Proposition 7

Let $b_1, \ldots, b_n$ be two-sided ideals of $A$ such that $b_i + b_j = A$ for $i \neq j$. Then $b_1 \cap b_2 \cap \cdots \cap b_n = \sum_{\sigma \in S_n} b_{\sigma(1)} b_{\sigma(2)} \cdots b_{\sigma(n)}$. In particular, if $A$ is commutative, $b_1 \cap b_2 \cap \cdots \cap b_n = b_1 b_2 \ldots b_n$ (cf. Exercise 2).

Suppose first $n = 2$. There exist $a_1 \in b_1, a_2 \in b_2$ such that $a_1 + a_2 = 1$. If $x \in b_1 \cap b_2$, then $x = x(a_1 + a_2) = x a_1 + x a_2 \in b_2 b_1 + b_1 b_2$. Hence $b_1 \cap b_2 = b_1 b_2 + b_2 b_1$.

Suppose now the equation of the proposition is established for all integers $< n$. By Proposition 6, $b_n + (b_1 b_2 \ldots b_{n-1}) = A$ and hence