Proposition 10

*Let $A$ be a ring, $A'$ its centre and $(b_i)_{i \in I}$ a finite family of two-sided ideals of $A$. The following conditions are equivalent:

(a) the family $(b_i)_{i \in I}$ is a direct decomposition of $A$; (b) there exists a family $(e_i)_{i \in I}$ of idempotents of $A'$ such that $e_i e_j = 0$ for $i \neq j$, $1 = \sum_{i \in I} e_i$ and $b_i = A(1 - e_i)$ for $i \in I$; (c) $b_i + b_j = A$ for $i \neq j$ and $\bigcap_{i \in I} b_i = \{0\}$; (d) $b_i + b_j = A$ for $i \neq j$ and $\prod_{i \in I} b_i = \{0\}$ for every total order on $I$; (e) there exists a direct decomposition $(b'_i)_{i \in I}$ of $A'$ such that $b_i = A b'_i$ for $i \in I$.*

(a) $\Rightarrow$ (b). If condition (a) holds, $A$ may be identified with the ring $\prod_{i \in I} (A/b_i)$ and $b_i$ with the kernel of $\mathrm{pr}_i$. The existence of the $e_i$ with the properties in (b) then follows from no. 10.

(b) $\Rightarrow$ (d). Suppose that the $e_i$ exist with the properties in (b). For $i \neq j$, $1 - e_i \in b_i, \ e_i = e_i(1 - e_j) \in b_j$, hence $1 \in b_i + b_j$ and $A = b_i + b_j$. On the other hand, if I is given a total ordering and $(x_i)_{i \in I}$ is a family of elements of A, then, since the $e_i$ are central,

hence $\prod_{i \in I} b_i = \{0\}$.

(d) $\Rightarrow$ (c). This follows from no. 9, Proposition 7.

(c) $\Rightarrow$ (a). This follows from Proposition 9.

Thus conditions (a), (b), (c) and (d) are equivalent. Suppose they hold. By (b) $\Rightarrow$ (a), the family of $b'_i = A'(1 - e_i)$ is a direct decomposition of $A'$. Then $b_i = A(1 - e_i) = Ab'_i$ for all $i \in I$. Hence condition (e) holds.

Finally, suppose condition (e) holds. By (a) $\Rightarrow$ (b), there exists a family $(e_i)_{i \in I}$ of idempotents of $A'$ such that $e_i e_j = 0$ for $i \neq j$, $1 = \sum_{i \in I} e_i$ and $b'_i = A'(1 - e_i)$ for $i \in I$. Then $b_i = Ab'_i = A(1 - e_i)$ for $i \in I$, hence condition (b) holds.