We restrict our attention to the R-R-bimodules in which the left action is the same as the right action. In effect we are dealing with (left) R-modules. The isomorphism classes of invertible modules constitute a subgroup of Pic R that is called the projective class group of the commutative ring R. The following result identifies the modules whose isomorphism classes constitute the projective class group.

THEOREM 7.10.   Let R be a commutative ring, M an R-module. Then M is invertible if and only if it is faithful finitely generated projective and rank M = 1.

Proof.     Suppose that M is invertible and M′ is an R-module such that M_RM′ ≅ R. Then M and M′ are finitely generated projective. Let P be a prime ideal in R. Then M_RM′ ≅ R implies that M_PRpM′ _P ≅ R_P. Since these modules are free of finite rank over the local ring R_P, this relation implies that rank M_P/_RP = 1. Since this holds for all P, we see that rank M = 1. Conversely, suppose that M satisfies the stated conditions and let M* = hom_R(M,R). Since M is faithful finitely generated projective, M is a progenerator by Theorem 3.22. Hence, as in the Morita theory, we have the isomorphism μ of M_RM* onto End_RM sending xy*, x ∈ _M, y* ∈ M*, into the map y y*(y)x. Since R is commutative, any r ∈ R determines an R-endomorphism y ry. We can identify R with this set of endomorphisms and so End_R M ⊃ R. Hence to show that M _R M* ≅ R, it suffices to prove that End_RM = R. This will follow if we can show that (End_RM)_P = R_P for every prime ideal P. Now (End_RM)_P ≅ R_PREnd_RM Since M is finitely generated projective, it is easily seen that R_PREnd_RM ≅ EndR_pM_p (see the proof of Proposition 3.14, p. 154). Now if M is a free module of rank n over a commutative ring R, then End_RM is a free module of rank n² over R (≅ M_n(R)). By hypothesis, M_P has rank 1 over R_P. Hence End_RpM_P has rank 1 over R_P. Thus (End_RM)_P has rank 1 over R_P and hence (End_RM)_P = R_P for every P. This completes the proof.

We shall not give any examples of projective class groups at this point. Later (section 10.6) we shall see that if R is an algebraic number field, then this group can be identified with a classical group of number theory.

7.10   NOETHERIAN RINGS

In the remainder of this chapter we shall be interested primarily in noetherian rings and modules. We recall that the noetherian condition that a module satisfy the ascending chain condition for submodules is equivalent to the maximum condition that every non-vacuous set of submodules contains submodules that are maximal in the set, and equivalent to the “finite basis condition” that every submodule has a finite set of generators (p. 103, exercise 1). We recall also that a (commutative) ring is noetherian if it is noetherian as a module with respect to itself, which means that every properly ascending chain of ideals in the ring terminates, that every non-vacuous set of ideals contains maximal ones, and that every ideal has a finite set of generators. Moreover, any one of these conditions implies the other two. In 1890 Hilbert based a proof of a fundamental theorem in invariant theory on the following theorem :

HILBERT’S BASIS THEOREM.   If R is a field or the ring of integers, then any ideal in the polynomial ring R[x₁,x₂, …, x_r], x_i indeterminates, has a finite set of generators.

Hilbert’s proof admits an immediate extension of his basis theorem to

HILBERT’S (GENERALIZED) BASIS THEOREM.   If R is a ring such that every ideal in R is finitely generated, then every ideal in R[x_x,… ,x_r] is finitely generated.

Proof.   Using induction it evidently suffices to prove the theorem for the case of one indeterminate x. We have to show that any ideal B of R[x] has a finite set of generators. Let j = 0, 1,2,… and let I_j = I_j(B) be the set of elements b_j ∈ R such that there exists an element of the form

It is evident that I_j is an ideal in R, and since f_j ∈ B implies that xf_j = b_jx^j+1 + ^… ∈ B, it is clear that I_j ⊂ I_{j+ 1} Hence I = I_j is an ideal in R. This has a finite set of generators, and we may assume that these are contained in one of the ideals I_m. If {b_m⁽¹⁾, … ,b^{_k)_m} are these generators, we have polynomials where deg g_m⁽ⁱ⁾ < m. Similarly, since every I_j is finitely generated for j < m, we have polynomials where deg g_j^(i_j) < j and 1 ≤ i_j ≤ k_j. We claim that the finite set of polynomials

generate B. Let f ∈ B. We prove by induction on n = deg f that f = for suitable h_ij ∈ R[x]. This is clear if f = 0 or deg f = 0, so we assume that f = b_nxⁿ + f₁ where b_n ≠ 0 and deg f₁ < n. Then b_n ∈ I_n ⊂ I. If n ≥ m, then b_n = and has the same leading coefficient as f. Hence deg (f – ∑a_ix^n–mf⁽ⁱ⁾_m) < n. Since this polynomial is in B, the result follows by induction. If n < m, we have b_n = . Again the result follows by induction.

An alternative version of Hilbert’s theorem is that if R is noetherian, then so is the polynomial ring R[x …, x_r]. We also have the following stronger result.

COROLLARY.   Let R be noetherian and R′ an extension ring of R, which is finitely generated (as algebra) over R. Then R′ is noetherian.

Proof.     The hypothesis is that R′ = R[u₁, …, u_r] for certain u ∈ R′ . Then R′ is a homomorphic image of R[x₁, …, x_r], x_i indeterminates. Since R[x₁, …, x_r] is noetherian, so is R′ .

