4. Primary Decomposition

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MATH 205B NOTES 2010 COMMUTATIVE ALGEBRA 25

4. Primary decomposition This chapter of Atiyah-MacDonald is different from other treatments of primary decomposition because it does not assume that the ring A is Noetherian. So, the definition of an associated prime is different. Definition 4.1. An ideal q in A is called primary if every zero divisor in A/q is nilpotent. This is equivalent to saying: xy q,x / q ym q for some m 1 ∈ ∈ ⇒ ∈ ≥ In other words, y r(q). ∈ Note that prime ideals are primary. Proposition 4.2. q primary implies r(q) is prime. Proof. If xy r(q) then xnyn q for some n.Ifx/r(q) then xn / q. So, (yn)m q∈which implies y ∈ r(q). So, r(q) is prime.∈ ∈ ∈ ∈ We say that q is p-primary if r(q)=p. Proposition 4.3. p/q contains all zero divisors of A/q. 4.1. primary decomposition. Definition 4.4. A primary decomposition of an ideal a is defined to be an expression of the form: n

a = qi i=1 where qi are primary and n is ﬁnite. If n is minimal, this is called a minimal primary decomposition.

Let pi =r(qi). The following lemma shows that, in a minimal primary decomposition, the pi are distinct.

Lemma 4.5. If q1, q2 are p-primary (for the same p) then q1 q2 is also p-primary. ∩

Proof. Suppose that xy q1 q2 but x/q1 q2. Then either x/q1 ∈ ∩ ∈ ∩ n m ∈ or x/q2. In either case we get y p. Then y q1 and y q2. So, ynm∈ q q making this primary.∈ And it is easy∈ to see that∈ the ∈ 1 ∩ 2 radical is p. Example 4.6. Here are two examples of primary decompositions. 2 (1) (12) = (4) (3) in A = Z. Here q1 =(4)=p1, p1 = (2) and ∩ q2 = p2 =(3). (2) (x2,xy)=(x) (x, y)2 in A = k[x, y] where k is a ﬁeld. Here ∩ 2 q1 =(x)=p1 and q2 = p2 with p2 =(x, y). 26 MATH 205B NOTES 2010 COMMUTATIVE ALGEBRA

This example is not typical since the primary ideals are powers of prime ideals. (1) pm might not be primary. (2) q being p-primary does not imply q = pm. The second statement is not surprising and it is easy to find examples, e.g., p = m =(x, y) in A = k[x, y]. There are many ideals q between m and m2 and they are all m-primary by Prop. 4.8 below. But the first statement is surprising. Example 4.7. Let A = k[x, y, z]/(xy z2) and let x, y, z A be the images of x, y, z. Then p =(x, z) is− prime since A/p = ∈k[y] is the polynomial ring in one generator y. (No power of y is in p.) Claim: p2 =(x2, z2, xz) is not primary. For the proof, note that x y = z2 p2 but x/p2 and ym / p2 for all m since p2 p which does not contain∈ any power∈ of y. So,∈ p2 is not primary. ⊂ Proposition 4.8. If r(a)=m is maximal then a is primary. In par- ticular, mn is primary. Note that in the examples I gave the primary ideals which were not prime were powers of maximal ideals. Proof. We have an epimorphism A/a A/ r(a)=A/m which is a field. We need to show that every zero→ divisor x A/a is nilpotent. Suppose not. Then x x =0 A/ r(a). Since this∈ is a field, x is not → ∈ a zero divisor. So, x is not a zero divisor. 4.2. associated primes. Before we do the associated primes, we need to review the prime avoidance lemma 1.2.2 which has two forms: n (1) If a i=1 pi then a pi for some i. ⊆ n ⊆ (2) If p ai then p ai for some i. And, if p = ai then ⊇ i=1 ⊇ p = ai for some i. Definition 4.9. Given an A-module M an associated prime1 is defined to be a prime ideal of the form r(Ann(x)) for some x M. Taking M = A/a, the associated primes for a are those of the form∈ p =r(a : x) for some x A. ∈ Theorem 4.10. If a = qi is a minimal primary decomposition then the prime ideals pi =r(qi) are exactly the associated primes for a. 1In the Noetherian case p is an associated prime if p =(a : x) for some x A. ∈ MATH 205B NOTES 2010 COMMUTATIVE ALGEBRA 27

