Prime avoidance lemma
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.
Statement and proof
The following statement and argument are perhaps the most standard.
Statement: Let E be a subset of R that is an additive subgroup of R and is multiplicatively closed. Let be ideals such that are prime ideals for . If E is not contained in any of 's, then E is not contained in the union .
Proof by induction on n: The idea is to find an element that is in E and not in any of 's. The basic case n = 1 is trivial. Next suppose n ≥ 2. For each i choose
where the set on the right is nonempty by inductive hypothesis. We can assume for all i; otherwise, some avoids all the 's and we are done. Put
Then z is in E but not in any of 's. Indeed, if z is in for some , then is in , a contradiction. Suppose z is in . Then is in . If n is 2, we are done. If n > 2, then, since is a prime ideal, some is in , a contradiction.
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