multiplicatively closed

Let R be a ring. A subset S of R is said to be multiplicatively closed if S, and whenever a,bS, then abS. In other words, S is a multiplicative set where the multiplicationPlanetmathPlanetmath defined on S is the multiplication inherited from R.

For example, let aR, the set S:={ai,ai+1,,an,} is multiplicatively closed for any positive integer i. Another simple example is the set {1}, if R is unital.

Remarks. Let R be a commutative ring.

  • If P is a prime idealMathworldPlanetmathPlanetmathPlanetmath in R, then R-P is multiplicatively closed.

  • Furthermore, an ideal maximal with respect to the being disjoint from a multiplicative set not containing 0 is a prime ideal.

  • In particular, assuming 1R, any ideal maximal with respect to being disjoint from {1} is a maximal idealMathworldPlanetmath.

A multiplicatively closed set S in a ring R is said to be saturatedPlanetmathPlanetmathPlanetmath if for any aS, every divisorMathworldPlanetmathPlanetmath of a is also in S.

In the example above, if i=1 and a has no divisors, then S is saturated.


  • In a unital ring, a saturated multiplicatively closed set always contains U(R), the group of units of R (since it contains 1, and therefore, all divisors of 1). In particular, U(R) itself is saturated multiplicatively closed.

  • Assume R is commutativePlanetmathPlanetmathPlanetmath. SR is saturated multiplicatively closed and 0S iff R-S is a union of prime ideals in R.


    This can be shown as follows: if let T be a union of prime ideals in R and a,bR-T. if abR-T, then abPT for some prime ideal P. Therefore, either a or bPT. This contradicts the assumptionPlanetmathPlanetmath that a,bT. So R-T is multiplicatively closed. If abR-T with aR-T, then aPT for some prime ideal P, which implies abPT also. This contradicts the assumption that abT. This shows that R-T is saturated. Of course, 0R-T, since 0 lies in any ideal of R.

    Conversely, assume S is saturated multiplicatively closed and 0S. For any rS, we want to find a prime ideal P containing r such that PS=. Once we show this, then take the union T of these prime ideals and that S=R-T is immediate. Let r be the principal idealMathworldPlanetmathPlanetmathPlanetmath generated by r. Since S is saturated, rS=. Let M be the set of all ideals containing r and disjoint from S. M is non-empty by construction, and we can order M by inclusion. So M is a poset and Zorn’s lemma applies. Take any chain C in M containing r and let P be the maximal elementMathworldPlanetmath in C. Then any ideal larger than P must not be disjoint from S, so P is prime by the second remark in the first set of remarks. ∎

  • The notion of multiplicative closureMathworldPlanetmath can be generalized to be defined over any non-empty set with a binary operation (multiplication) defined on it.


  • 1 I. Kaplansky, Commutative Rings. University of Chicago Press, 1974.
Title multiplicatively closed
Canonical name MultiplicativelyClosed
Date of creation 2013-03-22 17:29:15
Last modified on 2013-03-22 17:29:15
Owner CWoo (3771)
Last modified by CWoo (3771)
Numerical id 6
Author CWoo (3771)
Entry type Definition
Classification msc 13B30
Classification msc 16U20
Synonym saturated
Related topic MSystem
Defines saturated multiplicatively closed