In algebra, the prime avoidance lemma says that if an ideal I in a commutative ring R is contained in a union of finitely many prime ideals P_i's, then it is contained in P_i for some i.

There are many variations of the lemma (cf. Hochster); for example, if the ring R contains an infinite field or a finite field of sufficiently large cardinality, then the statement follows from a fact in linear algebra that a vector space over an infinite field or a finite field of large cardinality is not a finite union of its proper vector subspaces.

The following statement and argument are perhaps the most standard.

Theorem (Prime Avoidance Lemma): Let E be a subset of commutative ring R that is an additive subgroup of R and is multiplicatively closed. (In particular, E could be a subring or ideal of R.) Let ${\displaystyle I_{1},I_{2},\dots ,I_{n},n\geq 1}$ be ideals such that ${\displaystyle I_{i}}$ are prime ideals for ${\displaystyle i\geq 3}$. If E is not contained in any of the ${\displaystyle I_{i}}$, then E is not contained in the union ${\textstyle \bigcup I_{i}}$.

Proof by induction on n: The idea is to find an element of R that is in E and not in any of the ${\displaystyle I_{i}}$. The base case ${\displaystyle n=1}$ is trivial. Next suppose ${\displaystyle n\geq 2}$. For each i, choose

where each of the sets on the right is nonempty by the inductive hypothesis. We can assume ${\displaystyle z_{i}\in I_{i}}$ for all i; otherwise, there is some ${\displaystyle z_{k}}$ among them that avoids all of the ${\displaystyle I_{i}}$, and we are done. Put

Because E is closed under addition and multiplication, z is in E by construction. We claim that z is not in any of the ${\displaystyle I_{i}}$. Indeed, if ${\displaystyle z\in I_{i}}$ for some ${\displaystyle i\leq n-1}$, then ${\displaystyle z_{n}\in I_{i}}$, a contradiction. Next suppose ${\displaystyle z\in I_{n}}$. Then ${\displaystyle z_{1}\cdots z_{n-1}\in I_{n}}$. If ${\displaystyle n=2}$, this is already a contradiction. If ${\displaystyle n>2}$, then, since ${\displaystyle I_{n}}$ is a prime ideal, ${\displaystyle z_{i}\in I_{n}}$ for some ${\displaystyle i\leq n-1}$, again a contradiction. ${\displaystyle \square }$

There is the following variant of prime avoidance due to E. Davis.