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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