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Cancellative set in a semiring that is not multiplicatively closed

  1. Sep 2, 2010 #1
    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.
     
  2. jcsd
  3. Sep 6, 2010 #2
    As there may not be any connection between + and . of R, Can(R) need not be closed.
     
  4. Sep 6, 2010 #3
    But there is a connection: distributivity.
     
  5. Sep 8, 2010 #4
    Consider those r in Can(R) which satisfy the additional condition that ra=rb implies a=b. They form a semiring.
     
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