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 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.

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-zero-divisors 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 Definition. If S ⊆ R is a multiplicatively closed set then its saturation S̄ is the set of all divisors of elements of S: S̄ = {t ∈ R | ∃s ∈ S t | s}.

It is easy to see that the saturation of a multiplicatively closed set S is a saturated multiplicatively closed set that contains S.

10.4 Corollary. 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 = ∅.

Proof. 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. By the saturation of S, u ∉ S implies Ru ∩ S = ∅. By Lemma 10.2 there exists a prime ideal P with Ru ⊆ P and S ∩ P = ∅. □

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 and 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.

10.8 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. □

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

10.9 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}.

10.10 Corollary of 10.8. 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. Note that for a primary ideal Q, ab ∈ Q implies a ∈ Q or b ∈ \sqrt{Q} as well as b ∈ Q or a ∈ \sqrt{Q} . (R is commutative after all). Therefore, for an ideal Q ≠ R, being primary is equivalent to ab ∈ Q implies a ∈ Q or b ∈ Q or, if neither a nor b is in Q then a and b are

both in \sqrt{Q} .

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

Proof. Easy. □

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, particularly surjective ones:

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 primary then f^{−1}(Q) is 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 → S^{−1}R and with respect to the canonical projection onto a residue class ring:

10.21 Proposition. Let S^{−1}R be a ring of fractions of R and f_{S}: R → S^{−1}R the canonical homomorphism.

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

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

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

Q ∩ S = ∅ implies \sqrt{Q} ∩ S = ∅.)

∗ ∗ ∗ Symbolic powers ∗ ∗ ∗

10.22 Definition. Let P be a prime ideal of R. The n-th symbolic power of P is defined by P^{(n)} = R ∩ P^{n}R_{P}.

The n-th symbolic power of P is the contraction back to R of the extension of P^{n} into R_{P} and thus contains P^{n}.

10.23 Proposition. Let P be a prime ideal of R and n ∈ ℕ.

(1) P^{(n)} is P -primary.

(2) If Q is a primary ideal then P^{n} ⊆ Q ⟹ P^{(n)} ⊆ Q.

(3) P^{(n)} = P^{n} ⟺ P^{n} is primary.

10.24 Corollary. If M is a maximal ideal of R then M^{(n)} = M^{n} for all n ∈ ℕ.

∗ ∗ ∗ Primary decompostion of ideals in Noetherian rings ∗ ∗ ∗

10.25 Definition. An ideal I ≠ R of a commutative ring R is called irreducible if there do not exist ideals A, B of R, each properly containing I, such that I =A∩B.

Note that prime ideals are certainly irreducible: Suppose P = A ∩ B. If P is prime, then AB ⊆ A ∩ B ⊆ P implies A ⊆ P or B ⊆ P.

It is easy to see that I ≠ R being irreducible is equivalent to: there do not exist finitely many ideals A_{i} with A_{i} ⊃ I and I = A_{1} ∩ … ∩ A_{n}. It is however possible for an irreducible ideal to be the intersection of an infinite collection of ideals each properly containing it (just consider (0) in ℤ).

10.26 Lemma. Let R be a Noetherian ring. Then every proper ideal of R can be expressed as a finite intersection of irreducible ideals.

Proof. Let S be the set of proper ideals of R that are not finite intersections of irreducible ideals. If S ≠ then S contains a maximal ideal I. As I is certainly not irreducible, there exist ideals A, B, each properly containing I with I = A ∩ B. By maximalty of I, A, B ∉ S. Therfore A and B are each expressible as a finite intersection of irreducible ideals, and so is I, a contradiction. □

10.27 Lemma. Let R be a Noetherian ring. Then every irreducible ideal of R is primary.

Proof. Let I be irreducible and ab ∈ R. To show: b ∈ I or there exists n ∈ ℕ with a^{n} ∈ I.

Consider the ascending chain of ideals of R:

I ⊆ (I :_{R} a) ⊆ (I :_{R} a^{2}) ⊆ … ⊆ (I :_{R} a^{n}) ⊆ (I :_{R} a^{n+1}) ⊆ …,

Let N ∈ ℕ be such that (I :_{R} a^{n}) = (I :_{R} a^{N}) for all n ≥ N. We will show that I = (I + (b)) ∩ (I + (a^{N})). Once this is established, we are done, because irreducibility of I then implies either I + (b) = I (and hence b ∈ I) or I = I + (a^{N}) (and hence a^{N} ∈ I).

Clearly, I ⊆ (I + (b)) ∩ (I + (a^{N})). Now let r ∈ (I + (b)) ∩ (I + (a^{N})). Then r = i + sb = j + ta^{N} for some i, j ∈ I and s, t ∈ R. Multiplying by a we get ai − aj + sab = ta^{N+1}. The left hand side is in I, therefore t ∈ (I : a^{N+1}). By the choice of N, (I : a^{N=1}) = (I : a^{N}), and so ta^{N} ∈ I. This implies r = j + ta^{N} ∈ I. □

10.28 Corollary. Let R be a Noetherian ring. Then every proper ideal of R can be expressed as a finite intersection of primary ideals.

10.29 Definition. Let I be an ideal. A finite list of primary ideals Q_{1}, …, Q_{m} such that I = Q_{1} ∩ … ∩ Q_{m} is called a primary decomposition of I.

