# Prime avoidance lemma

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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 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 ${\displaystyle I_{1},I_{2},\dots ,I_{n},n\geq 1}$ be ideals such that ${\displaystyle I_{i}}$ are prime ideals for ${\displaystyle i\geq 3}$. If E is not contained in any of ${\displaystyle I_{i}}$'s, then E is not contained in the union ${\displaystyle \cup I_{i}}$.

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

${\displaystyle z_{i}\in E-\cup _{j\neq i}I_{j}}$

where the set on the right is nonempty by inductive hypothesis. We can assume ${\displaystyle z_{i}\in I_{i}}$ for all i; otherwise, some ${\displaystyle z_{i}}$ avoids all the ${\displaystyle I_{i}}$'s and we are done. Put

${\displaystyle z=z_{1}\dots z_{n-1}+z_{n}}$.

Then z is in E but not in any of ${\displaystyle I_{i}}$'s. Indeed, if z is in ${\displaystyle I_{i}}$ for some ${\displaystyle i\leq n-1}$, then ${\displaystyle z_{n}}$ is in ${\displaystyle I_{i}}$, a contradiction. Suppose z is in ${\displaystyle I_{n}}$. Then ${\displaystyle z_{1}\dots z_{n-1}}$ is in ${\displaystyle I_{n}}$. If n is 2, we are done. If n > 2, then, since ${\displaystyle I_{n}}$ is a prime ideal, some ${\displaystyle z_{i},i is in ${\displaystyle I_{n}}$, a contradiction.

## Notes

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.