Proposition 5

Let A, B be two rings, S a multiplicative subset of A, T a multiplicative subset of B and f a homomorphism from A to B such that $f(S) \subset T$. Let M be an A-module, N a B-module and u an A-linear mapping from M to N. Then there exists a unique $S^{-1}A$-linear mapping $u'$ from $S^{-1}M$ to $T^{-1}N$ such that $u'(m/1) = u(m)/1$ for all $m \in M$.

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 $S^{-1}u$.