There is another important class of examples of noetherian rings: rings of formal power series over noetherian rings. If R is an arbitrary ring, we can define the ring R[[x]] of formal power series over R as the set of unrestricted sequences

a_i ∈ R with addition defined component-wise, 0 = (0, 0, …), 1 = (1, 0,0, …), and product ab for a as above and b = (b₀,b₁, …) defined as c = (c₀, c_l, …) where

It is easily checked that R[[x]] is a ring (exercise 7, p. 127 of BAI). It is clear that R can be identified with the subring of R[[x]] of elements (a₀,0, 0, …) and R[x] with the subring of sequences (a₀, a₁, …, a_n,0, 0, …), that is, the sequence having only a finite number of non-zero terms (BAI, pp. 116–118).

In dealing with formal power series, the concept of order of a series takes the place of the degree of a polynomial. If a = (a₀, a_l, …), we define the order o(a) by

Then we have

and

If R is a domain, then (22) can be strengthened to o(ab) = o(a) + o(b). In any case if we define

(with the convention that 2^{– ∞} = 0), then we have the following properties of the map a |a| of R[[x]] into :

            (i) |a| ≥ 0 and |a| = 0 if and only if a = 0.

         (ii) |a + b| ≤ max (|a|, |b|).

(iii)   |ab| ≤ |a| |b|.

The second of these implies that |a + b| ≤ |a| + |b|. Hence (i) and (ii) imply that R[[x]] is a metric space with distance function d(a, b) =|a – b|. We can therefore introduce the standard notions of convergence of sequences and series, Cauchy sequences etc.

We say that the sequence {a^{i)}, a⁽ⁱ⁾ ∈ R[[x]], i = 1, 2, 3,…, converges to a ∈ R[[x]] if for any real ε > 0 there exists an integer N = N(ε) such that |a – a⁽ⁱ⁾| < ε for all i ≥ N. Then we write lim a⁽ⁱ⁾ = a. The sequence {a⁽ⁱ⁾} is called a Cauchy sequence if given any ε > 0 there exists an integer N such that |a⁽ⁱ⁾ – a^(j)| < ε for all i,j ≥ N. R[[x]] is complete relative to its metric in the sense that every Cauchy sequence of elements in R[[x]] converges. For, let {a⁽ⁱ⁾} be a Cauchy sequence in R[[x]]. Then for any integer n ≥ 0 we have an N such that o(a⁽ⁱ⁾ – a^(j)) > n or all i,j ≥ N. Let a_n be the entry in the (n + l)-st place of a^(N). Then every a⁽ⁱ⁾, i ≥ N, has this element as its (n + l)-st entry. It is readily seen that if we take a = (a₀,a₁,a₂, …), then lim a⁽ⁱ⁾ = a.

We can also define convergence of series in the usual way. We say that a⁽¹⁾ + a⁽²⁾+ ^… = a if the sequence of partial sums a⁽¹⁾,a⁽¹⁾ + a⁽²⁾,… converges to a.

If we put x = (0, 1,0,…), then xⁱ has 1 in the (i + l)th place, 0’s elsewhere. It follows that if a = (a₀,a_l,a₂, …) and we identify a_j with (a_j,0, 0, …), then we can write

It is clear that if R is a domain, then so is R[[x]]. It is also easy to determine the units of R[[x]], namely, ∑a_ixⁱ is a unit if and only if a₀ is a unit in R. The condition is clearly necessary. To see that it is sufficient, we write ∑^∞₀ a_ixⁱ = a₀(1 – z) where z = ∑^∞₁ b_jx^j and b_j = – a₀^{– 1}a_j. Then o(z) ≥ 1 and o(z^k) ≥ k. Hence 1 + z + z² + ^… exists. It is readily seen that

Hence ∑^∞₀ a_ixⁱ = a₀(l – z) is a unit with inverse a₀^{– 1} ( ∑^∞₀ zⁱ). It is clear also that the set of power series ∑^∞₁b_ixⁱ of order > 0 is an ideal in R[[x]]. An immediate consequence of these remarks is

PROPOSITION 7.23.   If F is a field, then F[[x]] is a local ring.

Proof.     The set of elements ∑^∞₀a_ixⁱ that are not units is the set for which a₀ = 0. This is an ideal. Thus the non-units constitute an ideal and hence F[[x]] is local.

It is easily seen also that if R is local, then so is R[[x]]. We leave the proof to the reader.

We shall prove next the following important

THEOREM 7.11.   If R is noetherian, then so is R[[x]].

Proof.   The proof is quite similar to the proof of the Hilbert basis theorem. Let B be an ideal. For any j = 0, 1,2,… let I_j be the set of b_j ∈ R such that there exists an element f_j = b_jx^j + g_j ∈ B where o(g_j) > j. Then I_j is an ideal in R and I₀ ⊂ I₁ ⊂ …, so I = I_j is an ideal in R. Since R is noetherian, I has a finite set of generators and all of these are contained in I_m for some m. Let these generators be b_m⁽¹⁾ …, b_m^(k). It is clear from (22) and (23) that the set of elements of R[[x]] of order ≥ m form an ideal. The intersection of this set with B is an ideal B_m in R[[x]] containing the elements f_m⁽ⁱ⁾, 1 ≤ i ≤ k. Now let f ∈ B_m and suppose f = b_nxⁿ + g where b_n ∈ R, n ≥ m and o(g) > n. Then b_n = ∑_ia_ib_n⁽ⁱ⁾ for a_i ∈ R and

Iteration of this process yields a sequence of integers n₁ = n – m < n₂ < n₃^… and elements a_ij ∈ R, 1 ≤ i ≤ k, j = 1, 2,… such that

r = 1, 2,…. Then a_i = ∑^∞_{j = 1} a_ijx^n_J is well defined and . Now consider the ideals I_j, 0 ≤ j < m. Choose a set of generators {b⁽¹⁾_j,…,b_j^(k_J)} for I_j and . Then, as in the polynomial case,

is a set of generators for B.

