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 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.^[1]

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 $I_{1},I_{2},\dots ,I_{n},n\geq 1$ be ideals such that $I_{i}$ are prime ideals for $i\geq 3$ . 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\neq 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 $i\leq n-1$ , 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.

Notes

↑ 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.

References

Mel Hochster, Dimension theory and systems of parameters, a supplementary note

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[1] 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.