Proof. We have two sets of prime ideals and we need to show that they are equal. To show that the ﬁrst set is contained in the second, suppose that p is an associated prime. Then

p =r(a : x)=r qi : x =r (qi : x)= r(qi : x) By prime avoidance we have p =r(qi : x). But (q : x)= a A ax q i { ∈ | ∈ i} is a set of zero divisors modulo q . So q (q : x) p . This implies i i ⊆ i ⊆ i that p = pi. So, the ﬁrst set is contained in the second. To show the converse, take p1 =r(q1). We need to show that p1 = n r(a : x) for some x A. Since n is minimal, a i=1 qi. So, there is n ∈ some x qi so that x/a and x/q1. ∈ i=1 ∈ ∈ Claim:r(a : x)=p1. This is all we have to show. It is clear that q (a : x) since x is 1 ⊆ already in all of the other qi and multiplying it by any element of q1 will put it in qi = a. So, all we have to do is to show that (a : x) p ⊆ 1 Suppose y (a : x). Then xy a q1 but x/q1. Since q1 is primary this implies∈ ym q which means∈ ⊆y r(q )=∈ p which is what we ∈ 1 ∈ 1 1 needed to show.

Deﬁnition 4.11. A prime pi as above is called a minimal associated prime if it does not contain any other associated prime. The others are called embedded primes. Exercise 4.12. If a =r(a) then a has no embedded primes.

Proposition 4.13. pi is equal to the set of zero divisors modulo a. This is a corollary of the proof of the theorem above. 4.3. saturation. To get the next theorem we needed to review what we learned last time. (1) If S is a multiplicative set in A, there is a 1-1 correspondence between the prime ideals in A which don’t meet S: p S = 1 e 1 ∩ ∅ and the prime ideals of S− A given by p p = S− p and in c 1→ the other direction, q = q A q S− A. e 1∩ ← 1 ⊂ (2) If p S = then p = S− p = S− A. ∩ ∅ Deﬁnition 4.14. If S A is a multiplicative set the saturation S(a) of an ideal a is the contraction⊆ of its extension: e 1 S(a)=a = S− a A ∩ 28 MATH 205B NOTES 2010 COMMUTATIVE ALGEBRA

1 (Actually, this notation isn’t technically correct since f : A S− A may not be injective. The intersection just means the inverse→ image in A.) With this notation we can rephrase the bijective correspondence: (1) S(p)=p if p S = . (2) S(p)=A if p∩ S =∅ ∩ ∅ Lemma 4.15. Suppose that a = qi is a minimal primary decomposition and S is a multiplicative set which is disjoint from the ﬁrst m primary ideals: qi S = for i =1, ,m and qj S = for j>m. Then ∩ ∅ ··· ∩ ∅ m

S(a)= qi i=1 Proof. First of all n n 1 1 1 S− a = S− qi = S− (qi) i=1 i=1 1 because S− is an exact functor (Prop 3.19 (i)). Contraction also com- mutes with intersection. So n n m 1 1 1 1 S(a)=f − S− a = f − S− (qi)= S(qi)= qi i=1 i=1 i=1 where at the last step we use the following easy facts.

(1) S(qi)=qi if qi S = (i.e., for i m) (2) S(q )=A if q ∩ S = ∅ (i.e., for i>m≤ ). i i ∩ ∅

Theorem 4.16. If a has a primary decomposition a = qi then the primary components qi corresponding to the minimal primes pi are uniquely determined by a. Proof. The minimal primes are uniquely determined by the previous theorem. If p is a minimal prime, let S = A p be the complement 1 1 − 1 of p1. Then S1 is disjoint from q1 but S1 meets qi for every i>1 since otherwise pi p1 contradicting the minimality of p1. By the lemma we have: ⊆ S1(a)=q1 Exercise 4.17. The nth symbolic power p(n) of a prime ideal p is deﬁned to be the saturation S(pn) where S = A p. Show that if pn has a primary decomposition then one of the components− is p(n).