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Here is the full question:

If $M$ is a commutative monoid, $S\subset M$ is a submonoid and there is a $z\in S$ that is absorbing in $M$ (i.e. $zm=z$ for all $m\in M$), then show that $M_S$ has only one element.

I know that this is a divide-by-zero type of question, and I'm just not sure how to word this. Here's what I've written so far.


We want to construct a monoid of fractions, $M_S=S^{-1}M$, where we can write the congruence classes of $M_S$ as $\frac{a}{s}=[(a,s)]$. From Theorem 1, part (2) of the handout, we need that for every $s \in S$, then $\phi(s)=[(s,e)]$ has an inverse $[(e,s])$ in $M_S$.

For $z \in S$, consider $\phi(z)$. Theorem says that $\phi(z)=[(z,e)]$ must have an inverse, $[(e,z)]$. So then we must have $[(z,e)][(e,z)]=[(e,e)]$. But $z$ is absorbing, so then $[(z,e)][(e,z)]=[(ze,ez)]=[(z,z)]$, and we have $[(e,e)]=[(z,z)]$.

I'm not sure how to proceed from here, or if what I wrote above is on the right track.


Additionally, here is the Theorem that I cite (which we are allowed to do):

Let $M$ be a commutative monoid, $S$ be a submonoid all of whose elements are cancellable in $M$ and $M_S$ be the localization of $M$ by $S$.

  1. The function $\phi:M\rightarrow M_S$ given by $\phi(m)=[(m,e)]$ is an injective monoid homomorphism.
  2. For every $s\in S$, $\phi(s)=[(s,e)]$ has inverse $[(e,s)]\in M_S$.
  3. If $\psi:M\rightarrow N$ is a homomorphism of commutative monoids such that $\psi(s)$ is invertible in $N$ for every $s\in S$, then there is a unique monoid homomorphism $\rho:M_S\rightarrow N$ such that $\psi=\rho\phi$.

$\rho$ is given by the formula $\rho([(m,s)])=\psi(m)*\psi(s)^{-1}$.

Thanks ahead of time for any feedback.

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    $\begingroup$ Doesn’t this follow directly from the definition of equivalence for $M_S$? If I recall correctly, it’s similar to the one for ring localization... Namely, isn’t $[(m,s)] = [(n,s’)]$ if and only if there exists $t\in S$ such that $tms’ = tns$? $\endgroup$ – Arturo Magidin Apr 6 at 21:07

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