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).

Suppose further that $T$ is contained in the multiplicative subset of $B$ generated by $f(S)$. Then, iff is surjective (resp. injective) so is $f'$.

The first assertion amounts to saying that there exists a unique homomorphism $f': A[S^{-1}] \to B[T^{-1}]$ giving a commutative diagram:

Now the relation $f(S) \subset T$ implies that $i_B^T(f(s))$ is invertible in $B[T^{-1}]$ for all $s \in S$ and it is sufficient to apply Proposition 1 to $i_B^T \circ f$. It follows easily from (2) that, for all $a \in A$ and $s \in \overline{S}$ (multiplicative subset of $A$ generated by $S$),

Suppose that $T$ is contained in the multiplicative subset generated by $f(S)$, which is precisely $f(\overline{S})$. Then it follows from (3) that, iff is surjective, so is $f'$.

Suppose now that $f$ is injective. Let $a/s$ be an element of the kernel off'. As the multiplicative subset generated by $T$ is $f(\overline{S})$, there is an element $s_1 \in \overline{S}$ such that $f(s_1)f(a) = 0$, whence $f(s_1a) = 0$ and consequently $s_1a = 0$ since $f$ is injective; then $a/s = 0$, which proves that $f'$ is injective.

Remark (6). If the elements of $T$ are invertible in $B$, $B[T^{-1}]$ is identified with $B$ by means of the isomorphism $i_B^T$ and $f'$ then becomes identical with the unique homomorphism $u'$ of $A[S^{-1}]$ to $B$ such that $u' \circ i_A^S = f$.

This is an immediate consequence of Proposition 2 and Remark 6.

For all $a \in A$, $i_A^{T,S}$ then maps the element $a/s$ in $A[S^{-1}]$ to the element $a/s$ in $A[T^{-1}]$.

Remark (7). Note that, if $i_A^T$ is injective, so is $i_A^{T,S}$ (Corollary 1). This is what happens if $T$ is the set $R$ of elements of $A$ which are not divisors of 0; then it is possible to identify $A[S^{-1}]$ with the subring of the total ring of fractions $A[R^{-1}]$ generated by $A$ and the inverses in $A[R^{-1}]$ of the elements of $S$.

This follows easily from the definitions.

In particular, if $S, T, U$ are three multiplicative subsets of $A$ such that $S \subset T \subset U$, then $i_A^{U,S} = i_A^{U,T} \circ i_A^{T,S}$.

The mapping $g$ is injective by Corollary 1; the ring $g(B[{S'}^{-1}])$ contains $i_A^S(A)$ and the inverse of the elements of $S'$; hence it is equal to $A[S^{-1}]$.

If $A$ is an integral domain and $0 \notin S$, the notation $A[S^{-1}]$ agrees with that of Algebra, Chapter IV, § 2, no. 1; also, if $S$ is multiplicative, $A[S^{-1}]$ coincides in this case with the set denoted by $S^{-1}A$ in Algebra, Chapter I, § 1, no. 1.

As an extension of notation, for any multiplicative subset $S$ of a ring $A$, we henceforth denote by $S^{-1}A$ the ring of fractions $A[S^{-1}]$. If $S$ is the complement of a prime ideal $p$ of $A$, we write $A$, instead of $S^{-1}A$.

If $A$ is an integral domain and $0 \notin S$, $S^{-1}A$ is always identified with a subring of the field of fractions of $A$, containing $A$ (Remark 7).

In other words, $(M', f)$ is a solution of the universal mapping problem (Set Theory, Chapter IV, § 3, no. 1) with the following conditions: the species of structure $\Sigma$ is that of an $A$-module in which the homotheties induced by the elements of $S$ are bijective, the morphisms are $A$-module homomorphisms and the $a$-mappings are also $A$-module homomorphisms.

