Lemma 33.42.5. Let $A$ be a ring, $I \subset A$ an ideal, $\mathfrak p_1, \ldots , \mathfrak p_ r$ primes of $A$, and $\overline{f} \in A/I$ an element. If $I \not\subset \mathfrak p_ i$ for all $i$, then there exists an $f \in A$, $f \not\in \mathfrak p_ i$ which maps to $\overline{f}$ in $A/I$.

Proof. We may assume there are no inclusion relations among the $\mathfrak p_ i$ (by removing the smaller primes). First pick any $f \in A$ lifting $\overline{f}$. Let $S$ be the set $s \in \{ 1, \ldots , r\}$ such that $f \in \mathfrak p_ s$. If $S$ is empty we are done. If not, consider the ideal $J = I \prod _{i \not\in S} \mathfrak p_ i$. Note that $J$ is not contained in $\mathfrak p_ s$ for $s \in S$ because there are no inclusions among the $\mathfrak p_ i$ and because $I$ is not contained in any $\mathfrak p_ i$. Hence we can choose $g \in J$, $g \not\in \mathfrak p_ s$ for $s \in S$ by Algebra, Lemma 10.15.2. Then $f + g$ is a solution to the problem posed by the lemma. $\square$

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