# Multiplicatively closed set

(Redirected from Multiplicatively closed subset)

In abstract algebra, a multiplicatively closed set (or multiplicative set) is a subset S of a ring R such that the following two conditions hold:[1][2]

• $1 \in S$.
• For all x and y in S, the product xy is in S.

In other words, S is closed under taking finite products, including the empty product 1.[3] Equivalently, a multiplicative set is a submonoid of the multiplicative monoid of a ring.

Multiplicative sets are important especially in commutative algebra, where they are used to build localizations of commutative rings.

A subset S of a ring R is called saturated if it is closed under taking divisors: i.e., whenever a product xy is in S, the elements x and y are in S too.

## Examples

Common examples of multiplicative sets include:

• the set-theoretic complement of a prime ideal in a commutative ring;
• the set $\{ 1, x, x^2, x^3, \dots \}$, where x is a fixed element of the ring;
• the set of units of the ring;
• the set of non-zero-divisors of the ring;
• 1 + I   for an ideal I.

## Properties

• An ideal P of a commutative ring R is prime if and only if its complement R\P is multiplicatively closed.
• A subset S is both saturated and multiplicatively closed if and only if S is the complement of a union of prime ideals.[4] In particular, the complement of a prime ideal is both saturated and multiplicatively closed.
• The intersection of a family of multiplicative sets is a multiplicative set.
• The intersection of a family of saturated sets is saturated.