Localization, Saturated Sets and Prime Ideals

Saturated Sets and Prime Ideals

Assume R is commutative. An application of zorn's lemma shows there is a maximal set S, without 0, and closed under multiplication. You can always start an ascending chain with 1.

The complement of S contains the ideal 0, and hence a maximal prime ideal P. Now the complement of P is multiplicatively closed and contains S, so it must be S. Thus the complement of S is a prime ideal. Furthermore, a smaller prime ideal inside P would imply a larger, multiplicatively closed set containing S, hence the complement of S is a minimal prime ideal.

Conversely, if S is the complement of a minimal prime ideal P then S is maximal, for any larger S has some prime ideal in its complement, and hence in P. Minimal prime ideals and maximal closed sets are in one to one correspondence.


A multiplicatively closed set S is saturated if S contains x and y whenever S contains xy. For instance, the set of elements that are not zero divisors is saturated. We will show that S is saturated iff its complement is the union of prime ideals in R. (Yes, we're still assuming R is commutative.)

Assume S is saturated and let x be in R-S. Since xy is not in S, the principal ideal generated by x misses S. What about the image of this ideal in R/S? It consists of x in the numerator, and all elements of S downstairs. Is x/S the entire fraction ring R/S? Only if x/v = w/w, the multiplicative identity. This means uwx = uwv for some u in S. An element in the ideal generated by x is in S, and this is a contradiction, so our ideal remains proper in R/S. Given this, we can drive x up to a maximal prime ideal missing S. This holds for all x, so R-S is the union of prime ideals.

Conversely, assume R-S consists of prime ideals. If xy is in R-S then x or y is also in R-S. Turning this around, if x and y are in S, so is xy, hence S is multiplicatively closed. Now let xy lie in S. If x is not in S it is in some prime ideal P disjoint from S. This means xy is in P and not in S, which is a contradiction. Therefore S is saturated.

The saturation of a multiplicatively closed set S is a minimal saturated set S′ containing S. We need to show this is well defined.

Let S′ be the complement of the union of the prime ideals in R that do not hit S. By the above, S′ is saturated, and it clearly contains S. Another saturated set T containing S, and strictly smaller than S′, has, in its complement, an element x not in the complement of S′. Thus x is in some prime ideal P, not seen in the complement of S′. This prime ideal misses T, and hence S, and contradicts the construction of S′. Thus S′ is indeed the smallest saturated set containing S.