Prime avoidance lemma

Equivalently (and often how it is used): if $I$ is an ideal not contained in any of the prime ideals $\mathfrak p_1,\dots,\mathfrak p_n$, then there exists an element $x\in I$ such that $x\notin \mathfrak p_1\cup\cdots\cup \mathfrak p_n$.

This “element-choosing” form underlies many constructions, for example when passing to a localization to avoid certain primes or when proving facts about basic open sets in the Zariski topology.

1. In $\mathbb Z$. Take $R=\mathbb Z$, $I=(6)$, and $\mathfrak p_1=(2)$, $\mathfrak p_2=(3)$ (both prime ideals). Then
   $$(6)\subseteq (2)\cup(3)$$
   because every multiple of $6$ is divisible by $2$ (and also by $3$). Prime avoidance correctly concludes that $(6)\subseteq (2)$ or $(6)\subseteq (3)$ (in fact both hold).

1. Choosing an element outside a finite union. In $R=k[x,y]$ over a field $k$, let $\mathfrak p_1=(x)$ and $\mathfrak p_2=(y)$, both prime. The ideal $I=(x,y)$ is not contained in $\mathfrak p_1$ and is not contained in $\mathfrak p_2$. Prime avoidance therefore guarantees an element of $I$ outside $\mathfrak p_1\cup\mathfrak p_2$; concretely, $x+y\in (x,y)$ but $x+y\notin (x)$ and $x+y\notin (y)$.

1. Why finiteness matters (failure for infinite unions). Let $R=k[x_1,x_2,\dots]$ be a polynomial ring in infinitely many variables over a field $k$, and let
   $$I=(x_1,x_2,\dots).$$
   For each $n$, the ideal $\mathfrak p_n=(x_1,\dots,x_n)$ is prime. Every element of $I$ involves only finitely many variables, so it belongs to $\mathfrak p_n$ for some $n$, hence
   $$I \subseteq \bigcup_{n\ge 1}\mathfrak p_n.$$
   But $I$ is not contained in any single $\mathfrak p_n$ (since $x_{n+1}\in I$ but $x_{n+1}\notin \mathfrak p_n$). This shows the lemma is genuinely a finite-union phenomenon.