Proposition 9

a) Order the set $\mathfrak{P}(\mathscr{S})$ of subsets of $\mathscr{S}$ and the set $\mathscr{B}_A$ of two-sided ideals of A by inclusion. The mapping $\Lambda \mapsto \mathfrak{a}_{\Lambda}$ is an isomorphism of ordered sets from $\mathfrak{P}(\mathscr{S})$ to $\mathscr{B}_A$.

b) The minimal two-sided ideals of A are the ideals $\mathfrak{a}_{\lambda}$.

c) We have $\mathfrak{b}_{\lambda}=\mathfrak{a}_{\mathscr{S}-\{\lambda\}}$ for every $\lambda \in \mathscr{S}$, and the ideals $\mathfrak{b}_{\lambda}$ are the maximal two-sided ideals of A.

d) For every $\lambda \in \mathscr{S}$, the canonical mapping from A to $A/\mathfrak{b}_{\lambda}$ induces an isomorphism of A-modules from $\mathfrak{a}_{\lambda}$ to $A/\mathfrak{b}_{\lambda}$.

Assertion a) follows from Proposition 8, d) of VIII, p. 87 applied to the A-module $A_s$. It follows that the minimal two-sided ideals of A are the $\mathfrak{a}_{\lambda}$ and that the maximal two-sided ideals are the ideals $\mathfrak{c}_{\lambda}=\mathfrak{a}_{\mathscr{S}-\lambda}$ (for $\lambda \in \mathscr{S}$).

It remains to show the equality of $\mathfrak{b}_{\lambda}$ and $\mathfrak{c}_{\lambda}$ for every $\lambda \in \mathscr{S}$. Let $\lambda$ and $\mu$ be distinct in $\mathscr{S}$. The A-submodule $\mathfrak{a}_\mu S_{\lambda}$ of $S_{\lambda}$ is the union of the images of the linear mappings $a\mapsto ax$ from $\mathfrak{a}_\mu$ to $S_{\lambda}$ for $x\in S_{\lambda}$. Consequently, it is zero, and we have $\mathfrak{a}_\mu\subset \mathfrak{b}_{\lambda}$. We therefore have $\mathfrak{c}_{\lambda}\subset \mathfrak{b}_{\lambda}$, and finally $\mathfrak{c}_{\lambda}=\mathfrak{b}_{\lambda}$ because $\mathfrak{c}_{\lambda}$ is a maximal two-sided ideal of A and $\mathfrak{b}_{\lambda}$ is distinct from A.