Proposition 9

Let $A$ be a ring and $(b_1, \ldots, b_n)$ two-sided ideals of $A$ such that $b_i + b_j = A$ for $i \neq j$. The canonical homomorphism of $A$ into $\prod_{i=1}^n (A/b_i)$ is surjective of kernel $\bigcap_{i=1}^n b_i = \sum_{\sigma \in S_n} b_{\sigma(1)} b_{\sigma(2)} \cdots b_{\sigma(n)}.$

Clearly the kernel is $\bigcap_{i=1}^n b_i$. To prove the surjectivity, it is necessary to show that, for every family $(x_i)_{1 \leq i \leq n}$ of elements of $A$, there exists $x \in A$ such that $x \equiv x_i (b_i)$ for all $1 \leq i \leq n$. We prove this assertion by induction on the cardinal $n$ of $I$, the case $n \leq 1$ being trivial. By the induction hypothesis, there exists $y \in A$ such that $y \equiv x_i (b_i)$ for $1 \leq i \leq n-1$. We look for an $x$ of the form $y + z$ with $z \in A$. Of necessity $z \equiv 0 (b_i)$ for $i < n$, that is $z \in b = \bigcap_{i=1}^{n-1} b_i$, and on the other hand $z \equiv x_n - y (b_n)$. Now $b_n + b = A$ by no. 9, Proposition 6, whence the existence of $z$. Finally, the second expression of the kernel follows from no. 9, Proposition 7.