Proposition 15

Let $A$ be a ring, $I$ a finite set and $E$ the set of orthogonal families $(e_i)_{i \in I}$ of idempotents $e_i \neq 0$ of $A$ such that $\sum_{i \in I} e_i = 1$. For all $(e_i)_{i \in I} \in E$, we set $\overline{\omega}((e_i)_{i \in I}) = (\mathrm{V}(A(1 - e_i)))_{i \in I}$, $\sigma((e_i)_{i \in I}) = (Ae_i)_{i \in I}$. Then $\overline{\omega}$ is a bijection of $E$ onto the set $P$ of partitions $(U_i)_{i \in I}$ of $X = \mathrm{Spec}(A)$ into open sets and $\sigma$ is a bijection of $E$ onto the set $S$ of families $(a_i)_{i \in I}$ of ideals $\neq 0$ of $A$ such that $A$ is the direct sum of the $a_i$.

Let $(e_i)_{i \in I}$ be an element of $E$ and set $Y_i = \mathrm{V}(A(1 - e_i))$; if $i \neq j$, then $1 = 1 - e_i + e_i(1 - e_j) \in A(1 - e_i) + A(1 - e_j)$, whence $Y_i \cap Y_j = \varnothing$ (formulae (1) and (2)). On the other hand,

(formula (3));

by hypothesis $\prod_{i \in I} (1 - e_i) = 1 - \sum_{i \in I} e_i = 0$, whence $\bigcup_{i \in I} Y_i = X$ (formula (I)). As the $Y_i$ are closed, they are also open, whence $\overline{\omega}(E) \subset P$. Also, obviously $A = \sum_{i \in I} Ae_i$; if $0 = \sum_{i \in I} a_i e_i$ where $a_i \in A$, we obtain, by multiplying by $e_i$, $O = a_i e_i^2 = a_i e_i$ for all $i$; this proves that $\sigma(E) \subset S$.