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Assume we are given an additive semigroup $M$ which we know it is non-trivial i.e. $M\neq \lbrace 0 \rbrace$. Let $R$ be the semiring obtained from adding a multiplication law to the semigroup. Under which circumstances can we guarantee that $R$ is non trivial i.e. ($ 1 \neq 0$)?

Edit: after Arturo's comments I rephrase the question: Consider the Grothendieck semigroup $SK_0(M)$ of some category $M$ (more precisley $M$ is the category of definable sets in some structure) that is $SK_0(M)$ is the free semigroup over symbols $[X]$ where $X$ is an object in $M$ and we quotient by the following relations: $ [X]=[Y]$ if $X$ and $Y$ are isomorphic in $M$, $[X\cup Y]+[X\cap Y]=[X]+[Y]$ for any two objects $X$ and $[Y]$ and also $[\varnothing]=0$. My question is: if we equip $SK_0(M)$ with a multiplication that is if quotient $SK_0(M)$ by relations of the form $[X \times Y] =[X] \cdot [Y]$ is the resulting semiring structure trivial? Thank you

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    Can it happen that the addition of the multiplication law forces the semigroup to "collapse"?2011-10-12
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    It's a strange question: if you are "adding a multiplication to the semigroup", then you still have the same underlying set, and as such you cannot "collapse" the semigroup. Of course, you might be asking whether it is *possible* to extend the semigroup structure into a semiring structure on that set, or something like that. Also, I would argue against using " **the** semiring" if you are adding " **a** multiplication". What multiplication? Is there a prefered choice?2011-10-12
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    Arturo: my motivation comes from motivic integration where the semigroup is some SK$_0$ of a category (Grothendieck semigroup) and I proved in some context that it is nontrivial. Then I want to add multiplication defined by relations $[X] \cdot [Y] := [X \times Y]$.2011-10-12
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    Then your question is not what you posed. Rather, your question is either (i) can one make that definition and get a semiring on the same underlying set? or (ii) If we consider the semiring presented by that condition, is the resulting semiring trivial? Because the way you phrased your question made it either trivially true or nonsense.2011-10-12
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    Crossposted to MO: http://mathoverflow.net/questions/77942/non-triviality-of-a-semigroup-and-a-semiring2011-10-12
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    @Ali: Your new question is still too fuzzy: you are trying to make your question too general. "using a (given) multiplication law" is a useless condition unless you *specify* what that "(given) multiplication law" **is**. Otherwise, the only possible answer is "Depends on the multiplication law". Since you have *specific* case in mind with a *specific* multiplication law, then just ask about **that**!2011-10-12
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    @ Arturo: thank you for your comments.2011-10-12
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    @Ali: Generally speaking, it is bad form to post the same question simultaneously (or nearly so) in both MO and here. If you believe it is a *research level* question, go ahead and post it in MO; worse thing that will happen is it will be closed and you can then come here. If you are unsure and don't want to risk it, post it here, **and then wait a while** to see if you get answers before posting in MO, where you should definitely *say* that you posted the question here as well.2011-10-13
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    I see that you have found an answer on MO, please add it here and accept it, as to signal that your problem was solved and to avoid further auto-bump of this question.2012-03-11

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