For every $A$-module $N$ and all $a \in A$, denote by $h_a$ the homothety $y \mapsto ay$ in $N$; $a \mapsto h_a$ is then a ring homomorphism from $A$ to $\operatorname{End}_A(N)$. To say that $h_a$ is bijective means that $h_a$ is an invertible element of $\operatorname{End}_A(N)$. Suppose that, for all $s \in S$, $h_s$ is invertible in $\operatorname{End}_A(N)$; the elements $h_a$, where $a \in A$, and the inverses of the elements $h_s$ where $s \in S$, then generate in $\operatorname{End}_A(N)$ a commutative subring $B$ and the homomorphism $a \mapsto h_a$ from $A$ to $B$ is such that the images of the elements of $S$ are invertible. Then it follows (no. 1, Proposition 1) that there exists a unique homomorphism $h'$ of $A[S^{-1}]$ to $B$ such that

we know (Algebra, Chapter II, , no. 14) that such a homomorphism defines on N an $A[S^{-1}]$-module structure such that $(a/s).y = h_s^{-1}(a.y)$; the A-module structure derived from this $A[S^{-1}]$-module structure by means of the homomorphism $i_A^S$ is precisely the structure given initially

Conversely, if N is an $A[S^{-1}]$-module and it is considered as an A-module by means of $i_A^S$, the homotheties $y \mapsto sy$, for $s \in S$, are bijective, for $y \mapsto (1/s)y$ is the inverse mapping of $y \mapsto sy$; and the $A[S^{-1}]$-module structure on N derived from its A-module structure by the process described above is the $A[S^{-1}]$-module structure given initially. Thus there is a canonical one-to-one correspondence between $A[S^{-1}]$-modules and A-modules in which the homotheties induced by the elements of S are bijective; moreover, if N, N' are two A-modules with this property every A-module homomorphism $u : N \to N'$ is also a homomorphism of the $A[S^{-1}]$-module structures of N and N', as, for ally $\in N$ and all $s \in S$, we may write $u(y) = u(s.((1/s)y)) = s.u((1/s)y)$, whence $u((1/s)y) = (1/s)u(y)$; the converse is obvious.

This being so, the statement of Proposition 3 is just the characterization of the module obtained from M by extending the scalars to $A[S^{-1}]$, taking account of the above interpretation (Algebra, Chapter II, § 5, no. 1, Remark 1).

If s' \in S is such that s'm = 0, clearly m|s = (s'm)/(ss') = 0. Conversely, suppose that m|s = 0. As 1/s is invertible in S^{-1}A, m/1 = 0. For every sub-A-module P of S^{-1}A containing 1, we denote by \beta(P, m) the image of (m, 1) under the canonical mapping of M \times P to M \otimes_A P; then \beta(S^{-1}A, m) = 0. We know (Algebra, Chapter II, § 6, no. 3, Corollary 4 to Proposition 7) that there exists a finitely generated submodule P of S^{-1}A containing 1 and such that \beta(P, m) = 0. For all t \in S we denote by A_t the set of a/t, where a \in A; as P is finitely generated, there exists t \in S such that P \subset A_t (no. 1, Remark 2), whence \beta(A_t, m) = 0. The mapping a \mapsto a/t from A to A_t is surjective; let B be its kernel. It defines a surjective mapping h : M \otimes_A A \to M \otimes_A A_t, whose kernel is BM (M being identified with M \otimes A); then \beta(A_t, m) = h(tm) and consequently tm can be expressed in the form $\sum_{i=1}^r b_i m_i$, where b_i \in B, m_i \in M (1 \leq i \leq r). As b_i/t = 0 for 1 \leq i \leq r, there exists t' \in S such that t'b_i = 0 for 1 \leq i \leq r, whence t'tm = 0, which proves Proposition 4.

Without any conditions on M, clearly the relation sM = 0 for some s \in S implies S^{-1}M = 0. Conversely, suppose that S^{-1}M = 0 and let (m_i)_{1 \leq i \leq n} be a system of generators of M; the m_i/1 generate the S^{-1}A-module S^{-1}M, hence to say that S^{-1}M = 0 amounts to saying that m_i/1 = 0 for 1 \leq i \leq n; by proposition 4 there exist $s_i \in S$ such that $s_i m_i = 0$ and, taking $s = s_1 s_2 \cdots s_n$, $s m_i = 0$ for all $i$ and hence $sM = 0$.

Clearly the relation $(1 + a)M = 0$ implies $M = aM$. To prove the converse we use the following lemma:

The first assertion is obvious, as well as the fact that $a'$ is an ideal of $S^{-1}A$. On the other hand, $(1/1) + (a/s) = (s + a)/s$ and $s + a \in S$ for all $s \in S$ and $a \in a$ by definition of S; hence $(1/1) + (a/s)$ is invertible in $S^{-1}A$ for all $a/s \in a'$, which completes the proof of the lemma (Algebra, Chapter VIII, § 6, no. 3, Theorem 1).

