# The monoid of fractions associated with the submonoid of cancellable elements of a commutative monoid E

Let $$E$$ be a commutative monoid, $$\Sigma$$ the submonoid of cancellable elements of $$E$$, $$E_{\Sigma}$$ the monoid of fractions of $$E$$ associated with $$\Sigma$$ and $$\varepsilon$$ the canonical homomorphism of $$E$$ into $$E_\Sigma$$.

Since every element of $$\Sigma$$ is cancellable, $$\varepsilon$$ is injective.

Logically speaking, what does "identifying the monoid $$E$$ with the submonoid $$\varepsilon(E)$$ of $$E_\Sigma$$" allow me to do? I mean: w.r.t the set-theoretic formulas, what can I formally replace with what? For example, does it allow me to write $$\varepsilon(a)=a$$ for $$a\in E$$ (I hope not...) ? What is gained by this "identification"?

I ask this because I would like to know how one can "identify" $$E_{\Sigma}$$ with the submonoid generated by $$E\cup\Sigma^*$$, where $$\Sigma^*$$ denotes the set of inverses of the elements of $$\Sigma$$.

EDIT: (You can ignore the details concerning this specific construction of the monoid of fractions)

General remarks

If you have algebraic structures $$A$$ and $$B$$, and an injective map $$f\colon A\to B$$ that is a morphism, then when we say that we "identify $$A$$ with a substructure of $$B$$" we mean that we can consider the function $$f$$ as an isomorphism onto its image, and then consider the inclusion map. That is, $$A\stackrel{f}{\to}B$$ factors as $$A\stackrel{f,\cong}{\longrightarrow } f(A)\stackrel{i}{\hookrightarrow} B.$$

We commonly "identify" two structures when they are isomorphic: the isomorphism just represents a "renaming" of elements. You identify $$a\in A$$ with $$f(a)\in f(A)$$; the fact that $$f$$ is one-to-one ensures that this is simply a relabeling of names. The fact that $$f$$ is a homomorphism guarantees that the algebraic structure is being preserved.

Once you accept that $$f(A)$$ is "essentially" just $$A$$, then you can follow it up by the embedding, thus we have recognized ("identified") $$a$$ with its image inside of $$B$$. Given that there is a bijection between $$A$$ and $$f(A)$$ that respects the structure, one may just as well "skip the intermediary" and simply work with $$f(A)$$ sitting inside of $$B$$, rather than with the three distinct objects $$A$$, $$f(A)$$, and $$B$$.

Thus, for example, even though $$\mathbb{N}$$ is not technically a subset of $$\mathbb{Z}$$ (as the latter is constructed as a set of equivalence classes of pairs of natural numbers), there is a natural embedding of $$\mathbb{N}$$ into $$\mathbb{Z}$$ and we treat the image as if it were $$\mathbb{N}$$ itself, thus "identifying" $$\mathbb{N}$$ with its canonical image inside of $$\mathbb{Z}$$ and treating $$\mathbb{N}$$ as just a subset/substructre of $$\mathbb{Z}$$.

Here you start with a monoid $$E$$. You construct a separate, distinct monoid $$E_{\Sigma}$$. But you would really like to think of $$E_{\Sigma}$$ as being an "extension" of $$E$$, something you get out of $$E$$ by "adding" stuff to it, just like we think of $$\mathbb{Q}$$ as being an extension of $$\mathbb{Z}$$, even though the latter are not "fractions".

So you have a morphism $$\varepsilon E\to E_{\Sigma}$$ that is one-to-one. That allows you, as per above, to think of $$E$$ as being a subset of $$E_{\Sigma}$$ via the embedding, and so ignore the technical fact that $$E$$ is not literally a submonoid of $$E_{\Sigma}$$ (it's not even a subset).

