Cancellative set in a semiring that is not multiplicatively closed

  • Context: Graduate 
  • Thread starter Thread starter icantadd
  • Start date Start date
  • Tags Tags
    Closed Set
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
3 replies · 3K views
icantadd
Messages
109
Reaction score
0
Definition: A semigroup is a pair (R,op) where R is a set an op is a binary operation that is closed and associative. A commutative semigroup is a semigroup where op satisfies for all a,b in R, op(a,b) = op(b,a). A monoid is a semigroup where with an identity,e, for op, satisfying for all r in R, op(r,e)=op(e,r)=r. A commutative monoid is a semigroup satisfying the monoid and commutative semigroup laws.

Definition:
A semiring (rig) is a triple (R,+,.) where (R,+) is a commutative monoid, and (R,.) is a semigroup, and where distributivity holds;i.e.
a(b+c) = ab+ac and (b+c)a = ba+ca.

Definition:
Let R=(R,+,.) be a semiring. A cancellable element r satisfies
r+a = r+b implies a = b.
The set of all cancellable elements in R is denoted Can(R).

It is an easy exercise to show that Can(R) is a submonoid of (R,+). However, it does not seem to be multiplicatively closed. If anyone knows or can sketch a proof that would be great! Are there any canonical examples of Rigs, where the cancellative elements are not multiplicatively closed.
 
Physics news on Phys.org
As there may not be any connection between + and . of R, Can(R) need not be closed.
 
But there is a connection: distributivity.
 
Consider those r in Can(R) which satisfy the additional condition that ra=rb implies a=b. They form a semiring.