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6. Zariski Topology Definition 6.1. A topological space X is Noetherian if it satisfies the descending chain condition for closed sets, that is for closed sets Yi if Y1 ⊃ Y2 ⊃ · · · , then for some R, we have Yr = Yr+1 for all r ≥ R. Example 6.2. (i) An (k) with the Zariski topology is Noetherian. For any √ descending chain Y1 ⊃ Y2 ⊃ · · · , we have Yi = V (Ji ) where Ji = Ji is a radical ideal, and so J1 ⊂ J2 ⊂ · · · ∈ k[x1 , . . . , xn ]. Since k[x1 , . . . , xn ] is Noetherian, the ascending chain of Ji ’s satisfies the ascending chain condition, and the result follows. (ii) Let X ⊂ An (k). Then X endowed with the subspace topology is Noetherian. (iii) Let X = Spec(A) where A is a Noetherian ring. Then the closed sets are V (I) = {P : I ⊂ P }. Recall an irreducible subset Y of X is a subset such that if Y = Y1 ∪Y2 and Y1 and Y2 are closed, then either Y = Y1√or Y = Y2 . Also remember that Y = V (J) is irreducible if and only if J is prime. Proposition 6.3. In a Noetherian topological space X, every nonempty closed subset Y ⊂ X can be written as a finite union Y = Y1 ∪ · · · ∪ Yr , where each Yi is irreducible. If Yi 6⊂ Yj for i 6= j, then this decomposition is unique. Each Yi is called an irreducible component of Y. Proof: Let S be the set of nonempty closed subsets Y ⊂ X that do not have such a decomposition. If S 6= ∅, the Noetherian hypothesis implies that S has a minimal element, Y . Since Y ∈ S, Y admits no finite decomposition into a union of irreducibles. In particular, Y is not irreducible. Thus we may write Y = Y1 ∪Y2 where Y1 , Y2 are closed and Y1 6= Y and Y2 6= Y . Since Y ∈ S is minimal, neither Y1 nor Y2 ∈ S. So Y1 = Y10 ∪ · · · ∪ Ys0 and Y2 = Y100 ∪ · · · ∪ Yt00 giving a decomposition for Y = Y10 ∪ · · · ∪ Ys0 ∪ Y100 ∪ · · · ∪ Yt00 and thus S = ∅. Now suppose Y = Y10 ∪ · · · ∪ Ys0 S is another such representation. Then 0 0 Y1 ⊂ Y = Y1 ∪ · · · ∪ Yr , so Y1 = (Y10 ∩ Yi ). But Y10 is irreducible, so Y10 ⊂ Yi for some i, say i = 1. Similarly, Y1 ⊂ Yj0 for some j. Then Y10 ⊂ Yj0 , so j = 1, and thus Y1 = Y10 . Now let Z = (Y − Y1 ). Then Z = Y2 ∪· · ·∪Yr , and also Z = Y20 ∪· · ·∪Ys0 . So we proceed by induction on r to obtain uniqueness. Definition 6.4. An ideal Q ⊂ A is primary if one of the equivalent conditions holds: (1) Every zero divisor of A/Q is nilpotent. (2) If xy ∈ Q, then x ∈ Q or y n ∈ Q for some n ≥ 1 (ie y ∈ Q). 1 2 If Q is primary, then √ Q = P is prime, and we say Q is P -primary. If Q1 and Q2 are P -primary, then Q1 ∩ Q2 is P -primary.√ If M is a maximal ideal, then M r is M -primary. More generally, if I = M is maximal, then I is M -primary. Observe that it is not true that P r is P -primary for any prime ideal P . Example 6.5. Let A = k[x, y, z]/(xy − z 2 ) = k[x, y, z]. Then P = (x, z) is prime because A/P ∼ = k[y] is an integral domain.√But P 2 is not P -primary because x · y = z 2 ∈ P 2 yet x ∈ / P and y ∈ / P2 = P. Definition 6.6. An irredundant primary decomposition of an ideal I ⊂ A is an expression √ I = Q1 ∩ · · · ∩ Qr with Qi primary and (1) the prime Pi = Qi are all distinct. (2) No Qi can be omitted from the decomposition, that is \ Qj 6⊂ Qi for any i. i6=j Theorem 6.7. Let A be a commutative Noetherian ring. (1) Every ideal I ⊂ A has an irredundant primary decomposition: I = Q1 ∩ · · · ∩ Qr . √ √ (2) The set of primes P1 = Q1 , . . . , Pr = Qr is determined by I. These are the associated primes of I, denoted Ass(I). (3) This decomposition of I need not be unique, but the Qi belonging to the minimal elements of the set {Pj } are uniquely determined by I. Non-minimal elements are called embedded √ √ primes. (4) If I = I, then I has a unique decomposition I = P1 ∩· · ·∩Pt where Pi is prime and Pi 6⊂ Pj for i 6= j. Example 6.8. This is an example of a non-unique primary decomposition. Let A = k[x, y]. Then let I = (x2 , xy). Now, I admits two 2 decompositions Q1 ∩ Q2 = (x) ∩ (x, y) Q1 ∩ Q02 = (x) ∩ (x2 , y). p and √ √ Then Q1 = P1 = (x) and Q2 = Q02 = (x, y) = P2 . Note that P1 ⊂ P2 , so P2 is an embedded prime.