We can iterate the process for forming formal power series to construct R[[x, y]]= (R[[x]]) [[y]], etc. We can also mix this construction with that of forming polynomial extensions. If we start with a noetherian R and perform these constructions a finite number of times, we obtain noetherian rings.

Another construction that preserves the noetherian property is described in the following

THEOREM 7.12.   Let R be noetherian and let S be a submonoid of the multiplicative monoid ofR. Then the localization R_s is noetherian.

Proof.     Let B′ be an ideal in R_s. As on p. 401 let j(B′ ) be the set of elements b ∈ R such that b/s ∈ B′ for some s ∈ S. Then j(B′ ) is an ideal in R and j(B′ )_s = B′ . Since R is noetherian, j(B′ ) has a finite set of generators {b₁, …, b_m}. Then the set {b₁/1, …, b_m/1} ⊂ = B′ and generates this ideal.

EXERCISES

        1. Show that if I is an ideal in a commutative ring R such that I is not finitely generated and every ideal properly containing I is finitely generated, then I is prime. Use this to prove that if every prime ideal in R is finitely generated, then R is noetherian.

        2. If R is a ring, we define the ring of formal Laurent series R((x)) over R as the set of sequences (a_i), – ∞ < i < ∞, a_i ∈ R such that there exists an integer n (depending on the sequence) such that a_i = 0 for i < n. Define addition and multiplication as for power series. Show that R((x)) is a localization of R[[x]] and hence that R((x)) is noetherian if R is noetherian. Show that if R is a field, so is R((x)), and R((x)) contains the field R(x) of rational expressions in x.

        3. (Emmy Noether’s finiteness theorem on invariants of a finite group.) Let E = F[u₁,…,u_m] be a finitely generated algebra over the field F and let G be a finite group of automorphisms of E/F. Let Inv G = {y ∈ _E|sy = y, s ∈ G}. Show that Inv G is a finitely generated algebra over F. (Sketch of proof. Let f_i(x) = ∏_s∈G(x – su_i) = xⁿ – P_i1x^{n – 1} + P_i2x^{n – 2 –…} , where x is an indeterminate. Then I = F[p₁₁,… ,p_mm] ⊂ Inv G and E is integral over I. Hence, by exercise 1, p. 411, E is a finitely generated I-module. Since I is noetherian, Inv G is finitely generated, say by v₁, …, v_r. Then Inv G = F[p_{1 1}, … p_mn,v₁, …, v_r].)

7.11   COMMUTATIVE ARTINIAN RINGS

The study of artinian rings constitutes a major part of the structure theory of rings as developed in Chapter 4. It is interesting to see how this theory specializes in the case of commutative rings, and to consider relations between the artinian and noetherian conditions. Our first result in this direction is valid for rings that need not be commutative.

THEOREM 7.13.   If R is a ring that is left (right) artinian, then R is left (right) noetherian. Moreover, R has only a finite number of maximal ideals.

Proof.     Let J = rad R, the Jacobson radical of R. By Theorem 4.3 (p. 202), J is nilpotent, so we have an integer n such that Jⁿ = 0. We have the sequence of ideals R ⊃ J ⊃ J² ⊃ ^… ⊃ Jⁿ = 0. We regard R as left R-module. Then we have the sequence of R-modules _i = Jⁱ/J^{i + 1} where J⁰ = R and all of these are annihilated by J, so they may be regarded as modules for the semi-primitive artinian ring = R/J. Since all modules of a semi-primitive artinian ring are completely reducible (Theorem 4.4, p. 208), this is the case for the modules _i. Moreover, any completely reducible artinian (noetherian) module is noetherian (artinian) and hence has a composition series. Accordingly, for each _i we have a sequence of submodules

such that every _ij/_{i,j + 1} is irreducible, and these can be regarded as R-modules. Corresponding to these we have a sequence of left ideals M_i1 = Jⁱ ⊃ M_i2 ⊃ ^… ⊃ M_{i, n +1} = Jⁱ⁺¹ such that M_ij/M_{i,j + 1} is an irreducible R-module. Putting together these sequences we obtain a composition series for R as left R-module. The existence of such a series implies that R is left noetherian as well as left artinian. The same argument applies to right artinian rings. This proves the first statement of the theorem.

To prove the second, we note that any maximal ideal I of R contains J and = I/J is a maximal ideal of the semi-primitive artinian ring . We have where the _i are minimal ideals. It follows that the only maximal ideals of are the ideals and that the only maximal ideals of R are the ideals M_j = R₁ + ^… + R _{j – i} + R_{j + 1} + ^… + R_s where R_j is the ideal in R such that _j = R_j/J.

For commutative rings we have the following partial converse to Theorem 7.13.

THEOREM 7.14.     If R is a commutative noetherian ring that has only a finite number of prime ideals and all of these are maximal, then R is artinian.

