Prime Avoidance Lemma

Statement

Let $R$ be a commutative ring and let $I, J_1,\dots,J_n$ be ideals of $R$. Assume that $J_2,\dots,J_n$ are prime ideals . If

then $I \subseteq J_k$ for some $k\in\{1,\dots,n\}$.

Equivalently (the “avoidance” form): if $I\not\subseteq J_i$ for every $i$, then there exists an element

This lemma is frequently used to choose elements that avoid finitely many primes, e.g. when forming a localization or proving existence of “good” elements in a Noetherian ring .

Examples

1. Avoiding two coordinate primes.
   Let $R=k[x,y]$, $J_1=(x)$, $J_2=(y)$. Both are prime.
   Take $I=(x,y)$. Then $I\not\subseteq (x)$ and $I\not\subseteq (y)$. Prime avoidance guarantees some $f\in (x,y)$ is in neither $(x)$ nor $(y)$.
   Concretely, $f=x+y\in (x,y)$, and $x+y\notin (x)$, $x+y\notin (y)$.

2. Why “finite union” matters (failure for infinite unions).
   Let $R=k[x_1,x_2,x_3,\dots]$ (infinitely many variables) and $I=(x_1,x_2,x_3,\dots)$.
   For each $n\ge 1$, set $P_n=(x_1,\dots,x_n)$. Each $P_n$ is prime, and

   $$I=\bigcup_{n\ge 1} P_n$$

   because any polynomial in $I$ involves only finitely many variables.
   But $I\not\subseteq P_n$ for any fixed $n$ (since $x_{n+1}\in I\setminus P_n$).
   So prime avoidance can fail for infinite unions.

3. Picking a non-nilpotent element.
   If $R$ has finitely many minimal prime ideals $P_1,\dots,P_r$ and an ideal $I$ is not contained in any $P_i$, then prime avoidance yields $x\in I\setminus \bigcup_i P_i$.
   In particular, $x$ is not contained in every prime ideal, so $x$ is not nilpotent (compare nilradical ).