Proposition 10

Let $A$ be a Noetherian ring and $p_1, \ldots, p_n$ the prime ideals associated with the A-module $A$, where $p_i \neq p_j$ for $i \neq j$.

(i) The set $S = \bigcap_{i=1}^n (A - p_i)$ is the set of elements which are not divisors of 0 in $A$.

(ii) If all the $p_i$ are minimal elements of $\operatorname{Ass}(A)$, the total ring of fractions $S^{-1}A$ of $A$ is Artinian.

(iii) If the ring $A$ is reduced, all the $p_i$ are minimal elements of $\operatorname{Ass}(A)$ (and therefore are the minimal elements of $\operatorname{Spec}(A)$) and each of the $A_{p_i}$ is a field; for each index $i$, the canonical homomorphism $S^{-1}A \to A_{p_i}$ (Chapter II, § 2, no. 1, Corollary 1 to Proposition 2) is surjective and its kernel is $S^{-1}p_i$; finally the canonical homomorphism from $S^{-1}A$ to $\prod_{i=1}^n (S^{-1}A/S^{-1}p_i)$ is bijective.

The fact that $S$ is the set of elements which are not divisors of 0 in $\mathbf{A}$ has already been seen (§ 1, no. 1, Corollary 3 to Proposition 2). The prime ideals of $S^{-1}A$ are of the form $S^{-1}p$, where $p$ is a prime ideal of $A$ contained in $\bigcup_{i=1}^n p_i$ (Chapter 11, § 2, no. 5, Proposition 10) that is contained in one of the $p_i$ (Chapter II, § 1, no. 1, Proposition 2). If $p_i$ is a minimal element of $\operatorname{Ass}(A)$, it is a minimal element of $\operatorname{Spec}(A)$ (§ 1, no. 3, Corollary to Proposition 7); if each of the $p_i$ is a minimal element of $\operatorname{Ass}(A)$, we then see that the prime ideals of $S^{-1}A$ are the $S^{-1}p_i$ and they are therefore all maximal, which proves that $S^{-1}A$ is Artinian (Proposition 9).

Suppose finally that the ring $A$ is reduced. Then $\bigcap_{i=1}^n p_i = \{0\}$ (§ 1, no. 3, Corollary 2 to Proposition 7). We deduce that $\{0\} = \bigcap_{i=1}^n p_i$ is a reduced primary decomposition of the ideal $\{0\}$ (no. 3, Corollary to Proposition 4); in particular, none of the $p_i$ can contain a $p_j$ of index $j \neq i$ and therefore the $p_i$ are all minimal elements of $\operatorname{Ass}(A)$. The ring $S^{-1}A$ is then Artinian by (ii). The $S^{-1}p_i$ are prime ideals associated with the $S^{-1}A$-module $S^{-1}A$ (§ 1, no. 2, Corollary to Proposition 5) and $\{0\} = S^{-1} \left( \bigcap_{i=1}^n p_i \right) = \bigcap_{i=1}^n S^{-1}p_i$ (Chapter II, § 2, no. 4); as the $S^{-1}p_i$ are distinct, $(S^{-1}p_i)_{1 \leq i \leq n}$ is a reduced primary decomposition of $\{0\}$ in $S^{-1}A$ (no. 3, Corollary to Proposition 4). Proposition 8 then shows that the canonical homomorphismg,: $S^{-1}A \to (S^{-1}A)_{p_i}$ is surjective and has kernel $S^{-1}p_i$ and the canonical homomorphism $S^{-1}A \to \prod_{i=1}^n (S^{-1}A/S^{-1}p_i)$ is bijective. We know moreover that the canonical homomorphism $S^{-1}A \to A$, is composed of g, and an isomorphism $(S^{-1}A)_{S^{-1}p_i} \to A$, (Chapter II, § 2, no. 3, Proposition 7). Finally, it follows from Proposition 8 that $(S^{-1}A)_{S^{-1}p_i}$ is isomorphic to $S^{-1}A/S^{-1}p_i$ and hence is a field since $S^{-1}p_i$ is a maximal ideal.