Proof.     Since the Jacobson radical J is the intersection of the maximal ideals of R and the nil radical N is the intersection of the prime ideals of R, the hypothesis that the prime ideals are maximal implies that J = N. Hence J is a nil ideal and since R is noetherian, J is finitely generated. As is easily seen, this implies that J is nilpotent. Since R contains only a finite number of maximal ideals, = R/J is a subdirect product of a finite number of fields. Then is a direct sum of a finite number of fields (Lemma on p. 202) and hence is artinian. As in the proof of Theorem 7.11, we have the chain of ideals R ⊃ J ⊃ ^… ⊃ J^{n – 1} Jⁿ = 0 and every Jⁱ/J^{i + 1} is a completely reducible noetherian module for the semi-primitive artinian ring . Then Jⁱ/J^{i + 1} has a composition series and R has a composition series as R-module. Then R is artinian.

It is easy to determine the structure of commutative artinian rings. This is given in the following

THEOREM 7.15.   Let R be a commutative artinian ring. Then R can be written in one and only one way as a direct sum R = R₁ R₂ … R_s where the R_i are artinian and noetherian local rings and hence the maximal ideal of R_i is nilpotent. Conversely, if R = R₁ … R_s where the R_i are noetherian local rings with nilpotent maximal ideals, then R is artinian.

Proof.     We base the proof of the first part on the results on modules that we obtained in connection with the Krull-Schmidt theorem (pp. 110–115) and the fact that for a commutative ring R we have the isomorphism R End_RR. We have seen that R has a composition series as R-module. Hence R = R₁ … R_s where the R_i are indecomposable R-modules with composition series. Hence End_RR_i is a local ring. Now R_i End_RiR_i = End_RR_i, so R_i is a local ring. Now there is only one decomposition of a ring into indecomposable ideals (p. 204). Hence the R_i are unique. The artinian property of R carries over to the R_i. Hence, by Theorem 7.13, every R_i is also noetherian. It is clear also that the maximal ideal, rad R_i, of R_i is nilpotent.

To prove the converse, suppose first that R is a commutative local ring whose maximal ideal J is nilpotent. Let P be a prime ideal in R. Then P ⊂ J and Jⁿ ⊂ P for some n. Since P is prime, this implies that J = P. Hence P is the only prime ideal of R. Since R is noetherian, R is artinian by Theorem 7.14. The general case in which R = R₁ … R_s follows immediately from this special case.

If R is an artinian ring and M is a finitely generated R-module, then M has a composition series since R is also noetherian and hence M is artinian and noetherian. We can therefore define the length l(M) of M as the length of any composition series for M. The use of the length provides a tool that is often useful in proving results on modules for artinian rings.

7.12   AFFINE ALGEBRAIC VARIETIES.THE HILBERT NULLSTELLENSATZ

Let F be an algebraically closed field, F⁽ⁿ⁾ the n-dimensional vector space of n-tuples (a₁,a₂, … ,a_n), a_i ∈ F, and F[x₁,x₂, … ,x_n] the ring of polynomials in n indeterminates x_i over F. If S is a subset of F[x₁, … ,x_n] we let V(S) denote the set of points (a₁, … ,a_n) ∈ F(n) such that f(a₁, … , a_n) = 0 for every f ∈ S and we call V(S) the (affine algebraic) variety defined by S. It is clear that V(S) = V(I) where I = (S), the ideal generated by S, and also V(I) = V(nilrad I). Hence if I₁ and I₂ are two ideals such that nilrad I₁ = nilrad I₂, then V(I₁) = V(I₂). A fundamental result, due to Hilbert, is tha t conversely if V(I₁) = V(I₂) for two ideals in F[x₁, …, x_n], F algebraically closed, then nilrad I₁ = nilrad I₂. There are a number of ways of proving this result. In this section we shall give a very natural proof, due to Seidenberg, which is based on Krull’s theorem that the nil radical of an ideal is the intersection of the prime ideals containing it and on a general theorem of elimination theory that we proved in a completely elementary fashion in BAI, pp. 322–325. For convenience we quote the theorem on elimination of unknowns:

Let K = or /(p), p a prime, and let A = K[t₁, …, t_r], B = A[x₁, …, x_n] where the t’s and x’s are indeterminates. Suppose F_l, …, F_m, G ∈ B. Then we can determine in a finite number of steps a finite collection where such that for any extension field F of K and any (c₁, …, c_r), c_i ∈ F, the system of equations and inequations

is solvable for x’s in some extension field E of F if and only if the c_i satisfy one of the systems Γ_j:

Moreover, when the conditions are satisfied, then a solution of (27) exists in some algebraic extension field E of F.

For our present purposes the important part of this result is the last statement. This implies the following theorem, which perhaps gives the real meaning of the algebraic closedness property of a field.

THEOREM 7.16.   Let F be an algebraically closed field and let f₁, …, f_m, g ∈ F[x₁, …, x_n] where the x_i are indeterminates. Suppose that the system of equations and inequation

has a solution in some extension field E of F. Then (29) has a solution in F.

