multiplicatively closed

multiplicatively closed

Let

R

be a ring. A subset

S

of

R

is said to be

multiplicatively closed

if

S

≠

∅

, and whenever

a

,

b

∈

S

, then

a

⁢

b

∈

S

. In other words,

S

is a

multiplicative set

where the

multiplication

defined on

S

is the multiplication inherited from

R

.

For example, let

a

∈

R

, the set

S

:=

{

a

i

,

a

i

+

1

,

⋯

,

a

n

,

⋯

}

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 ideal

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

1

∈

R

, any ideal maximal with respect to being disjoint from

{

1

}

is a

maximal ideal

.

A multiplicatively closed set

S

in a ring

R

is said to be

saturated

if for any

a

∈

S

, every

divisor

of

a

is also in

S

.

In the example above, if

i

=

1

and

a

has no divisors, then

S

is saturated.

Remarks

.

•

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

commutative

.

S

⊆

R

is saturated multiplicatively closed and

0

∉

S

iff

R

-

S

is a union of prime ideals in

R

.

Proof.

This can be shown as follows: if let

T

be a union of prime ideals in

R

and

a

,

b

∈

R

-

T

. if

a

⁢

b

∉

R

-

T

, then

a

⁢

b

∈

P

⊆

T

for some prime ideal

P

. Therefore, either

a

or

b

∈

P

⊆

T

. This contradicts the

assumption

that

a

,

b

∉

T

. So

R

-

T

is multiplicatively closed. If

a

⁢

b

∈

R

-

T

with

a

∉

R

-

T

, then

a

∈

P

⊆

T

for some prime ideal

P

, which implies

a

⁢

b

∈

P

⊆

T

also. This contradicts the assumption that

a

⁢

b

∉

T

. This shows that

R

-

T

is saturated. Of course,

0

∉

R

-

T

, since

0

lies in any ideal of

R

.

Conversely

, assume

S

is saturated multiplicatively closed and

0

∉

S

. For any

r

∉

S

, we want to find a prime ideal

P

containing

r

such that

P

∩

S

=

∅

. 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 ideal

generated by

r

. Since

S

is saturated,

⟨

r

⟩

∩

S

=

∅

. 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 element

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

closure

can be generalized to be defined over any non-empty set with a binary operation (multiplication) defined on it.

References

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

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