What is gained is purely conceptual: you can now think of $$E_{\Sigma}$$ as being an extension of $$E$$ obtained by adding stuff to it. Just like thinking of $$\mathbb{Z}$$ as a subset of $$\mathbb{Q}$$: you can formally do everything via a one-to-one morphism, but it is clearer and less cluttered to simply think of $$\mathbb{Z}$$ as a subset of $$\mathbb{Q}$$, rather than to think of the map that sends $$\mathbb{Z}$$ into $$\mathbb{Q}$$ and respects the operations.

The final paragraph actually shows you why this identification is useful: you have $$S\subseteq \Sigma$$. You can construct two different monoids, $$E_S$$ and $$E_{\Sigma}$$. Formally, they are completely separate objects, with different underlying sets, and with a different equivalence relation defined on them. But you want to establish a relation between $$E$$, $$E_S$$, and $$E_{\Sigma}$$.

By thinkin of $$E$$ not as a separate object distinct from $$E_{\Sigma}$$, but rather as a subobject of $$E_{\Sigma}$$, you can also view $$S$$ as a subset of $$E_{\Sigma}$$; now you have the set $$E\cup S$$ sitting inside $$E_{\Sigma}$$, and so you can consider the submonoid it generates. Then one obtains a morphism from $$E_S$$ to this submonoid and proves that this is one-to-one, so that you can actually think of these three monoids as "sitting inside one another", $$E \subseteq E_S\subseteq E_{\Sigma}$$ instead of thinking of them as three completely separate objects, $$E$$, $$(E\times S)/R_S$$, and $$(E\times \Sigma)/R_{\Sigma}$$ (where $$R_S$$ is the corresponding equivalence relation for the construction of $$E_S$$ and $$R_{\Sigma}$$ for the construction of $$E_{\Sigma}$$.

Just like you can construct a ring starting from $$\mathbb{Z}$$ and formally adding a multiplicative inverse of $$2$$; and then consider the rationals as being obtained by inverting all nonzero elements. Formally, three different rings ($$\mathbb{Z}$$, $$\mathbb{Z}[\frac{1}{2}]$$, and $$\mathbb{Q})$$, but you'd rather think of them as sitting insider one another, $$\mathbb{Z}\subseteq \mathbb{Z}[\frac{1}{2}]\subseteq \mathbb{Q}$$. Otherwise, you need to keep track of all the embedding morphisms and you will make talking about these objects as they related to one another extremely cumbersome.

• Superb! Thank you for taking the time to write this detailed answer.---I am sorry about the misunderstanding that occurred earlier; the mistake was mine. Dec 16, 2019 at 21:20

Whether or not $$\epsilon(a) = a$$ depends on the set theoretic construction of the monoid of fractions, but in practice the answer is either no or it's irrelevant.

More formally you can say the monoid $$E$$ is isomorphic to the monoid $$\epsilon(E)$$. Informally it means that the monoid structures of $$E$$ and $$\epsilon(E)$$ are the same, so properties of the monoid structure that don't depend on the precise set-theoretic construction are the same in both $$E$$ and $$\epsilon(E)$$. This is not a precise answer. I believe there is a precise answer, but it would be a very long answer.

I don't understand the second part of the question. $$E \cup\Sigma^{*} = E$$, since $$\Sigma^{*} \subseteq E$$

• For the second part, let's pick a subset $S$ of $\Sigma$ (i.e. some arbitrary set of cancellable elements of $E$). Then, how can we identify $E_S$ (same notation as before) with the submonoid of $E_\Sigma$ generated by $E\cup S^*$? Dec 16, 2019 at 18:01
• Again $S^{*} \subseteq E$, so $E \cup S^{*} = E$. Unless you mean $\epsilon(S)^{*}$ Dec 16, 2019 at 18:36
• Yes the whole point is to use the identification between $\varepsilon(E)$ and $E$ in order to identify $E_S$ with $E\cup S^*$. I want to know how it's done. Dec 16, 2019 at 18:45
• I have made an edit to the post; please take a look. Dec 16, 2019 at 18:49