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 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] 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.

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

The basic case n = 1 is trivial.

For each i, choose where the set on the right is nonempty by inductive hypothesis.

is a prime ideal, some

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

Theorem — [2] Let A be a ring,

prime ideals, x an element of A and J an ideal.

Proof:[3] We argue by induction on r. Without loss of generality, we can assume there is no inclusion relation between the

's; since otherwise we can use the inductive hypothesis.

for each i, then we are done; thus, without loss of generality, we can assume

By inductive hypothesis, we find a y in J such that

is a prime ideal, we have: Hence, we can choose

, the element

has the required property.

Let A be a Noetherian ring, I an ideal generated by n elements and M a finite A-module such that

= the maximal length of M-regular sequences in I = the length of every maximal M-regular sequence in I.

; this estimate can be shown using the above prime avoidance as follows.

We argue by induction on n. Let

be the set of associated primes of M. If

, then, by prime avoidance, we can choose for some

= the set of zero divisors on M. Now,

elements and so, by inductive hypothesis,