Proof.     The field F has one of the rings K = or /(p) as its prime ring. Now by choosing enough additional indeterminates t₁, …, t_r we can define polynomials F₁, …, F_m, G ∈ B as in the elimination theorem such that F_i(c₁, …, c_r; x₁, …, x_n) = f_i(x₁, …, x_n), G(c₁, …, c_r; x₁, …, x_n) = g(x₁, …, x_n). Then it follows from the result quoted that, if (29) has a solution in some extension field E of F, then we have a j for which (28) holds, and this in turn implies that (29) has a solution in an algebraic extension field of F. Since F is algebraically closed, the only algebraic extension field of F is F itself (BAI, p. 216, or p. 460 below). Hence (29) is solvable in F.

Hilbert’s theorem, the so-called Nullstellensatz, is an immediate consequence of this result and the characterization of the nil radical as intersection of prime ideals.

NULLSTELLENSATZ.   Let I be an ideal in the polynomial ring F[x₁, …, x_n] in n indeterminates x_i over an algebraically closed field F and let g∈F[x₁, …, x_n]. Suppose g(a_l, …, a_n) = 0 for all (a₁, …, a_n) on the variety V(I) defined by I. Then g ∈ nilrad I.

Proof.     Suppose g nilrad I. Then there exists a prime ideal P ⊃ I such that g P. Consider the domain D = F[x₁, …, x_n]/P = F[₁, …, _n] where _i = x_i + P. Then for f = f(x₁, …, x_n) ∈ I, f ∈ P and hence f(₁, …, _n) = 0. On the other hand, g P, so g(₁, …, _n) ≠ 0. Now F[x₁, …, x_n] is noetherian, so I has a finite set of generators f₁, …, f_m. Let E be the field of fractions of D. Then E is an extension field of F containing the elements ₁, …, _n such that f_i(₁, …, _n) = 0, 1 ≤ i ≤ m, g(₁, …, _n) ≠ 0. Hence by Theorem 7.14, there exist a_l, …, a_n ∈ F such that f_i(a₁, …, a_n) = 0, g(a₁, … , a_n) ≠ 0. Since the f_i generate I, we have f(a₁, …, a_n) = 0 for all f ∈ I. This contradicts the hypothesis and proves that g ∈ nilrad I.

Evidently the Nullstellensatz implies that if I₁ and I₂ are ideals in F[x₁, …, x_n] such that V(I₁) = V(I₂), then I₁ ⊂ nilrad I₂ so nilrad I₁⊂ nilrad (nilrad I₂) = nilrad I₂. By symmetry, nilrad I₂ ⊂ nilrad I₁ and so nilrad I₁ = nilrad I₂. An important special case of the Nullstellensatz is

THEOREM 7.17.   If I is a proper ideal in F[x₁, …, x_n], F algebraically closed, then V(I) ≠ Ø.

Proof.     If V(I) = Ø, then g = 1 satisfies the condition of the Nullstellensatz so 1 ∈ nilrad I and hence 1 ∈ I, contrary to the hypothesis that I is proper.

The Nullstellensatz permits us also to determine the maximal ideals in F[x₁, …, x_n]. If (a₁, …, a_n) ∈ F⁽ⁿ⁾, the ideal M_{a1, …, an} = (x₁ – a₁, …, x_n – a_n) is maximal in F[x₁, …, x_n] since F[x₁, …, x_n]/M_{a1, …, an} F. Evidently V(M_{a1, …, an}) consists of the single point (a₁, …, a_n). If I is any proper ideal, the foregoing result and the Nullstellensatz imply that I ⊂ nilrad M_{a1, …, an} for some point (a₁, …, a_n). Since M_{a1, …, an} is maximal, it coincides with its nilradical, so I ⊂ M_{a1, …, an}. Hence we see that the only maximal ideals of F[x₁, …, x_n] are those of the form M_{a1, …, an}. Moreover, since V(M_{a1, …, an}) = {(a₁, …, a_n)}, we have the following result.

COROLLARY.   If F is an algebraically closed field, then the map (a₁, …, a_n) M_a1, …, a_n = (x₁ – a₁, …, x_n – a_n) is a bijection of F⁽ⁿ⁾ onto the set of maximal ideals of F[x₁, …, x_n].

We now suppose that F is any infinite field and we consider again F⁽ⁿ⁾ and F[x₁, …, x_n], and define the variety V(S) for a subset S of F[x₁, …, x_n] as in the algebraically closed case. Also if V₁ is a variety, we let i(V₁) denote the ideal of polynomials f ∈ F[x₁, …, x_n], which vanish for every (a₁, …, a_n) ∈ V₁. The following properties follow directly from the definition:

(1) V(F[x₁, …, x_n]) = Ø,

(2) V(Ø) = F⁽ⁿ⁾,

(3) V(S_α) = V(S_α),

(4) V(S) = V(I) if I = (S), the ideal generated by S,

(5) V(I₁I₂) = V(I₁) ∪ V(I₂) for any ideals I₁ and I₂,

(6) V(S₁) ⊃ V(S₂) if S₁ ⊂ S₂,

(7) i(Ø) = F[x₁, …, x_n]

(8) i(V₁) ⊃ i(V₂) if V₁ ⊂ V₂,

(9) i(V(I₁)) ⊃ I₁ for any ideal I₁,

(10) V(i(V₁)) ⊃ V₁ for any variety V₁.

Relations (6), (8), (9), and (10) have the immediate consequences that

(11) V(i(V(I₁))) = V(I₁),

(12) i(V(i(V₁))) = i(V₁).

