Proposition 7

Let A be a ring and S, T two multiplicative subsets of A. Set T' = i(T_A^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.