This being so, if we set $N = S^{-1}M$, clearly N is a finitely generated $S^{-1}A$-module; if $aM = M$, then $a'N = N$ and it follows that $N = 0$ by Nakayama's Lemma (Algebra, Chapter VIII, § 6, no. 3, Corollary to Proposition 6); the corollary then follows from Corollary 2.

The mapping $i_N^T \circ u$ from M to $T^{-1}N$ is A-linear. Moreover, if $s \in S$, then $f(s) \in T$, hence the homothety induced by s on $T^{-1}N$ is bijective. The existence and uniqueness of $u'$ then follow from Proposition 3. Then, for $m \in M$ and $s \in S$,

With the Same notation, let C be a third ring, U a multiplicative subset of C, g a homomorphism from B to C such that $g(T) \subset U$, P a C-module, v a B-linear mapping from N to P and $v'$ the $T^{-1}B$-linear mapping from B to $U^{-1}P$ associated with v. Then

where the left hand side is the A-linear mapping $S^{-1}M \to U^{-1}P$ associated with $v \circ u$. Similarly, if $u_1$ is a second A-linear mapping from M to N, then

the left-hand side being the A-linear mapping $S^{-1}M \to T^{-1}N$ associated with $u + u_1$.

Remark (5). If, in Proposition 5, we take $B = A, T = S$ and $f = 1_A$, it is easily seen that $u'$ is just the mapping $u \otimes 1 : M \otimes S^{-1}A \to N \otimes S^{-1}A$. We shall henceforth denote it by $S^{-1}u$; if S is the complement of a prime ideal $p$ of A, we write $u_p$ instead of $S^{-1}u$.

If $b, b'$ are in B, $s, s'$ in S, the conditions $b|s = b'|s'$ and $b|f(s) = b'|f(s')$ are equivalent, as follows from Corollary 1 to Proposition 4, which establishes the existence of j and shows that j is bijective; the uniqueness of j is obvious. Clearly j is an additive group isomorphism. If $a \in A, b \in B, s \in S, t \in S$, then

from which its follows that j is $(S^{-1}A)$-linear. Clearly $j \circ f^1 = S^{-1}f$. Finally, if $b \in B, b' \in B, s \in S$, then $j(b.(b'|f(s))) = j(bb'|f(s)) = bb'|s = b.(b'|s)$, which proves the last assertion.

The mapping j of Proposition 6 is called the canonical isomorphism of $(fS)^{-1}B$ onto $S^{-1}B$. These two sets are in general identified by means of $f'$; then $f' = S^{-1}f, i_B^S = i_B^{f(S)}$.

(i) There exists a unique isomorphism j from the ring (ST)^{-1}A onto the ring T'^{-1}(S^{-1}A) such that the diagram

is commutative.

(ii) Let M be an A-module. There exists an (ST)^{-1}A-isomorphism k from the (ST)^{-1}A-module (ST)^{-1}M onto the T'^{-1}(S^{-1}A)-module T'^{-1}(S^{-1}M) such that the diagram

is commutative.

(i) We use the definition of (ST)^{-1}A as the solution of a universal mapping Problem. Let B be a ring and f a homomorphism from A to B such that f(ST) consists of invertible elements. As f(S) consequently consists of invertible elements, there exists a unique homomorphism f': S^{-1}A \to B such that f = f' \circ i_A^S (no. 1, Proposition 1). For all t \in T, f'(i_A^S(t)) = f(t) is invertible in B by hypo-\&\&. hence f'(T') consists of invertible elements; then there exists, by no. 1, Proposition 1, a unique homomorphism f'' from T'^{-1}(S^{-1}A) to B such that f'' = f'' \circ i_{S^{-1}A}^{T'}, whence f = f'' \circ u, setting u = i_{S^{-1}A}^{T'} \circ i_S^A.

Moreover, iff $f''_1 : {T'}^{-1}(S^{-1}A) \to B$ is a second homomorphism such that $f''_1 \circ u = f$, then $(f''_1 \circ i^{T'}_{S^{-1}A}) \circ i^S_A = (f'' \circ i^{T'}_{S^{-1}A}) \circ i^S_A$, whence $f''_1 \circ i^{T'}_{S^{-1}A} = f'' \circ i^{T'}_{S^{-1}A}$ and consequently $f''_1 = f''$.