We now recall an important theorem, proved in BAI on p. 136, that if F is an infinite field and f(x₁, …, x_n) is a non-zero polynomial with coefficients in F, then there exist a_i ∈ F such that f(a₁, …, a_n) ≠ 0. Evidently this implies

(13) i(F⁽ⁿ⁾) = 0.

Relations (1)–(5) show that the set of varieties satisfy the axioms for closed subsets of a topological space. The resulting topology is called the Zariski topology of the space F⁽ⁿ⁾. The open subsets in this topology are the sets F⁽ⁿ⁾ – V, V a variety. If V = V(S), then F⁽ⁿ⁾ – V = ∪_{f ∈ S} (F⁽ⁿ⁾ – V({f})). Thus the open sets 0_f = (F⁽ⁿ⁾ – V({f})) form a base for the open subsets of F⁽ⁿ⁾. Evidently 0_f is the set of points (a₁, …, a_n) such that f(a₁, …, a_n) ≠ 0.

We now observe that if F is algebraically closed (hence infinite), then F⁽ⁿ⁾ with its Zariski topology is the same thing as the maximum spectrum of the ring F[x₁, …, x_n]. More precisely, we have a canonical homeomorphism between these two spaces. This is the map (a₁, …, a_n) M_{a1, …, an} given in the corollary above. Since the condition that f(a₁, …, a_n) = 0 for a polynomial f(x₁, …, x_n) is equivalent to f ∈ M_{a1, …, an}, the map we have defined induces a bijective map of the set of closed sets in F⁽ⁿ⁾ with the set of closed sets in Maxspec (see p. 403). Hence we have a homeomorphism.

When F = or , the Zariski topology on F⁽ⁿ⁾ has strikingly different properties from the usual Euclidean topology. Let us now consider some of these properties for F⁽ⁿ⁾, F any infinite field. Since any point (a₁, …, a_n) is the variety defined by the ideal M_{a1, …, an} = (x₁ – a₁, …, x_n – a_n), it is clear that F⁽ⁿ⁾ is a T₁-space. However, it is not a Hausdorff space. On the contrary, F⁽ⁿ⁾ is irreducible in the sense that any two non-vacuous open subsets of F⁽ⁿ⁾ meet. In other words, any non-vacuous open subset is dense in F⁽ⁿ⁾. Since the sets O_f constitute a base, it suffices to see that if O_f ≠ Ø and O_g ≠ Ø, then O_f ∩ O_g ≠ Ø. The theorem on non-vanishing of polynomials implies that O_f ≠ Ø if and only if f ≠ O. It is clear also that O_f ∩ O_g = O_fg. Since F[x₁, …, x_n] is a domain, it follows that O_f ≠ Ø, O_g ≠ Ø imply O_f ∩ O_g ≠ Ø.

We note next that the space F⁽ⁿ⁾ with the Zariski topology is noetherian in the sense that the ascending chain condition holds for open subsets of F⁽ⁿ⁾. Equivalently the descending chain condition holds for varieties. Suppose that V₁ ⊃ V₂ ⊃ … is a descending chain of varieties. Then we have the ascending chain of ideals i(V₁) ⊂ i(V₂) ⊂ …, so there exists an m such that i(V_j) = i(V_{j + 1}) for all j ≥ m. Since V_j = V(i(V_j)), we have V_j = V_{j + l} for j ≥ m.

Let us now look at the simplest case: n = 1. Here the variety V({f}) defined by a single polynomial f is a finite set f ≠ 0 and is the whole space F if f = 0. Moreover, given any finite set {a_i | 1 ≤ i ≤ r} we have V({f}) = {a_i} for f = ∏(x – a_i). It follows that the closed sets in the Zariski topology are the finite subsets (including the vacuous set) and the whole space. The open subsets of F are the vacuous set and the complements of finite sets. A subbase for the open sets is provided by the complements of single points, since any non-vacuous open set is a finite intersection of these sets.

It should be observed that the Zariski topology provides more open sets than one gets from the product topology obtained by regarding F⁽ⁿ⁾ as product of n copies of F. For example, the open subset of F⁽ⁿ⁾ defined by x₁ + … + x_n ≠ 0 cannot be obtained as a union of open subsets of the form O₁ × 0₂ × … × O_n, O_i open in F.

Any polynomial f(x₁, …, x_n) defines a polynomial function

of F⁽ⁿ⁾ into F. We have considered such functions in BAI, pp. 134–138. The theorem we quoted on non-vanishing of polynomials shows that the map sending the polynomial f(x₁, …, x_n) into the function defined by (30) is an isomorphism of F[x₁, …, x_n] onto the ring of polynomial functions. More generally, consider a second space F^(m). Then a sequence (f₁, …, f_m), f_i = f_i(x₁, …, x_n), defines a polynomial map

of F⁽ⁿ⁾ into F^(m) Such maps are continuous of F⁽ⁿ⁾ into F^(m), both endowed with the Zariski topology. To see this we take any neighborhood O_g of (f₁(a₁, …, a_n), …, f_m(a₁, …, a_n)). Then we have a polynomial g(x₁, …, x_m) such that g(f₁(a₁, …, a_n), …, f_m(a₁, …, a_n)) ≠ 0. Then h(x₁, …, x_n) = g(f₁(x₁, …, x_n), …, f_m(x₁, …, x_n) ≠ 0 in F[x₁, …, x_n] and O_h is mapped into O_g by (31). Thus (31) is continuous. We remark that since non-vacuous open subsets are dense, to prove that a polynomial map is 0 (that is, sends every element into 0), it suffices to show this for a non-vacuous open subset.

