10. Primary ideals and the radical.

∗ ∗ ∗ Multiplicatively closed setes. ∗ ∗ ∗

10.1 Definition. A non-empty subset S of a commutative ring R is called multiplicatively closed, or multiplicative, if s, t ∈ S implies st ∈ S.

A multiplicatively closed set S is called saturated if it contains every divisor of each of its elements, i.e., if s = ab ∈ S implies a ∈ S.

The set S̄ = {t ∈ R | ∃s ∈ S t | s} of divisors of elements of a multiplicatve set S – clearly the smallest saturated multiplcative set containing S – is called the saturation of S.

Examples of multiplicatively closed sets (1–3 are saturated):

1) R \ P, where P is a prime ideal of R,

2) R \ ⋃_{α∈A} P_{α}, where all P_{α} are prime ideals of R

3) the set of all non-zerodivisors in R

4) the set of powers of a fixed element {r^{n} | n ∈ ℕ_{0}}.

10.2 Lemma. Let S be a multiplicatively closed subset of R and I an ideal of R with I ∩ S = ∅. Then there exists a prime ideal P of R with I ⊆ P and P ∩ S = ∅.

Proof. Apply Zorn’s Lemma to J = {J ⊴ R | I ⊆ J and S ∩ J = ∅}. Since I ∈ J, J ≠ ∅. Every chain of ideals in J has an upper bound in J: its union is in J. Let P be a maximal element of J. We will show that P is prime. S ≠ ∅ implies P ≠ R. Suppose ab ∈ P, a ∉ P and b ∉ P. By maximailty of P there exists s ∈ S ∩ (P + Ra) and t ∈ S ∩ (P + Rb). But then st ∈ S ∩ P, a contradiction. □

10.3 Corollary.

(i) Let S be a saturated multiplicatively closed subset of R. Then S =R\⋃_{P ∈P} P, where P is the set of all prime ideals P of R with S ∩ P = ∅.

(ii) The saturated multiplicative subsets of R are precisely the complements of unions of prime ideals. (The trivial case S = R corresponds to the empty union of prime ideals).

Proof. Ad (i). Clearly, S ⊆ R \ ⋃_{P ∈P} P. For the reverse inclusion, we show that for every u ∉ S there exists some prime ideal P with S ∩ P = ∅ and u ∈ P. S being saturated, u ∉ S implies Ru ∩ S = ∅. By Lemma 10.2 there exists a prime ideal P with Ru ⊆ P and S ∩ P = ∅. Since the complement of a union of prime ideals is a saturated multiplicateve set, (ii) follows. □

10.4 Corollary. The set Z(R) of zero-divisors of a commutative ring R is a union of prime ideals of R. More generally, the set Z(M) of zero-divisors of any R-module M is a union of prime ideals of R.

Proof. Verify that R \ Z(R), and more generally, R \ Z(M), is a saturated multplicative set. □

10.5 Lemma. Let I be an ideal of R and P a prime ideal containing I. Then there exists a minimal prime P′ of I with I ⊆ P′ ⊆ P.

Proof. Let S = R \ P and consider the set S of multiplicatively closed sets T with S ⊆ T and T ∩I = ∅. Then S is non-empty because S ∈ S. Every chain of elements of S has an upper bound in S (its union). Let S_{0} be a maximal element of S (which exists by Zorn’s Lemma), and set U = R \ S_{0}. By Lemma 10.2, there exists a prime ideal P′ containing I and disjoint from S_{0}, i.e., I ⊆ P′ ⊆ U. It follows that P′ = U and P′ is a minimal prime of I, since for any prime ideal Q with I ⊆ Q ⫋ U, its complement R \ Q would be an element of S stricltly containing S_{0}. □

∗ ∗ ∗ The radical. ∗ ∗ ∗

10.6 Definition. Let I be an ideal of R. The radical of I is defined by

\sqrt{I} = \{r ∈ R | ∃n ∈ ℕ r^{n} ∈ I\}.

Convince yourself of the following easy facts:

10.7 Remark:

(1) I ⊆ \sqrt{I}

(2) \sqrt{\sqrt{I}} = \sqrt{I}

(3) \sqrt{I} = R ⟺ I = R

(4) If Q is an intersection of prime ideals then \sqrt{Q} = Q.

Recall that an element r ∈ R is called nilpotent if there exists a natural number n with r^{n} = 0.

10.8 Definition. The radical of (0), consisting of all nilpotent elements of R, is called the nilradical of R:

Nil(R) = {r ∈ R | ∃n ∈ ℕ r^{n} = 0}.

10.9 Proposition.

\sqrt{I} = ⋂ ; {P prime \atop I ⊆ P} P.

