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