Proposition 11

Let A be a ring and S a multiplicative subset of A. For every ideal b' of S^{-1}A, let b = (iA^S)^{-1}(b') be such that b' = S^{-1}b.*

(i) Let f be the canonical homomorphism A \to A/b. The homomorphism from S^{-1}A to (f(S))^{-1}(A/b) canonically associated with f (no. 1, Proposition 2) is surjective and its kernel is b', which defines, by taking quotients, a canonical isomorphism of (S^{-1}A)/b' onto (f(S))^{-1}(A/b). Moreover, the canonical homomorphism from A/b to (f(S))^{-1}(A/b) is injective.

(ii) The mapping b' \mapsto b = (iM^S)^{-1}(b'), restricted to the set of maximal (resp. prime) ideals of S^{-1}A, is an isomorphism (with respect to inclusion) of this set onto the set of ideals of A which are maximal among those which do not meet S (resp. the set of prime ideals of A not meeting S).*

(iii) If q' is a prime ideal of S^{-1}A and q = (iA^S)^{-1}(q'), there exists an isomorphism of the ring of fractions A, onto the ring (S^{-1}A)_{q'}, which maps a/b to (a/1)/(b/1), where a \in A, b \in A - q.*

(i) (f(S))^{-1}(A/b) can be identified with S^{-1}(A/b) by means of the canonical isomorphism between these two modules (no. 2, Proposition 6). The exact sequence 0 \to b \to A \to A/b \to 0 then induces an exact sequence

(no. 4, Theorem 1) whose existence proves the first assertion of (i), taking account of the fact that b' = S^{-1}b. Since b is saturated with respect S, the conditions a \in A, s \in S, as \in b imply a \in b; the homothety of ratio s on A/b is then injective, which proves the second assertion of (i).

(ii) We note first that the relation b' = S^{-1}A is equivalent to the relation b \cap S \neq \varnothing, the latter expressing the fact that b' contains invertible elements of S^{-1}A. It follows from no. 4, Proposition 10 (iii) that b' \mapsto b = (i_A^S)^{-1}(b') is an isomorphism (with respect to inclusion) of the set of ideals of S^{-1}A distinct from S^{-1}A onto the set \mathscr{F} of ideals of A not meeting S and satisfying condition (MS) of Proposition 10. If b' is maximal (resp. prime), clearly b' is maximal in \mathscr{F} (resp. prime) and conversely (by (i)). On the other hand, if r is an ideal of A disjoint from S, its saturation r_1 with respect to S is an ideal of A containing t and disjoint from S : for no element $a \in S$ can satisfy $sa \in t$ for some $s \in S$, since it would follow that $sa \in t \cap S$. We conclude that, if $r$ is maximal among the ideals of $A$ meeting $S$, it is maximal in $\mathscr{F}$. Similarly, if $r$ is a prime ideal not meeting $S$, it satisfies condition (MS) of no. 4, Proposition 10 by definition of prime ideals and hence belongs to $\mathscr{F}$. This completes the proof of (ii).

(iii) Suppose that $q'$ is prime and such that $q$ is also prime. The set $T = A - q$ is a multiplicative subset of $A$ which contains $S$, whence $ST = T$. We write $T' = i_A^S(T)$; it follows from no. 3, Proposition 7 (i) that there exists a unique isomorphism $j$ of $T^{-1}A = A$, onto ${T'}^{-1}(S^{-1}A)$ such that

where $a \in A$ and $b \in T$. On the other hand $T'$ obviously does not meet $q'$; conversely, let $a/s \in S^{-1}A$; since $1/s$ is invertible in $S^{-1}A$, the condition $a/s \notin q'$ is equivalent to $i_A^S(a) = a/1 \notin q'$ and hence to $a \notin q$; it follows that $S^{-1}A - q' = S^{-1}T'$ and hence, by Proposition 8 of no. 3, ${T'}^{-1}(S^{-1}A) = (S^{-1}A)_{q'}$.

The isomorphism defined in (iii) is called canonical. If $A$ is an integral domain, the canonical isomorphisms of $A$, and $(S^{-1}A)_{q'}$ onto subrings of the field of fractions $K$ of $A$ have the same image.