Proposition 3

Let $A$ be a ring, $p$ a prime ideal of $A$ and $S = A - p$. For every ideal $b'$ of $A_p$ distinct from $A_p$, let $b$ be the ideal $(i_A^S)^{-1}(b')$ of $A$ so that $b' = bA_p$.

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

(ii) The mapping $b' \to b = (i_A^S)^{-1}(b')$, restricted to the set of prime ideals of $A_p$, is an isomorphism (with respect to inclusion) of this set onto the set of prime ideals of $A$ contained in $p$. If $b'$ is prime in $A_p$, there exists an isomorphism of the ring $A_b$ onto the ring $(A_p)_{b'}$, which maps $a/s$ to $(a/1)/(s/1)$ for all $a \in A, s \in A - b$.

This is just a particular case of $§ 2$, no. 5, Proposition 11.

Remarks (3) If $a$ is an ideal of $A$ not contained in $p$, then $aA_p = A$, and $(A/a)_p = 0$ ($§ 2$, no. 5, Remark). (4) Let $A, B$ be two rings, $\rho : A \to B$ a homomorphism, $q$ a prime ideal of $B$ and $p$ the prime ideal $\rho^{-1}(q)$ of $A$. As $\rho(A - p) \subset B - q$, a canonical homomorphism $\rho_q : A \to B_q$ is derived from $\rho$ ($§ 2$, no. 1, Proposition 2) and it is immediate that $\rho_q(pA_p) \subset qB_q$, hence $\rho$ is a local homomorphism.