Primary Ideals, An Introduction

Introduction

Throughout this topic, rings are assumed commutative.

You need to be somewhat familiar with prime ideals and radicals. In particular, the radical of H, written rad(H), is the intersection of prime ideals containing H, or, equivalently, the set of elements x such that some power of x lies in H.

Primary ideals are a generalization of prime ideals; hence prime ideals are primary. In a laskerian ring, every proper ideal is the finite intersection of primary ideals. In some cases an ideal is the product of primary ideals. Let's start by looking at primary rings, then we'll define a primary ideal.

Primary Ring

A ring R is primary if each zero divisor is nilpotent.

A subring of a primary ring is primary.

An integral domain, with no zero divisors, is primary.

Let R have one prime ideal P. Let xy = 0. Suppose the powers of x do not reach 0, and drive 0 up to a prime ideal Q missing x. With one prime ideal, Q = P. thus P does not contain x. Now x is a zero divisor, and is not a unit, hence x is contained in a maximal ideal that cannot equal P. This is impossible, hence x is nilpotent and R is primary.

By the above, the ring of integers mod p^k, with one maximal ideal, is primary.

Primary Ideal

A proper ideal H is primary if, whenever it contains xy and not x, it contains a power of y.

Every prime ideal is primary.

Let H be a power of a prime ideal in a pid. Say H = {p^k}. If xy lies in H, then xy collectively contains k or more instances of p. If x is not in H then p^k does not divide x, and y takes up the slack. In other words, p generates y. This means p^k divides y^k, and y^k lies in H. This makes H a primary ideal.

Quotient Ring

You could probably see this coming a mile away, but here it is. A quotient ring is primary iff its kernel is a primary ideal. Assume R/H is a primary ring, and x is an element not in H, and xy lies in H. If y is in H then we're done, so assume y is not in H, whence x and y are zero divisors in R/H. This means the image of y, raised to the n^th power in R/H, equals 0. Thus yⁿ lies in H, and H is a primary ideal. Conversely, if H is primary then select x and y outside of H, with xy in H, representing zero divisors in R/H. Since yⁿ lies in H, the image of y is nilpotent in R/H.

Contraction

Let f be a ring homomorphism from R into S, and let H be a primary ideal of S. Thus S/H = T, a primary ring. Let G be the preimage of H in R. Elements of R/G map forward, through f, to elements of S/H, which forms a subring of T, which is a primary ring. Therefore G is primary, and the contraction of a primary ideal is primary.

A similar proof shows the contraction of a prime ideal is prime. It works because a subring of an integral domain is an integral domain, making G a prime ideal.