Math 221 : Algebra Notes for Oct. 5

Math 221 : Algebra notes for Oct. 5

Alison Miller

1 More on Primary Ideals

Note on deﬁnition of primary ideal from last time: we should require that primary ideals be proper (that is, not the whole ring), just as we’ve done for prime ideals. Tn Proposition 1.1. If q1, ... , qn are p-primary, then k=1 qi is p-primary. pTn Tn √ Proof. First, k=1 qi = k=1 qi = p. Tn Now suppose that xy ∈ k=1 qi, x ∈/ p, then y must be in all the qi, so also in their intersection.

2 Primary Ideal Decomposition

Deﬁnition. A primary ideal decomposition of an ideal I is an expression of I as a ﬁnite intersection of primary ideals n \ I = qk. k=1

It is said to be minimal if the qk have distinct radicals, and if none of the qk contain in any of the others.

(Note that any primary ideal decomposition can be made minimal by replacing any primary ideals having the same radical with their intersection, and then throwing away any redundant elements.) Example. Here’s an example of minimal primary ideal decomposition that shows that it is not unique. Let A = k[x, y] and let I = (x2, xy). Then I = (x) ∩ (x2, xy, y2), or I = (x) ∩ (x2, y), or I = (x) ∩ (x2, x + y) or I = (x) ∩ (x2, xy, yn). We’ll now show that, in a Noetherian ring, any ideal has a primary decomposition.

Deﬁnition. An ideal I of a ring A is irreducible if for any ideals I1, I2 with I1 ∩ I2 = I we have either I1 = I or I2 = I.

1 Proposition 2.1. Let A be a Noetherian ring. Any ideal I of A is an intersection of ﬁnitely many irreducible ideals.

Proof. Suppose A is a Noetherian ring for which the above is not true. Then the set of ideals of A that are ﬁnite intersections of irreducible ideals must have a maximal element I. But I is not irreducible, so write I = I1 ∩ I2. Then both I1 and I2 are ﬁnite intersections of irreducible ideals, so their intersection is also.

Proposition 2.2. If I is irreducible then I is primary.

First noting a deﬁnition we’ll need here and later:

Deﬁnition. If M is an A-module and m ∈ M, AnnA(m) = {a ∈ A | am = 0}. Observe that AnnA(M) is an ideal of A. Also, if b ∈ A, AnnA(b) 6= 0 if and only if b is a zero-divisor. We’ll drop the subscript when it’s clear what ring we mean.

Proof. Both the hypothesis and conclusion are true for the ideal I ⊂ A if and only if they are true for the ideal 0 ⊂ A/I. Therefore, without loss of generality, we may assume that I = 0. So assume 0 is irreducible. We must show than any zero-divisor in A is nilpotent. Suppose that x ∈ A is an arbitrary element. We’ll show that either x is a non-zero-divisor or x is nilpotent. Consider the ascending chain of ideals Ann(x) ⊂ Ann(x2) ⊂ (where here annihilators are in the A-module A.) This chain must terminate, so there exists n such that Ann(xn) = Ann(xn+1). We claim that Ann(x) ∩ (xn) = 0. Indeed, suppose that b = axn ∈ (xn) annihilates x, so axn+1 = 0. Then a ∈ Ann(xn+1) = Ann(xn) so axn = b = 0. Now we use the fact that 0 is irreducible. Since Ann(x) ∩ (xn) is 0, we must either have Ann(x) = 0, and x is not a zero-divisor, or (xn) = 0 and x is nilpotent.

Example. An example of a non-irreducible primary ideal is (x2, xy, y2) ⊂ k[x, y], which can be written as (x, y2) ∩ (x2, y). √ We’ll now show that if I = ∩kqk is a primary decomposition, the set { qk} of primes occuring as radicals depends only on I. To do this we give another characterization of this set. p Deﬁnition. A prime p of A is associated to an A-module M if p = AnnA(m) for some m ∈ M. If I is an ideal of A, by abuse of notation we say that p is associated to I (or is an associated prime of I) if p is associated to the A-module A/I. Notation for this case: deﬁne(a : I) = {b ∈ A | ab ∈ I}. Then ifa ¯ denotes the image of a ∈ A/I we have AnnA(a¯) = (a : I) Let AssA(M) (or AssA(I)) denote the set of associated primes of M (or of I).

2 We’ll eventually show:

Theorem 2.3. If I = T q is a minimal primary decomposition of I, then the set of associated i i √ primes of I is precisely the set qi

First let’s see how this works out in the case where I = q is primary.

Proposition 2.4. Let q be a p-primary ideal. Let a be an arbitrary element of A. Then a) If a ∈ q, then (a : q) = A. b) If a ∈/ p then (a : q) = q. c) If a ∈/ q, then AnnA(a¯) is p-primary.

Proof. Part a) is clear, and b) is just the statement that q is p-primary. p For c), First, we have q ⊂ (a : q) ⊂ p, so (a : q) = p. Now we show that (a : q) is p-primary. Suppose bc ∈ AnnA(a¯ and b ∈/ p. Then abc ∈ q, so since q is p-primary and b ∈/ p we must have ac ∈ p and c ∈ AnnA(a¯) as desired.