(c)Karen E. Smith 2019 UM Math Dept licensed under a Creative Commons By-NC-SA 4.0 International License. Worksheet on Primary Decomposition: Uniqueness Let R be a commutative ring with 1.

Definition. A proper ideal q is primary if every zero-divisor in R/q is nilpotent.

Terminology. A primary ideal q always has prime radical p; we say q is p-primary.

Theorem on Uniqueness of Primary Decomposition. Suppose an ideal J in an arbitrary ring admits a primary decomposition

J = q1 ∩ q2 ∩ · · · ∩ qt. Then J admits a minimal primary decomposition, meaning that the intersection can be assumed irredundant and that the qi are pi-primary for distinct primes pi. In this case, the set √ √ √ { q1, q2,..., qt} = {p1, p2,..., pt} is independent of the choice of minimal primary decomposition. Furthermore, the minimal primes among the set {p1, p2,..., pt} are precisely the minimal primes of J, and for these minimal primes, the corresponding primary component is uniquely determined by qi = JRpi ∩ R.

Noetherian Case: Every ideal in a Noetherian ring admits a minimal primary decomposition J = q1 ∩ q2 ∩ · · · ∩ qt. In this case, the set {p1, p2,..., pt} = Ass(R/J).

Caution: In the non-Noetherian case, primary decompositions do not always exist for a given J, and the radicals of the primary components do not have to be associated primes.

(1) Lemma 1. Prove that finite intersection commutes with taking radicals, localization, and computing colons. That is, for any finite set of ideals J1,...,Jt in an arbitrary ring R, prove that p √ √ √ (a) (J1 ∩ J2 · · · ∩ Jt) = J1 ∩ J2 · · · ∩ Jt; −1 −1 −1 −1 (b)( J1 ∩ J2 · · · ∩ Jt)U R = J1U R ∩ J2U R · · · ∩ JtU R for any multiplicative set U ⊂ R; and (c)( J1 ∩ J2 ∩ · · · ∩ Jt): x = (J1 : x) ∩ (J2 : x) ∩ · · · ∩ (Jt : x) for arbitrary x ∈ R.

(2) Lemma 2. Prove that in any ring, if P1 ∩ P2 ∩ · · · ∩ Pn is an intersection of mutually incomparable prime ideals, then minimal primes of this intersection are precisely the Pi. In particular, when is such an intersection prime? [Hint: P1P2 ··· Pn ⊂ P1 ∩ P2 ∩ · · · ∩ Pn.]

(3) Minimal primes. Let J be an ideal in an arbitrary ring which admits a primary decom- T T position q1 q2 ··· qt. (a) By grouping together primary ideals with the same radical, explain why we can assume the qi have√ distinct radicals pi. [Hint: A finite intersection of p-primary ideals is p-primary.] Tt (b) Prove that J = i=1 pi. (c) Prove the minimal primes among {p1, p2,..., pt} are precisely the min primes of J. (d) Observe that (a) and (c) together establish part of the Theorem on Uniqueness of Primary decomposition. 2

(4) Lemma 3. Let q be any p-primary ideal in an arbitrary ring R. p (a) Prove that p ⊂ (q : x). [Hint: First show q ⊂ (q : x).] (b) Prove that ( p, for x∈ / q p(q : x) = R for x ∈ q. (c) Prove that for x∈ / p, we have (q : x) = q. T T (5) Uniqueness of the pi. Let J = q1 q2 ··· qt be any primary decomposition of J in which the radicals pi of qi are distinct. (a) Use Lemmas 1 and 3 to show that for any x ∈ R, √ \ J : x = pi.

x6∈qi

(b) Fix i. Explain why, if the decomposition is irredundent, we can find x in every qj except qi. √ (c) With x as in (b), show that J : x = pi. 0 0 0 (d) Show that if q1 ∩ q2 ∩ · · · ∩ qt and q1 ∩ q2 ∩ · · · ∩ qm are two different minimal primary decompositions of an ideal J, then √ √ √ p 0 p 0 p 0 { q1, q2,..., qt} = { q1, q2,..., qm}. In particular t = m. [Hint: Use (a), (c) and Lemma 2.] (e) Conclude that part of the Theorem on Uniqueness of Primary decomposition is proven.

(6) Minimal components. Let J = q1 ∩ q2 ∩ · · · ∩ qt be a primary decomposition. (a) Let q be p-primary and P be any prime ideal such that p 6⊂ P . Show that qR = R . √ √ √ P P (b) Suppose that pi is minimal among { q1, q2,..., qt}. Prove that JRpi ∩ R = qi. [Hint: Use Lemma 1(b) and (a).] (c) Complete the proof of the Uniqueness Theorem for Primary Decomposition.

(7) Let R be the ring of all sequences of real numbers. Show that R has infinitely many minimal prime ideals, and therefore the zero ideal does not admit a primary decomposition.

(8) Lemmas on Associated Primes. Let M and N be arbitrary R-modules. (a) Show Ass(M ⊕ N) = Ass(M) ∪ Ass(N). [Hint: Consider 0 → M → M ⊕ N → N → 0.]

(b) Show that Ass(R/J1 ∩ J2) ⊂ Ass(R/J1) ∪ Ass(R/J2). [Hint: Find R/J ,→ R/J1 ⊕ R/J2.] (c) For all multiplicative sets U ⊂ R,, we have the following inclusion of sets in Spec U −1R: {PU −1R | P ∈ Ass(M) and P ∩ U = ∅} ⊂ Ass(U −1M). (d) * Prove the converse to (c) when R is Noetherian. (e) * For R Noetherian, show that every minimal prime of J is in Ass(R/J).

(9) Assume that R is Noetherian. Let J = q1 ∩ · · · ∩ qt be a minimal primary decomposition. (a) Prove that if q is p-primary, then Ass(R/q) = {p}. [Hint: If Q ∈ Ass(R/q), then Q = (q : x) √ for some x∈ / q. Show that q ⊂√Q ⊂ q.]√ (b) Show that Ass(R/J) ⊂ { q1,..., qt}. Thus all associated primes contribute to the primary components. √ (c) * Show that every q is in Ass(R/J). [Hint: Take x ∈ q for all i > 1 but not q . Then √ i i 1 p1 = J : x. So p1 is the only minimal prime of J : x and hence an associated prime of J : x. Note

that R/p1 ,→ R/(J : x) ,→ R/J.]