# Building a commutative monoid in which every cancellative element is invertible

I think that for every commutative monoid $M$, there should be a corresponding commutative monoid $F(M)$ satisfying the following implication.

$$\forall a(\forall x,y(ax=ay \rightarrow x=y) \rightarrow \exists z(za =1))$$

Anyway, I'm having trouble describing this construction explicitly. Bit of help, anyone?

The idea is that to build $F(M)$, we go through every element $a$ of $M$, and ask whether or not $a$ satisfies the cancellation law. If it does, then we ask whether it has an inverse. If it does not, we formally adjoin an inverse element $a^{-1}$ satisfying $a^{-1}a=1$. This process should be left adjoint $F$ to the forgetful functor $U:\mathbf{C} \rightarrow \mathbf{CMon},$ where $\mathbf{C}$ is the category of commutative monoids satisfying the above implication. The motivation is that the unit $\eta$ of the adjunction $F \dashv U$ should have the property that $\eta_M : M \rightarrow UFM$ is injective, for every commutative monoid $M$.

• Didn't see this question when it was asked 2 days ago. Next time, just add the "category theory" tag. ;) – Martin Brandenburg Nov 4 '14 at 22:26
• @MartinBrandenburg, :). – goblin Nov 4 '14 at 22:52

The usual localization theory makes sense for arbitrary commutative monoids, not just commutative rings (in fact, for arbitrary commutative monoid objects in well-behaved symmetric monoidal categories; see Toen-Vaquié's "Au-dessous de Spec(Z)"). So you may localize $M$ at the set of cancellative elements. Explicitly, its elements are fractions $m/a$, where $m \in M$ and $a \in M$ is cancellative, where $m/a=n/b$ iff $bm=an$. If $m/a$ is cancellative, then $m$ is cancellative in $M$ and therefore $m/a$ is invertible with inverse $a/m$. In analogy to the ring case, one might call this monoid the "total fraction monoid".