Proof. Clearly, \sqrt{I} is contained in all prime ideals containing I. Conversely, if r ∉ \sqrt{I} , then S = \{r^{n} | n ∈ ℕ\} is a multiplicatively closed set with S ∩ I = ∅ and by Lemma 10.2, there exists a prime ideal P with I ⊆ P and r ∉ P. □

10.10 Corollary. The nilradical of R is the intersection of all prime ideals of R.

10.11 Lemma. Some rules of arithmetic for radicals:

(1) \sqrt{IJ} = \sqrt{I ∩ J} = \sqrt{I} ∩ \sqrt{J}

(2) \sqrt{I + J} = \sqrt{\sqrt{I} + \sqrt{J}}

(3) \sqrt{I} + \sqrt{J} = R ⟺ I + J = R

Proof. Exercise. □

10.12 Lemma. Let P be a prime ideal of R and I an ideal of R.

(i) P^{n} ⊆ I ⊆ P for some n ∈ ℕ implies \sqrt{I} = P. (ii) If R is Noetherian, then \sqrt{I} = P implies for some n P^{n} ⊆ I ⊆ P.

Proof. (i) is easy. Ad (ii), P = (p_{1}, …, p_{m}) = \sqrt{I} implies for some k ∈ ℕ, p_{1}^{k}, …, p_{m}^{k} ∈ I. Every element q of P is an R-linear combination of the p_{i}. By the multinomial theorem, (r_{1}p_{1} + … + r_{m}p_{m})^{m}k is a sum of terms each of which is divisible by p^{k} for some i, therefore q^{n} ∈ I for every q ∈ P, n = km. □

∗ ∗ ∗ Primary ideals ∗ ∗ ∗

10.13 Definition. An ideal Q of R is called primary iff Q ≠ R and for all a, b ∈ R with ab ∈ Q either a ∈ Q or ∃n ∈ ℕ b^{n} ∈ Q.

The definition of primary tends to be confusing at first. If Q is primary, then ab ∈ Q implies a ∈ Q or b ∈ \sqrt{Q} , as well as, by symmetry, b ∈ Q or a ∈ \sqrt{Q} . Therefore, for an ideal Q ≠ R, being primary is equivalent to

ab ∈ Q ⇒ a ∈ Q ∨ b ∈ Q ∨ a, b ∈ \sqrt{Q}

10.14 Easy exercise. If Q is a primary ideal then \sqrt{Q} is a prime ideal.

10.15 Definition. A primary ideal Q with \sqrt{Q} = P is called P-primary.

Not every ideal whose radical is prime is primary, however. For instance, \sqrt{P^{n}} = P for every prime ideal P and n ∈ ℕ, but not every power of a prime ideal is primary. In a more positive vein, we have the following facts:

10.16 Lemma. Let M be a maximal ideal of R and Q an ideal with \sqrt{Q} = M. Then Q is M -primary.

Proof. Suppose ab ∈ Q and a ∉ M; we must show b ∈ Q. We have M + aR = R, i.e., \sqrt{Q} + aR = R. This implies Q + aR = R, so 1 = q + ar for some q ∈ Q and r ∈ R, and finally b = qb + abr ∈ Q. □

10.17 Corollary. Let M be a maximal ideal of R and I ≠ R an ideal with M^{n} ⊆ I for some n ∈ ℕ. Then I is M -primary.

Proof. I is not contained in any prime ideal P other than M: if I ⊆ P then for every m ∈ M, m^{n} ∈ I ⊆ P implies m ∈ P, hence M = P. Since I ≠ R, this shows \sqrt{I} = M and we are done. □

Just like prime ideals, primary ideals behave nicely with respect to ring homomorphisms:

10.18 Lemma. Let f: R → T be a ring homomorphism (with f(1) = 1, as always) and I, Q ideals of T. Then

(1) f^{−1}(\sqrt{I} ) = \sqrt{f^{−1}(I)} .

(2) If Q is P -primary then f^{−1}(Q) is f^{−1}(P)-primary.

Again, like prime ideals, primary ideals can be characterized by a property of their quotient ring.

10.19 Proposition. An ideal Q is primary if and only if R/Q is a non-zero ring in which every zero-divisor is nilpotent.

Proof. Easy. □

Together with the previous lemma this implies that a canonical projection induces a bijection between primary ideals containing the kernel and primary ideals in the quotient ring. (Note that everything we show for canonical projections holds mutatis mutandis for any surjective ring homomorphism f: R → T, since f is just the canonical projection π: R → R/Kerf followed by the isomorphism f̄: R/Kerf → T.)

10.20 Proposition. Let I be an ideal of R and π: R → R/I the canonical projection. Then a bijection between all primary ideals Q of R with I ⊆ Q and all primary ideals of R/I is given by Q ↦ π(Q). Its inverse is Q̃ ↦ π^{−1}(Q̃) (for Q̃ ⊴ R/I).

Primary ideals also behave just like prime ideals with respect to the canonical homomorphism into a ring of fractions f_{S}: R → R_{S} and with respect to the canonical projection onto a residue class ring:

10.21 Proposition. Let R_{S} be a ring of fractions of R and f_{S}: R → R_{S} the canonical homomorphism.

(1) If Q ⊴ R is primary and \frac{r}{s} ∈ Q_{S} then r ∈ Q.

(2) A bijection between the primary ideals Q of R with S ∩ Q = ∅ and all primary ideals of R_{S} is given by Q ↦ Q_{S} (for Q a primary ideal of R). Its inverse is Q̃ ↦ f_{S}^{−1}Q̃ (for Q̃ a primary ideal of R_{S}).

This is shown just like the analogous statement for prime ideals. (Note that

Q ∩ S = ∅ implies \sqrt{Q} ∩ S = ∅.)