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.
E. Davis' prime avoidance
There is the following variant of prime avoidance due to E. Davis.
Also, if 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 . If is not in , we are done. Otherwise, note that (since ) and since is a prime ideal, we have:
Hence, we can choose in that is not in . Then, since , 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 . Also, let = the maximal length of M-regular sequences in I = the length of every maximal M-regular sequence in I. Then ; 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 for each i. If , then, by prime avoidance, we can choose
for some in such that = the set of zerodivisors on M. Now, is an ideal of generated by elements and so, by inductive hypothesis, . The claim now follows.
- 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.
- Matsumura, Exercise 16.8.
- Adapted from the solution to Matsumura, Exercise 1.6.