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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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    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.2010-12-01
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    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).2010-12-01
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    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.2010-12-02
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    @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.2010-12-02
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    @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.2010-12-02
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    @Rajesh D: Knowing about them may or may not give it the structure that you would like. But *for certain*, **not** knowing about them will **not** help you. If you think it's better to waste your time trying to reinvent the wheel (and perhaps failing, perhaps succeeding), or worse, trying to construct a perpetual motion machine, then all power to you, stop asking whether the structure has a formalism or not, and just go bang your head against the wall on your own. The advantage of the formalism is simply that *others* have already broken through the wall, so that might save you some headaches.2010-12-02

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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.