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$S$ is a collection of disjoint sets. $(S,\cdot)$ is a commutative monoid and $(S,*)$ is a commutative semigroup. The identity element $e$ of $(S,\cdot)$ is the zero element of $(S,*)$. The monoid $(S,\cdot)$ does not have a zero element. The binary operator $'*'$ is distributive over the binary operator $'\cdot'$. What can be said about such an algebraic structure ? What could be its usefulness ?

EDIT

There is a third binary operator $\otimes$ with which $(S,\otimes)$ is a commutative semigroup and the operator $'\otimes'$ is distributive over $'\cdot'$. $(S,\otimes)$ does not have a zero (absorption) element.

Motivation behind the question :

I had something and wanted to check where it fits in a formalism and it happened to turn out like this. I have a naive question, why should i even bother about such a formalism like semigroup,semiring etc., how is it useful.

PS: I thought it is OK/appropriate to edit a question to add to ask more on the same subject. Please let me know if it is not OK.I can make it as another question.

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  • $\begingroup$ What's the difference between "identity element" and "zero element" in your usage of terminology? Notice that, by standard definition, a monoid is required to have an identity element while a semigroup isn't. Otherwise they are both just sets with an associative binary operation. $\endgroup$
    – Rasmus
    Dec 1, 2010 at 23:19
  • $\begingroup$ zeros for monoids are often "sinks" as is 0*a = a*0 = 0. Identities satisfy 1*a = a*1 = a. In other words he has a non-unital, associative "ring" without subtraction (and so has included 0*a=0 as an additional axiom, since the standard proof needs subtraction, I believe). $\endgroup$ Dec 1, 2010 at 23:36
  • $\begingroup$ Well, the example that I have in mind is the natural numbers including $0$. Also the power set of a given set seems to fit your axioms. So, I reckon most people find those two useful. But certainly your axioms do not characterize these. $\endgroup$ Dec 2, 2010 at 0:34
  • $\begingroup$ @Rajesh D: The reason you "bother about formalism" is that the structures have been studied before, many of their properties have been elucidated before, and so the conclusions that others have been able to draw will apply to your structure, if you are interested in such a structre. Why bother knowing something is a 'group'? Because then you automatically know all the theorems about group theory, just because you know it's a group. $\endgroup$ Dec 2, 2010 at 16:59
  • $\begingroup$ @Arturo Magidin : I am interested to study and know such properties. But i don't know exactly what to expect. How could it help to give a shape to a concept so that it is useful to create a mathematical model and hopefully find a physical system which is described by such a model. $\endgroup$
    – Rajesh D
    Dec 2, 2010 at 17:08

1 Answer 1

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I think what you are describing is a semirings without identity: a semiring would be aset $R$ with two operations, $+$ and $\cdot$, such that $(R,+)$ is a commutative monoid, $(R,\cdot)$ is a commutative monoid, $\cdot$ distributes over $+$, and the neutral element relative to $+$ is a zero element relative to $\cdot$. The only difference between this and your structure is the existence of an identity for the operation *. Some authors allow semirings to not have a multiplicative identity, which would be exactly what you have.

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