Lemma 10.15.3. Let $R$ be a ring. Let $x \in R$, $I \subset R$ an ideal, and $\mathfrak p_ i$, $i = 1, \ldots , r$ be prime ideals. Suppose that $x + I \not\subset \mathfrak p_ i$ for $i = 1, \ldots , r$. Then there exists an $y \in I$ such that $x + y \not\in \mathfrak p_ i$ for all $i$.

Proof. We may assume there are no inclusions among the $\mathfrak p_ i$. After reordering we may assume $x \not\in \mathfrak p_ i$ for $i < s$ and $x \in \mathfrak p_ i$ for $i \geq s$. If $s = r + 1$ then we are done. If not, then we can find $y \in I$ with $y \not\in \mathfrak p_ s$. Choose $f \in \bigcap _{i < s} \mathfrak p_ i$ with $f \not\in \mathfrak p_ s$. Then $x + fy$ is not contained in $\mathfrak p_1, \ldots , \mathfrak p_ s$. Thus we win by induction on $s$. $\square$

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