## Characterization of primary ideals

Let be a commutative ring. An ideal is called *primary* if whenever and , then for some . Prove the following about primary ideals.

- The proper primary ideals of are and where is a prime and .
- Every prime ideal or is primary.
- An ideal is primary if and only if every zero divisor in is nilpotent.
- If is a primary ideal then is a prime ideal.

- Suppose is a nonzero primary ideal. Suppose further that is composite, where and are both less than with . Now and , since does not divide . However, does not divide any power of since, in particular, does not divide any power of . Thus we have a contradiction, and if is primary then either or for some prime and .
Now we will verify that these ideals are in fact primary. is primary because if , then either or . Now suppose . Then divides , and in particular divides either or because is prime. If , then some power of must divide . Thus . Hence is primary.

- Suppose is a prime ideal. If and , then . Thus is primary.
- Suppose is primary. Let be a zero divisor. Then for some , where we have . Since , we have for some ; thus , and so is nilpotent in .
Suppose every zero divisor in is nilpotent. Suppose with . If , then is a zero divisor in , so that for some ; in this case, . Thus is primary in .

- Let be a primary ideal. Suppose , with . Now for some . Since , . Since is primary, , so that . Thus is a prime ideal in .

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## Comments

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)

I am not so sure this is primary. For instance, and , but no power of is in .