Localization, The Non Zero Divisors

The Non Zero Divisors

Let R be a commutative ring. Let D₀ be the set of 0 divisors in R, and let S₀ be its complement. Verify that S₀ is multiplicatively closed, and saturated. Apply an earlier theorem, and D₀ is the union of prime ideals.

Remember that S₀ includes 1, and all the other units.

Let P be any minimal prime ideal. The complement is a maximal closed set S. If S does not contain S₀, then the product set S*S₀ is properly larger than S₀, and does not contain 0. This is a contradiction, hence P lies in D₀.

In summary, D₀ is the union of prime ideals, and contains all the minimal prime ideals. Furthermore, a minimal prime ideal consists entirely of zero divisors.

Embedding R into a Fraction Ring

We will show that S₀ is the largest set S for which the map from R to R/S is injective.

We clearly have a 0 kernel, for there are no zero divisors in S₀. Thus the embedding of R → R/1 is injective.

Suppose S contains a zero divisor x, such that xy = 0. Now y/1 = 0/x, and the mapping is not 1-1. The set S₀ is the largest set, and includes all other sets S, that allow an embedding from R into R/S.

Let W be the fraction ring R/S₀. Every nonzero element of W is either a unit or a zero divisor. Such a ring, units and 0 divisors, always equals its fraction ring over S₀. This because S₀ is all units.