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Saturday, September 4, 2021

Condensation of commutative cancellative semigroups

The class of commutative cancellative semigroups is subalgebra closed, but it is not quotient closed. The free commutative monoid F(x) is commutative and cancellative but \frac{F(x)}{x^2=x} is not. If it is in the case that the condensation \frac{S}{H} is cancellative, it is therefore not an inherent property of the class of commutative cancellative semigroups.

Theorem. let S be a commutative cancellative semigroup then \frac{S}{H} is cancellative.

Proof. suppose a =H a' and ax =_H a'y then a = ba' and a' = ca for some b,c. Furthermore, ax \lambda = a'y ad ax = a'y \mu. By plugging in we get (ba')\lambda x = a'y \Rightarrow a'(b\lambda x) = a'y ax = (ca)\mu y \Rightarrow a(x) = a(c \mu y) By the cancellative property this yields ax = a(c\mu y) \Rightarrow x = (c\mu)y a'(b\lambda x) = a'y \Rightarrow y=(b \lambda)x Thusly, x =_H y. We can conclude that a =H a' and ax =_H a'y implies that x =H y. This is our first step. Let H_1,H_2,H_3 be three H classes and suppose that H_1 H_2 = H_1 H_3 then we have: \exists h_1,h_1' \in H_1, h_2 \in H_2,h_3 \in H_3 : h_1 h_2 =_H h_1' h_3 We have that h_1 =_H h_1' as well ast that h_1 h_2 =_H h_1' h_3 but as we previously concluded that this means that h_2 =_H h_3. If h_2 \in H_2 and h_3 \in H_3 then H_2 = H_3 because H classes are disjoint.

Then H_1 H_2 = H_1 H_3 \Rightarrow H_2 = H_3 implies that the condensation semigroup \frac{S}{H} is a commutative cancellative J-trivial semigroup. \square

This is an infinite source of commutative cancellative J-trivial semigroups. For example, it follows from this theorem that the ideal multiplication semigroup of a principal ideal domain is a commutative cancellative J-trivial semigroup with zero.

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