Suppose in fact that $p$ contains none of the $a_i$. For $1 \leq i \leq n$ there exists then an element $s_i \in a_i \cap p$; then $s = s_1 s_2 \ldots s_n$ is contained in $a_1 a_2 \ldots a_n$, and is not contained in $p$, which is absurd.

Such an ideal $p$ must contain $m$ by Proposition 1 applied to $a_i = m$ for $1 \leq i \leq n$; as $m$ is maximal, $p = m$.

We argue by induction on $n = \mathrm{Card}(I)$; the proposition is trivial if $n = 1$. Suppose that $n \geq 2$; if there exists an index $j$ such that $a \cap p_j \subset \bigcup_{i \neq j} p_i$, the set $a$, which is the union of the $a \cap p_i$ where $i \in I$, is contained in $\bigcup_{i \neq j} p_i$ and hence in one of the $p_i$ by the induction hypothesis. Suppose then that such an index does not exist; for every $j \in I$ let $y_j$ be an element of $a \cap p_j$ not belonging to any $p_i$ for $i \neq j$. Let $k$ be an element of $I$ chosen in such a way that $p_k$ is prime if $n > 2$ and chosen arbitrarily if $n = 2$; let $z = y_k + \prod_{i \neq k} y_i$. Then $z \in a$, since $a$ is closed under addition and multiplication; if $j \neq k$, $\prod_{i \neq k} y_i$ belongs to $p_j$, but $y_k \notin p_j$, whence $z \notin p_j$. On the other hand, $\prod_{i \neq k} y_i$ does not belong to $p_k$, as none of the factors $y_i \ (i \neq k)$ belongs to it and $p_k$ is prime if $n - 1 > 1$; as $y_k \in p_k$, $z$ does not belong to $p_k$ and the proposition is established.

Under the given hypothesis, every prime ideal which contains $a'$ contains $a$ and every prime ideal which contains $b'$ contains $b$. If a prime ideal contains $a'$ and $b'$, then it contains $a$ and $b$, which is absurd, since $a$ and $b$ are relatively prime; hence $a'$ and $b'$ are relatively prime.

Let $p$ be a prime ideal of $A$. If $p$ contains $a$ and $b_1 b_2 \ldots b_n$ it contains one of the $b_i$ (no. 1, Proposition 1), which is absurd since $a$ and $b_i$ are relatively prime.

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.

Clearly the canonical mapping of $M$ to $\prod_{i \in I} (M/a_i M)$ is zero on $aM$; then, by taking quotients, it defines a homomorphism $A : M/aM \to \prod_{i \in I} (M/a_i M)$. On the other hand, by Proposition 5, the canonical homomorphism

is bijective. Then so is $1_M \otimes \psi : M \otimes (A/a) \to M \otimes \prod_{i \in I} (A/a_i)$. Now $M \otimes (A/a)$ is identified with $M/aM$ and $M \otimes \prod_{i \in I} (A/a_i)$ with $\prod_{i \in I} M \otimes (A/a_i)$, which is itself identified with $\prod_{i \in I} (M/a_i M)$. It is immediately verified that the above identifications transforms $1_M \otimes \psi$ into $\lambda$, whence the proposition.