Proposition 6

Let $a, b_1, \ldots, b_n$ be two-sided ideals of $A$. If $A = a + b_i$ for all $i$, then $A = a + b_1 b_2 \ldots b_n = a + (b_1 \cap b_2 \cap \cdots \cap b_n)$.

By (21) it suffices to prove that $A = a + b_1 b_2 \ldots b_n$. By induction, it suffices to consider the case where $n = 2$. By hypothesis, there exist $a, a' \in a, b_1 \in b_1, b_2 \in b_2$ such that $1 = a + b_1 = a' + b_2$. Then

whence $A = a + b_1 b_2$.