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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"? –  user17090 Oct 12 '11 at 18:30
    
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? –  Arturo Magidin Oct 12 '11 at 18:48
    
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 $[X] \cdot [Y] := [X \times Y]$. –  user17090 Oct 12 '11 at 19:05
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Crossposted to MO: mathoverflow.net/questions/77942/… –  Zev Chonoles Oct 12 '11 at 19:10
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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. –  Arturo Magidin Oct 13 '11 at 3:53

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If a semiring $M$ has a multiplicative identity $1$ and more than one element, then $1$ cannot be equal to $0$. Your hypothesis on $M$, that it be a non-trivial semigroup, implies it has more than one element...

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