The following was a homework exercise in an undergraduate abstract algebra course, and I was unable to solve it. It's possible that the problem statement or my interpretation of it is in error. I am looking either for a solution or a counterexample. Here is the exercise, as I interpreted it (a direct quotation would require too much context to be reproduced):
Let $\left(M,+\right)$ be a commutative semigroup and $\left(N,+\right)$ a commutative monoid. Let $m \in M$ be $q$-divisible (meaning that for every positive integer, $d$, there is an $l\in M$ such that $m = q^d l$) and let $n\in N$ be such that $q^d n = 0_N$ for some positive integer $d$. Based on the definition of the semigroup tensor product [given below], show that $m\otimes n = 0_{M\otimes N}$?
Here is the definition of the semigroup tensor product I am working under. Let $\left(M,+\right)$, $\left(N,+\right)$, and $\left(P,+\right)$ be commutative semigroups. Define a biadditive pairing $\phi \colon M,N \rightarrow P$ to be a mapping that takes any pair $m,n$, with $m\in M$ and $n\in N$, to an element $p\in P$ such that $\phi\left(m+m^\prime,n\right) = \phi\left(m,n\right) + \phi\left(m^\prime,n\right)$ for all $m,m^\prime \in M$ and $n\in N$ and $\phi\left(m,n+n^\prime \right) = \phi\left(m,n\right) + \phi\left(m,n^\prime \right)$ for all $m\in M$ and $n,n^\prime \in N$. Define a semigroup tensor product $M \otimes N$ as the range of a biadditive pairing, $\upsilon$, such that for any biadditive pairing $\phi \colon M,N \rightarrow P$ there is a unique semigroup homomorphism $f \colon M\otimes N \rightarrow P$ such that $\phi = f\circ \upsilon$ (In other words, $\upsilon$ is an initial object in a particular category of biadditive pairings). Denote $\upsilon\left(m,n\right)$ as $m\otimes n$.
The seemingly obvious first steps are as follows. Since $m$ is $q$-divisible, biadditivity of the tensor product gives that $m\otimes n$ = $l \otimes 0_N$ for some $l\in M$. If $M$ were a group, it would be easy to show that $l \otimes 0_N = 0_M \otimes 0_N$, and from there to show that $0_M \otimes 0_N$ must be an additive identity in $M \otimes N$. However, $M$ is only assumed to be a semigroup, so I am not sure how to show that $l \otimes 0_N = 0_{M\otimes N}$, or even that $M\otimes N$ necessarily has an identity.
As a counterexample, let $M=\{0,1\}$ with addition defined by $0+0=0$ and $0+1=1+1=1$, and let $N=\{0\}$. Note that $\phi:M\times N\to M$ defined by $\phi(m,0)=m$ is biadditive. It follows that in $M\otimes N$, $1\otimes 0\neq 0\otimes 0$. This gives a counterexample to your exercise, with $m=1$ and $n=0$.