These rudimentary algebraic geometric ideas are often useful in “purely algebraic” situations. Some illustrations of the use of the Zariski topology will be given in the following exercises.

EXERCISES

In all of these we assume F algebraically closed (hence infinite). The topologies are Zariski.

        1. Let a ∈ M_n(F) and let f_a(λ) = λⁿ – tr(a) λ^{n – 1} + … +(– l)ⁿ det a be the characteristic polynomial of a. Show that the maps a tr a, …, a det a are polynomial functions on the n²-dimensional vector space. Show that the set of invertible matrices is an open subset of M_n(F).

        2. Let {ρ₁, …, ρ_n } be the characteristic roots of a (in some order) and let g(x₁, …, x_n) be a symmetric polynomial in the x_i with coefficients in F. Show that the map a g(ρ₁, …, ρ_n) is a polynomial function on M_n(F). Show that the set of matrices, which are similar to diagonal matrices with distinct diagonal entries, is an open subset of M_n(F).

        3. Let f, g ∈ F[x₁, …, x_n] and suppose that g(a₁, …, a_n) = 0 for every (a₁, …, a_n) such that f(a₁, …, a_n) = 0. Show that every prime factor of f is a factor of g.

        4. If a ∈ M_n(F), let U_a denote the linear transformation y aya of M_n(F) into itself. Let d(a) = det U_a. Use the Hilbert Nullstellensatz and Theorem 7.2 of BAI, p. 418, to prove that d(a) = (det a)²ⁿ.

        5. Give an alternative proof of the result in exercise 4 by noting that it suffices to prove the relation for a in the open subset of matrices that are similar to diagonal matrices with non-zero diagonal entries. Calculate d(a) and de a for a diagonal matrix.

        6. Show that the nil radical of any ideal in F[x₁, …, x_n] is the intersection of the maximal ideals containing it.

        7. Let V be a variety in F⁽ⁿ⁾. A polynomial function on V is defined to be a map p|V where p is a polynomial function on F⁽ⁿ⁾. These form an algebra A(V) over F under the usual compositions of functions. Show that A(V) F[x₁, …, x_n]/i(V) = F[₁, …, _n], _i = x_i + i(V). The latter algebra over F is called the coordinate algebra of the variety V. V is called irreducible if i(V) is prime. In this case the elements of the field of fractions of the coordinate algebra are called rational functions on V. If V′ is a variety in F^(m), a map of V into V′ is called regular if it has the form g|V where g is a polynomial map of F⁽ⁿ⁾ into F^(m). Show that if p′ is a polynomial function on V′, then p′g is a polynomial function on V and the map η(g) : p′ p is an algebra homomorphism of A(V′) into A(V). Show that g η(g) is a bijection of the set of regular maps of V into V′ and the set of algebra homomorphisms of A(V′) into A(V).

7.13   PRIMARY DECOMPOSITIONS

In this section we consider the classical Lasker-Noether decomposition theorems for ideals in noetherian rings as finite intersections of primary ideals. It is easy to see by using the ascending chain condition that any proper ideal I in a noetherian ring R can be written as an intersection of ideals that are indecomposable in the sense that I ≠ I₁ ∩ I₂ for any two ideals I_j ≠ I. Moreover, in any noetherian ring, indecomposable ideals are primary in a sense that we shall define in a moment. This type of decomposition is a weak analogue of the decomposition of an element as a product of prime powers. Associated with every primary ideal is a uniquely determined prime ideal. However, primary ideals need not be prime powers and not every prime power is primary. Although the decomposition into primary ideals is not in general unique, it does have some important uniqueness properties. The establishment of these as well as the existence of the primary decomposition are the main results of the Lasker-Noether theory. We shall begin with the uniqueness questions, since these do not involve the noetherian property. The classical results can be generalized to modules and the passage to modules makes the arguments somewhat more transparent. In our discussion we shall consider the general case of modules first and then specialize these to obtain the results on ideals.

If M is a module for R and a ∈ R, then as in section 3.1, we denote the map x ax, x ∈ M, by a_M. Since R is commutative, this is an endomorphism of M as R-module. We have the homomorphism ρ_M : a a_M of R into the ring of endomorphisms of M and ker ρ_M is the set of a ∈ R such that ax = 0 for all x. As usual we write ann_Rx for the ideal of elements b ∈ R such that bx = 0. Evidently ker ρ_M = ∩_{x ∈ M} ann_Rx = ann_RM. If ax = 0 for some x ≠ 0, then we shall call this element a of the ring R a zero divisor of the module M. The elements a that are not zero divisors are those for which a_M is injective. We now look at the nil radical of the ideals ker ρ_M and ann_Rx for a particular x. By definition, the first of these is the set of elements a ∈ R for which there exists an integer m such that a^m ∈ ker ρ_M, that is, a^mx = 0 for every x or, equivalently, a_M^m = (a^m)_M = 0. Likewise a ∈ nilrad (ann_Rx) if and only if there exists an m such that a^mx = 0 or, equivalently, the restriction of a_M to the submodule Rx is nilpotent.

The case of primary interest is that in which M = R/I where I is an ideal in R. Since we shall need to consider simultaneously such a module and the modules R/I′ for I′ an ideal of R containing I and since , we need to formulate our definitions and results in terms of a module and a submodule. The style is set in the following

DEFINITION 7.4.   A submodule Q of an R-module M is called primary if Q ≠ M and for every a ∈ R either a_M/Q is injective or it is nilpotent.

