Lemma 6

Let $R$ be a commutative ring and $p_i$ ($1 \leq i \leq n$) distinct prime ideals of $R$.

(i) For $1 \leq i \leq n$, let $H_i$ be a subset of $R/p_i$ satisfying the following condition: there exists no element $\alpha_i \in R/p_i$ such that $\alpha_i + H_i$ contains an ideal $\neq 0$ of $R/p_i$. Then there exists $a \in R$ such that, for $1 \leq i \leq n$, the canonical image of $a$ in $R/p_i$ does not belong to $H_i$.

(ii) If $\operatorname{Card}(H_i) < \operatorname{Card}(R/p_i)$, the $H_i$ satisfy the condition in (i).

(i) We argue by induction on $n$, the case $n = 0$ being trivial. Therefore let $n \geq 1$. We may make a permutation on the indices $i$ and may assume that $p_1$ is minimal among the $p_i$ and therefore, for $2 \leq i \leq n$, there exists $c_i \in p_i$ such that $c_i \notin p_1$. By the induction hypothesis there exists $b \in R$ such that the canonical image of $b$ in $R/p_i$ does not belong to $H_i$ for $2 \leq i \leq n$. For all $x \in R$ we write $a_i = b + x c_2 c_3 \ldots c_n$; as $c_i \in p_i$, obviously $a_i \equiv b \pmod{p_i}$ for $2 \leq i \leq n$. It therefore suffices to prove that there exists $x \in \mathbf{R}$ such that the canonical image of $a_i$ in $\mathbf{R}/p_1$ does not belong to $H_1$. Now, the set of canonical images of the $a_i$ in $\mathbf{R}/p_1$, where $x$ runs through $\mathbf{R}$, is just $\beta + c$, where $\beta$ is the canonical image of $b$ and $c$ the ideal of $\mathbf{R}/p_1$ generated by the canonical image of $c_2 c_3 \ldots c_n$; by virtue of the choice of the $c_i$, $c \neq 0$ since $\mathbf{R}/p_1$ is an integral domain and the hypothesis on $H_1$ implies the existence of an $x$ which solves the problem.

(ii) As $\mathbf{R}/p_i$ is an integral domain, every ideal $\neq 0$ of $\mathbf{R}/p_i$ has cardinal equal to that of $\mathbf{R}/p_i$ and the same is true of any translation of an ideal by an element of $\mathbf{R}/p_i$, whence the conclusion.