Lemma 2

If $e, f$ are two idempotents of $A$ such that $Ae$ and $Af$ have the same radical then $e = f$.

There exists by hypothesis integers $m \geq 0, n \geq 0$ such that $e = e^m \in Af$ and $f = f^n \in Ae$; let $x, y$ be elements of $A$ such that $e = xf, f = ye$; then $ef = x^2 = x = e$ and similarly $ef = ye^2 = ye = f$, whence $e = f$.

Lemma 2 and Corollary 2 to Proposition 11 show that the mappings $\overline{\omega}$ and $\sigma$ are injective.

Let us show that $\sigma$ is surjective. If $(a_i)_{i \in I}$ is an element of $S$, there are elements $e_i \in a_i$ such that $1 = \sum_{i \in I} e_i$; if $i \neq j$, then $e_i e_j \in a_i \cap a_j = \{0\}$, whence

finally, $Ae_i \subset a_i$, for all $i \in I$ and $\sum_{i \in I} Ae_i = A$, whence $Ae_i = a_i$.

It remains to prove that $\overline{\omega}$ is surjective. Let $(U_i)_{i \in I}$ be an element of $P$ and set $Z_i = \bigcup_{j \neq i} U_j$; as $U_i$ and $Z_i$ are closed, there exist ideals $a_i, b_i$ of $A$ such that $U_i = \mathrm{V}(a_i), Z_i = \mathrm{V}(b_i)$. We now show that it is possible to suppose further that $a_i \cap b_i = 0$. Now $U_i \cap Z_i = \varnothing$, whence $a_i + b_i = A$; let $a_i, b_i \in a_i, b_i$ be such that $a_i + b_i = 1$. Then $X = U_i \cup Z_i = \mathrm{V}(a_i b_i)$ (formula (3)); every element of $a_i b_i$ is therefore nilpotent (Corollary 2 to Proposition 11); let $p$ be an integer such that $a_i^p b_i^p = 0$. Then $U_i \subset \mathrm{V}(Aa_i) = \mathrm{V}(Aa_i^p)$,

and $V(Aa_i) \cap V(Ab_i) = V(Aa_i + Ab_i) = \varnothing$, hence $U_i = V(Aa_i^p)$ and $Z_i = V(Ab_i^p)$, which establishes our assertion by replacing $a_i$ by $Aa_i^p$ and $b_i$ by $Ab_i^p$. The ideals $a_i$ and $b_i$ thus chosen, it follows from the fact that $\sigma$ is bijective that there exist two idempotents $f_i \in a_i, e_i \in b_i$ such that $1 = e_i + f_i, e_i f_i = 0, a_i = Af_i, b_i = Ae_i$. If $i \neq j$, then $X = Z, \cup Z_j = V(Ae_i e_j)$, and as $e_i e_j$ is idempotent, Lemma 2 shows that $e_i e_j = 0$. Finally $e = \sum_{i \in I} e_i$ is idempotent and $e_i \in Ae$ for all $i \in I$, whence $V(Ae) \subset Z$, for all $i$; it follows that

and Lemma 2 shows also that $e = 1$.