Semirings embedded in a completely regular semiring

M. K. Sen; S. K. Maity

Displaying similar documents to “Semirings embedded in a completely regular semiring”

A note on orthodox additive inverse semirings

M. K. Sen, S. K. Maity (2004)

Acta Universitatis Palackianae Olomucensis. Facultas Rerum Naturalium. Mathematica

Similarity:

We show in an additive inverse regular semiring $(S, +, \cdot)$ with $E^{•} (S)$ as the set of all multiplicative idempotents and $E^{+} (S)$ as the set of all additive idempotents, the following conditions are equivalent: (i) For all $e, f \in E^{•} (S)$ , $e f \in E^{+} (S)$ implies $f e \in E^{+} (S)$ . (ii) $(S, \cdot)$ is orthodox. (iii) $(S, \cdot)$ is a semilattice of groups. This result generalizes the corresponding result of regular ring.

Decomposition of a semiring into division semirings

Giraldes, E. (1980)

Portugaliae mathematica

Similarity:

The Clifford semiring congruences on an additive regular semiring

A.K. Bhuniya (2014)

Discussiones Mathematicae - General Algebra and Applications

Similarity:

A congruence ρ on a semiring S is called a (generalized)Clifford semiring congruence if S/ρ is a (generalized)Clifford semiring. Here we characterize the (generalized)Clifford congruences on a semiring whose additive reduct is a regular semigroup. Also we give an explicit description for the least (generalized)Clifford congruence on such semirings.

Semirings whose additive endomorphisms are multiplicative

Tomáš Kepka (1993)

Commentationes Mathematicae Universitatis Carolinae

Similarity:

A ring or an idempotent semiring is associative provided that additive endomorphisms are multiplicative.

Notes on commutative parasemifields

Vítězslav Kala, Tomáš Kepka, Miroslav Korbelář (2009)

Commentationes Mathematicae Universitatis Carolinae

Similarity:

Parasemifields (i.e., commutative semirings whose multiplicative semigroups are groups) are considered in more detail. We show that if a parasemifield $S$ contains $ℚ^{+}$ as a subparasemifield and is generated by $ℚ^{+} \cup {a}$ , $a \in S$ , as a semiring, then $S$ is (as a semiring) not finitely generated.