Evidently this means that if a is a zero divisor of M/Q, then there exists an integer m such that a^mx ∈ Q for every x ∈ M. The set P of elements satisfying this condition is the ideal nilrad (ann_R M/Q). This ideal is prime since if a P and b P, then a_M/Q and b_M/Q are injective. Then (ab)_M/Q = a_M/Q b_M/Q is injective and hence ab P. We shall call P the prime ideal associated with the primary submodule Q. If M = R, the submodules are the ideals. Then the condition a^mx ∈ Q for every x ∈ R is equivalent to a^m ∈ Q. Thus in this case an ideal Q is primary if and only if ab ∈ Q, and b Q implies that there exists an m such that a^m ∈ Q. The associated prime of Q is the nilradical of this ideal.

EXAMPLES

1. Let R be a p.i.d. and Q = (p^e) = (p^e), p a prime. It is readily seen that if a ∈ R and a P = (p), then a_R/Q is invertible, hence injective. On the other hand, if a ∈ P, then a^e_R/Q = 0. Hence Q is primary and P is the associated prime ideal. It is easy to see also that 0 and the ideals (p^e) are the only primary ideals in R.

2. Let R = and let M be a finite abelian group written additively and regarded as a -module in the usual way. Suppose every element of M has order a power of a prime p. Then 0 is a p-primary submodule of M since if a ∈ and (a, p) = 1, then a_M is injective and if a is divisible by p, then a_M is nilpotent. The associated prime ideal of 0 is (P).

3. Let R = F[x, y, z]/(xy – z²) where F is a field and x, y , z are indeterminates. If a ∈ F[x, y, z], let be its image in R under the canonical homomorphism. Let P = (, ), the ideal in R generated by , . Then the corresponding ideal in F[ x, y, z] is (x, z), since this contains xy – z². Since F[x, v, z]/(x, z) F [y], it follows that P is prime in R. Since it is clear from Krull’s theorem that the nilradical of any power of a prime ideal is this prime, the ideal P² of R has P as its nilradical. However, this is not primary since P², P but .

We can obtain many examples of primary ideals by using the following

PROPOSITION 7.24.   If P is a maximal ideal in R, any ideal Q between P and P^e, e ≥ 1, is primary with P as associated prime ideal.

Proof.     If a ∈ P, then a^e ∈ P^e ⊂ Q. If a P, then we claim that a + Q is a unit in R/Q. For, since P is maximal, R/P is a field. Hence we have an element a′ ∈ R and a z ∈ P such that aa′ = 1 – z. Since z + Q is nilpotent in R /Q, (1 – z) + Q is a unit in R/Q and hence a + Q is a unit in R/Q. It is now clear that if ab ∈ Q, then b ∈ Q. Thus we have shown that if a ∈ P, then a_R/Q is nilpotent and if a P, then a_R/Q is injective. Then it is clear that Q is primary and P is the associated prime.

This result can be used to construct examples of primary ideals that are not prime powers. For instance, let P = (x, y) in F[x, y]. This is maximal, since F[x, y]/P F. We have . Hence (x², y) is a primary ideal that is not a prime power.

It is clear that if Q is a submodule of M and P is a subset of R such that 1) if a ∈ P, then there exists an integer m such that a^mx ∈ Q for all x ∈ M and 2) if a ∈ R – P, then a_M/Q is injective, then Q is primary with P as associated ideal. We use this to prove

PROPOSITION 7.25.   The intersection of a finite number of primary submodules of M having the same associated prime ideal P is primary with P as associated prime ideal.

Proof.     It suffices to prove this for two submodules Q₁ and Q₂. Then Q₁ ∩ Q₂ is a proper submodule since the Q_i are proper. Let a ∈ P. Then we have an integer m_i such that for all x ∈ M. If we take m = max(m₁, m₂), then a^mx ∈ Q = Q₁ ∩ Q₂. Next let a ∈ R – P and let x ∈ M – Q. We may assume that x Q₁. Then ax Q₁ so ax Q. Thus a_M/Q is injective. Hence Q is primary with P as associated prime ideal.

We have noted that for any x the set of a ∈ R such that there exists an integer m such that a^mx = 0 is the ideal nilrad(ann x). Hence if Q is a submodule and = x + Q in M/Q, then the set of a for which there exists an m such that a^mx ∈ Q is the ideal nilrad(ann_R). If Q is primary, we have the following important result.

PROPOSITION 7.26.   Let Q be a primary submodule of M, P the associated prime, and let x ∈ M. Let I_x = {a|a^mx ∈ Q for some m}. Then I_x = P if x Q and I_x = R if x ∈ Q.

Proof.   Let x Q. Evidently I_x ⊃ P, since a_M/Q is nilpotent for every a ∈ P. On the other hand, if a P, then a_M/Q is injective. This precludes a^mx ∈ Q for any m. Hence I_x = P. The other assertion that if x ∈ Q, then I_x = R is clear.

Now let N be a finite intersection of primary submodules Q_i of M: N = Q_i. It may happen that some of these are redundant, that is, that N is an intersection of a proper subset of the Q_i. In this case we can drop some of these until we achieve an irredundant decomposition N = Q_i, which means that for every j, N ≠ _{i ≠ j} Q_i. This is equivalent to assuming that for every j,