On a set which is a monoid and also a semigroup under 2 various binary drivers with some buildings

$S$ is a collection of disjoint collections. $(S,\cdot)$ is a commutative monoid and also $(S,*)$ is a commutative semigroup. The identification component $e$ of $(S,\cdot)$ is the absolutely no component of $(S,*)$. The monoid $(S,\cdot)$ does not have an absolutely no component. The binary driver $'*'$ is distributive over the binary driver $'\cdot'$. What can be claimed concerning such an algebraic framework? What could be its efficiency?

EDIT

There is a 3rd binary driver $\otimes$ with which $(S,\otimes)$ is a commutative semigroup and also the driver $'\otimes'$ is distributive over $'\cdot'$. $(S,\otimes)$ does not have an absolutely no (absorption) component.

Inspiration behind the inquiry :

I had something and also intended to examine where it suits a formalism and also it took place to end up similar to this. I have an ignorant inquiry, why need to i also trouble concerning such a formalism like semigroup, semiring etc, just how is it valuable.

PS : I assumed it is OK/appropriate to modify an inquiry to include in ask extra on the very same topic. Please allow me recognize if it is not OK.I can make it as an additional inquiry.

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2019-12-02 02:53:03
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Answers: 1

I assume what you are defining is a semirings without identity : a semiring would certainly be aset $R$ with 2 procedures, $+$ and also $\cdot$, such that $(R,+)$ is a commutative monoid, $(R,\cdot)$ is a commutative monoid, $\cdot$ disperses over $+$, and also the neutral component about $+$ is an absolutely no component about $\cdot$. The only distinction in between this and also your framework is the presence of an identification for the procedure *. Some writers permit semirings to not have a multiplicative identification, which would certainly be specifically what you have.

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2019-12-03 01:13:07
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