Characterization of primary ideals

Let R be a commutative ring. An ideal Q \subseteq R is called primary if whenever ab \in Q and a \notin Q, then b^n \in Q for some n \geq 1. Prove the following about primary ideals.

  1. The proper primary ideals of \mathbb{Z} are 0 and (p^k) where p is a prime and k \geq 1.
  2. Every prime ideal or R is primary.
  3. An ideal Q \subseteq R is primary if and only if every zero divisor in R/Q is nilpotent.
  4. If Q \subseteq R is a primary ideal then \mathsf{rad}(Q) is a prime ideal.

  1. Suppose (n) \subseteq \mathbb{Z} is a nonzero primary ideal. Suppose further that n = ab is composite, where a and b are both less than n with \mathsf{gcd}(a,b) = 1. Now ab \in (n) and a \notin (n), since n does not divide a. However, n does not divide any power of b since, in particular, a does not divide any power of b. Thus we have a contradiction, and if I \subseteq \mathbb{Z} is primary then either I = 0 or I = (p^k) for some prime p and k \geq 1.

    Now we will verify that these ideals are in fact primary. 0 is primary because if ab = 0, then either a = 0 or b = 0. Now suppose ab \in (p^k). Then p^k divides ab, and in particular p divides either a or b because p is prime. If a \notin (p^k), then some power of p must divide b. Thus b^k \in (p^k). Hence (p^k) is primary.

  2. Suppose P is a prime ideal. If ab \in P and a \notin P, then b = b^1 \in P. Thus P is primary.
  3. (\Rightarrow) Suppose Q \subseteq R is primary. Let x+Q \in R/Q be a zero divisor. Then (x+Q)(y+Q) = 0 for some y, where we have x,y \notin Q. Since y \notin Q, we have x^n \in Q for some n \geq 1; thus (x+Q)^n = 0, and so x+Q is nilpotent in R/Q.

    (\Leftarrow) Suppose every zero divisor in R/Q is nilpotent. Suppose xy \in Q with x \notin Q. If y \notin Q, then y+Q is a zero divisor in R/Q, so that (y+Q)^n = 0 for some n; in this case, y^n \in Q. Thus Q is primary in R.

  4. Let Q \subseteq R be a primary ideal. Suppose xy \in \mathsf{rad}(Q), with x \notin \mathsf{rad}(Q). Now (xy)^n = x^ny^n \in Q for some n \geq 1. Since x \notin \mathsf{rad}(Q), x^n \notin \mathsf{rad}(Q). Since Q is primary, y^n \in Q, so that y \in \mathsf{rad}(Q). Thus \mathsf{rad}(Q) is a prime ideal in R.
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  • Ori  On June 3, 2011 at 7:38 am

    There is a problem with the equivalence in 3
    P = (x^2,xy) in R = K[x,y] is primary, because if fg is in P then x divides either f or g, and if f and g are not already in P, then at least one of them, f^2 or g^2 is in P.
    On the other hand y is a zero-divisor in R/P because xy is in P, but y is not nilpotent in R/P ( I know that x is, but y is also a zero-divisor, and is not nilpotent as the proposition states)

    • nbloomf  On June 3, 2011 at 8:53 am

      I am not so sure this P is primary. For instance, xy \in P and x \notin P, but no power of y is in P.

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