Bourbaki

Definition 7

Tag 0576, permanent. Cite it as [Bourbaki, Tag 0576].

Algebra, chapter I, § 2. Identity element; cancellable elements; invertible elements

Let EE be a commutative monoid, SS a subset of EE and S′S' the submonoid of EE generated by SS. The quotient monoid (E×S′)/R(E \times S')/R, where the equivalence relation RR is as described in Lemma 1, is denoted by ESE_S and is called the monoid of fractions† of EE associated with SS (or with denominators in SS).

For a∈Ea \in E and p∈S′p \in S' the class of (a,p)(a, p) modulo RR is in general denoted by a/pa/p and called the fraction with numerator aa and denominator pp. Then by definition (a/p).(a′/∣p′)=aa′/∣pp′(a/p).(a'/|p') = aa'/|pp'. The fractions a/pa/p and a′/∣p′a'/|p' are equal if and only if there exists ss in S′S' with spa′=sp′aspa' = sp'a; if so, there exist σ\sigma and σ′\sigma' in S′S' with aσ=a′σ′a\sigma = a'\sigma' and pσ=p′σ′p\sigma = p'\sigma'. In particular, a/p=sa/spa/p = sa/sp for a∈Ea \in E and s,ps, p in S′S'. The identity element of ESE_S is the fraction e/ee/e.

Let a/e=ε(a)a/e = \varepsilon(a) for all a∈Ea \in E. The above shows that ε\varepsilon is a homomorphism of EE into ESE_S, called canonical. For all p∈S′p \in S', (p/e).(e/p)=e/e(p/e).(e/p) = e/e and hence e/pe/p is the inverse of ε(p)=p/e\varepsilon(p) = p/e; every element of ε(S′)\varepsilon(S') is therefore invertible. Then a/p=(a/e).(e/p)a/p = (a/e).(e/p), whence

a/p=ε(a)ε(p)∗ a/p = \varepsilon(a)\varepsilon(p)^*
for a∈Ea \in E and p∈S′p \in S', the monoid ESE_S is therefore generated by ε(E)∪ε(S)∗\varepsilon(E) \cup \varepsilon(S)^*.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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Transcribed from the printed edition, 1998, Springer. Not translated.