Proposition 5

*Let $(a_i)_{i \in I}$ be a non-empty finite family of ideals of a ring $A$. The following properties are equivalent = (a) For $i \neq j$, $a_i$ and $a_j$ are relatively prime. (b) The canonical homomorphism $\phi : A \to \prod_{i \neq I} (A/a_i)$ (Algebra, Chapter 11, § 1, no. 7) is surjective. If these hold, the intersection $a$ of the $a_i$ is equal to their product and the canonical homomorphism $\phi : A/a \to \prod_{i \in I} (A/a_i)$ (Algebra, Chapter II, § 1, no. 7) is bijective.*

We argue by induction on $n$ the number of elements in $I$, the case $n = 1$ being trivial. Consider first the case $n = 2$. Then the equivalence of (a) and (b) follows from the exactness of the sequence

(Algebra, Chapter II, § 1, no. 7, formula (30)). Moreover, there exist $e_1 \in a$, and $e_2 \in a_2$ such that $1 = e_1 + e_2$; then, for all $x \in a = a_1 \cap a_2$, $x = xe_1 + xe_2$; but by definition $xe_1 \in a_1 a_2$ and $xe_2 \in a_1 a_2$, hence $x \in a_1 a_2$; whence $a \subset a_1 a_2$ and the converse inclusion is obvious.

In the general case, suppose that condition (a) holds and let $k$ be an element of $I$ and $b_k = \bigcap_{i \neq k} a_i$; the induction hypothesis implies that $b_k = \prod_{i \in I} a_i$, and it follows from Proposition 4 that $a$, and $b$, are relatively prime; then

by the first part of the argument and for the same reason the canonical homomorphism $A/a \twoheadrightarrow (A/a_k) \times (A/b_k)$ is bijective; by the induction hypothesis the canonical homomorphism $A/b_k \to \prod_{i \neq k} (A/a_i)$ is bijective and so then is the composite homomorphism

which is precisely $\psi$; that is, (b) holds. Conversely, suppose that (b) holds. We show that the $a_i$ are necessarily relatively prime in pairs. In the contrary case, there would exist an ideal $c \neq A$ containing $a_i$ and $a_j$, for $i \neq j$. We set $a_h' = a_h$ for $h$ not equal to $i$ or $j$ and $a_i' = a_j' = c$; the canonical homomorphism $\phi': A \to \prod_{i \in I} (A/a_i')$ can be written as the composite mapping

$f$ being the product of the canonical homomorphisms $A/a_i \to A/a_i'$; clearly $\phi'$ is not surjective, the projection of $+(A)$ onto $(A/a_i') \times (A/a_i')$ being the diagonal of the product $(A/c) \times (A/c)$, which is distinct from this product since $c \neq A$. As $f$ is surjective, this shows that $\phi$ is not surjective.