A primary decomposition of I is called reduced if

(1) there doesn’t exist j such that Q_{j} ⊇ ⋂_{i≠j} Q_{i} and

(2) \sqrt{Q_{i}} ≠ \sqrt{Q_{j}} for i ≠ j.

10.30 Lemma. If Q_{1} and Q_{2} are P -primary ideals then Q_{1} ∩ Q_{2} is again P -primary.

Proof. Exercise. □

10.31 Corollary. Any ideal that has a primary decomposition has a reduced primary decomposition.

Proof. Given a primary decomposition I = Q_{1} ∩ … ∩ Q_{m} we can create a reduced primary decomposition by grouping together primary ideals with the same radical and taking their intersection (applying 10.30) and then deleting one by one those primary ideals that contain the intersection of the remaining ones (as long as any such redundant primary ideals are left). Since our list of Q_{i} is finite, this process terminates with a reduced decomposition. □

10.32 Remark. By 10.26, 10.27 and 10.31, every proper ideal in a Noetherian ring has a reduced primary decomposition. This is in general no longer a decomposition into irreducible ideals. (We have taken the intersections of irreducible components sharing the same radical.)

10.33 Lemma. Let Q be a P -primary ideal of R, and r ∈ R.

(1) If r ∉ Q then (Q: r) is P -primary.

(2) If r ∉ P then (Q: r) = Q. Proof. Easy. □

10.34 Lemma. Suppose I = Q_{1} ∩ … ∩ Q_{m} is a reduced primary decompostion

with \sqrt{Q_{i}} = P_{i} for 1 ≤ i ≤ m.

For any prime ideal P of R, the following are equivalent

(1) P = P_{i} for some i.

(2) There exists r ∈ R such that (I : r) is P -primary.

(3) There exists r ∈ R such that \sqrt{(I : r)} = P. Furthermore, if R is Noetherian, then the following is equivalent to the previous

three conditions:

(4) There exists r ∈ R such that (I : r) = P. Proof. (1 ⇒ 2) Suppose P = P_{i} and let a_{i} ∈ (⋂_{j≠i} Q_{i}) \ Q_{i} (which exists because the primary decomposition is reduced). Then

(I : a_{i}) = ( ⋂_{j=1}^{m} Q_{j} : a_{j}) = ⋂_{j=1}^{m} (Q_{j} : a_{j}) = (Q_{i} : a_{i}),

which is P_{i}-primary, i.e. P -primary, by 10.33.

(2 ⇒ 3) is trivial; (3 ⇒ 1)

P = \sqrt{(I : r)} = ⋂_{j=1}^{m} \sqrt{(Q_{j} : r)} = ⋂_{{j \atop a∉Q_{j}}} P_{j}

P is equal to a finite intersection of prime ideals (necessarily there exists some j such that a ∉ Q_{j}, otherwise P = R), therefore P coincides with one of the P_{j}.

(4 ⇒ 2) is trivial; for Noetherian R we show (2 ⇒ 4): The set of P -primary ideals of the form (I : r) contains a maximal element (I : s). We claim that (I : s) = P. It suffices to show that (I : s) is prime. Let ab ∈ (I : s), and suppose a, b ∉ (I : s). Then (I : s) is properly contained in (I : as) (witness b), and (I : as) is a proper ideal of R (witness a).

The radical of (I : as) is P: if c^{n} ∈ (I : as) then ac^{n} ∈ (I : s) while a ∉ (I : s) implies c^{n} ∈ P, and further c ∈ P. Therefore \sqrt{(I : as)} ⊆ P and the reverse inclusion follows from (I : s) ⊆ (I : as).

Let cd ∈ (I : as), c ∉ (I : as). Then cda ∈ (I : s), while ca ∉ (I : s) implies d ∈ \sqrt{(I : s)} = P = \sqrt{(I : as)} . Therefore (I : as) is P -primary, a contradiction to the maximality of (I : s). □

10.35 Theorem. Suppose I = Q_{1} ∩ … ∩ Q_{m} is a reduced primary decompostion, and P_{i} = \sqrt{Q_{i}} , for 1 ≤ i ≤ m. Then for every reduced primary decomposition of I, the same set of prime ideals {P_{1}, …, P_{m}} occurs as the set of radicals of the primary components.

Proof. By 10.34, the criterion for P to occur among the radicals of the Q_{i} does not depend on the primary decomposition but only on I. □

10.36 Theorem. Suppose I = Q_{1} ∩ … ∩ Q_{m} is a reduced primary decompostion, and P_{i} = \sqrt{Q_{i}} , where P_{1}, …, P_{k} are the isolated primes and P_{k+1}, …, P_{m} the embedded primes of I. Given any other reduced primary decompostion

I = Q′_{1} ∩ … ∩ Q′_{m} with P_{i} = \sqrt{Q′_{i}} , then Q′_{i} = Q_{i} for 1 ≤ i ≤ k.

Proof. Given an isolated prime P_{i}, let r ∈(⋂_{j≠i} P_{j})\P_{i}. Such an r exists, for, if P_{i} contained ⋂_{j≠i} P_{j}, it would contain some P_{j} with j ≠ i, contrary to assumption. Let n ∈ ℕ such that r^{n} ∈ ⋂_{j≠i} Q_{j} as well as r^{n} ∈ ⋂_{j≠i} Q′_{j}. By 10.33,

(I : r^{n}) = ⋂_{j=1}^{m}(Q_{j} : r^{n}) = (Q_{i} : r^{n}) = Q_{i}

and by the same token, (I : r^{n}) = Q′_{i}. □