Proposition 1

Let A be a ring and S a subset of A. There exists a ring A' and a homomorphism h of A to A' with the following properties: (1) the elements of h(S) are invertible in A ; (2) for every homomorphism u of A to a ring B such that the elements of u(S) are invertible in B, there exists a unique homomorphism u' of A to B such that u = u' ∘ h.

In other words, (A', h) is a solution of the universal mapping problem (Set Theory, Chapter IV, § 3, no. 1) with the following conditions: the species of structure Σ considered is that of a ring, the morphisms are ring homomorphisms and the a-mappings are homomorphisms of A to a ring such that the image of S under such a homomorphism consists of invertible elements. Recall (loc. cit.) that, if (A', h) and (A'_1, h_1) are both solutions of this problem, there exists a unique isomorphism j : A' → A'_1 such that h_1 = j ∘ h.

Let $\overline{S}$ be the multiplicative subset of A generated by S. Clearly every solution of the above universal mapping problem is also a solution of the universal mapping problem obtained by replacing S by $\overline{S}$ and conversely.

Consider, in the set A × S, the following relation between elements (a, s), (a', s') :

(1) "There exists t ∈ $\overline{S}$ such that t (sa' − s'a) = 0".

This relation is an equivalence relation: it is clearly reflexive and symmetric; it is transitive, for if t (sa' − s'a) = 0 and t'(s'a'' − s''a') = 0, then tt's'(sa' − s'a) = 0 and tt's' ∈ $\overline{S}$. Let A' be the quotient set of A × $\overline{S}$ under this equivalence relation; for every ordered pair (a, s) ∈ A × $\overline{S}$ we denote by a/s the canonical image of (a, s) in A and we set h(a) = a/1 for all a ∈ A. We shall see that A' can be given a ring structure such that the ordered pair (A', h) solves the problem.

Let x = a/s and y = b/t be two elements of A'. The elements (ta + sb)/st and $ab/st$ depend only on $x$ and $y$; for if $x = a'/s'$, there exists by hypothesis $r \in \overline{S}$ such that $r(s'a - sa') = 0$, whence

and $r(s'tab - sta'b) = 0$. It is easily verified that the laws of composition $(x, y) \mapsto x + y = (ta + sb)/st$ and $(x, y) \mapsto xy = ab/st$ define a commutative ring structure on $\mathbf{A}'$, under which $0/1$ is the identity element under addition and $1/1$ is the unit element. Moreover, it is immediate that $h$ is a ring homomorphism and that, for all $s \in S$, $s/1$ is invertible in $\mathbf{A}'$, its inverse being $1/s$. Finally let $B$ be a ring and $u : \mathbf{A} \to B$ a homomorphism such that the elements $u(S)$ are invertible in $B$; there exists a unique mapping $u' : \mathbf{A}' \to B$ such that

(2)

If $a/s = a'/s'$, there exists $t \in \overline{S}$ such that $t(sa' - s'a) = 0$, whence $u(t)(u(s)u(a') - u(s')u(a)) = 0$ and, as $u(t)$, $u(s)$ and $u(s')$ are invertible, $u(a)(u(s))^{-1} = u(a')(u(s'))^{-1}$. It is easily verified that $u'$ is a homomorphism with respect to addition and multiplication; finally, clearly $u' \circ h = u$ and $u'$ is the only homomorphism satisfying this relation, as it implies

and $1 = u'(1/1) = u'(s/1)u'(1/s) = u(s)u'(1/s)$, whence formula (2).