As the images under $u$ of the elements of ST in ${T'}^{-1}(S^{-1}A)$ are invertible, the ordered pair $({T'}^{-1}(S^{-1}A), u)$ is a solution of the universal mapping problem (relative to A and ST) considered in no. 1. This shows the existence and uniqueness of $j$.

(ii) The proof is completely analogous with that of (i), using in this case no. 2, Proposition 3, and is left to the reader.

(a) The homomorphism $i^{T,S}_A : S^{-1}A \to T^{-1}A$ is bijective. (b) For every A-module M, the homomorphism $i^{T,S}_A : S^{-1}M \to T^{-1}M$ is bijective. (c) For all $t \in T$, there exists $a \in A$ such that $at \in S$ (in other words, every element of T divides an element of S). (d) Every prime ideal which meets T meets S.

It has been seen above that $i^{T,S}_A = 1_M \otimes i^{T,S}_A$, which immediately proves the equivalence of (a) and (b). Set $T' = i^S_A(T)$; then (Proposition 7) $T^{-1}A$ is identified with ${T'}^{-1}(S^{-1}A)$ and (a) is equivalent to saying that the elements of $T'$ are invertible in $S^{-1}A$ (no. 1, Remark 5). Now, to say that $(t/1)(a/s) = 1/1$ ($t \in T$, $a \in A, s \in S$) means that there exists $s' \in S$ such that $tas' = ss'$, which shows the equivalence of (a) and (c). We show that (d) implies (c). Let $t$ be an element of T and suppose that $t/1$ is not invertible in $S^{-1}A$; then there exists a maximal ideal $m'$ of $S^{-1}A$ containing $t/1$ (Algebra, Chapter I, § 8, no. 7, Theorem 2) and $p = (i^S_A)^{-1}(m')$ is a prime ideal of A containing $t$ and not meeting S (since the image under $i^S_A$ of an element of S is invertible). Conversely, if there exists a prime ideal $p$ which meets T without meeting S, then no element of $p \cap T$ can divide an element of S; this proves that (c) implies (d) and completes the proof.

It follows from Proposition 8 that, amongst the multiplicative subsets T of A containing S and satisfying the equivalent conditions of Proposition 8, there exists a greatest, consisting of all the elements of A which divide an element of S (cf. Exercise 1).

For $\alpha \leq \beta \leq \gamma$, $\rho_{\gamma \alpha} = \rho_{\gamma \beta} \circ \rho_{\beta \alpha}$ (no. 1, Corollary 3 to Proposition 2), hence $(S_\alpha^{-1}A, \rho_{\beta \alpha})$ is a direct system. We write $A' = \lim \rightarrow S_\alpha^{-1}A$; as $\rho_\alpha = \rho_\beta \circ \rho_{\beta \alpha}$ for $\alpha \leq \beta$ (no. 1, Corollary 3 to Proposition 2), $(\rho_\alpha)$ is a direct system of homomorphisms and $j = \lim \rightarrow \rho_\alpha$ is the unique homomorphism from $A'$ to $S^{-1}A$ such that $j \circ \rho'_\alpha = \rho_\alpha$ for all $a \in I$. The homomorphisms $\rho'_\alpha \circ i^{S}_{A^\alpha} : A \to A'$ are all equal, for $\rho_{\beta \alpha} \circ i^{S}_{A^\alpha} = i^{S}_{A^\beta}$ for $a \leq \beta$; let $u$ be their common value. Clearly the elements of $u(S)$ are invertible in $A'$, which shows that there exists a homomorphism $h : S^{-1}A \to A'$ such that $h \circ i^{S}_{A} = u$ (no. 1, Proposition 1). Then

for all $\alpha \in I$ and consequently $j \circ h$ is the identity automorphism of $S^{-1}A$. On the other hand, for all $\alpha \in I$,

whence $h \circ j \circ \rho'_\alpha = \rho'_\alpha$ for all $\alpha \in I$; it follows that $h \circ j$ is the identity automorphism of $A'$ and consequently $j$ is an isomorphism.