Prime avoidance lemma

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 Pi's, then it is contained in Pi 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 cardinarity), then the statement follows from a fact in linear algebra that a vector space over an infinite field is not a finite union of its proper vector subspaces. (Proof of the fact: suppose the vector space is a finite union of proper subspaces. Consider a finite product of linear functionals, each of which vanishes on a proper subspace that appears in the union; then it is a nonzero polynomial vanishing identically, a contradiction.)

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 I_1, I_2, \dots, I_n, n \ge 1 be ideals such that I_i are prime ideals for i \ge 3. If E is not contained in any of I_i's, then E is not contained in the union \cup I_i.

Proof by induction on n: The idea is to find an element that is in E and not in any of I_i's. The basic case n = 1 is trivial. Next suppose n ≥ 2. For each i choose

z_i \in E - \cup_{j \ne i} I_j

where the set on the right is nonempty by inductive hypothesis. We can assume z_i \in I_i for all i; otherwise, some z_i avoids all the I_i's and we are done. Put

z = z_1 \dots z_{n-1} + z_n.

Then z is in E but not in any of I_i's. Indeed, if z is in I_i for some i \le n - 1, then z_n is in I_i, a contradiction. Suppose z is in I_n. Then z_1 \dots z_{n-1} is in I_n. If n is 2, we are done. If n > 2, then, since I_n is a prime ideal, some z_i, i < n is in I_n, a contradiction.

References