Prime avoidance lemma

From Justapedia, unleashing the power of collective wisdom
(Redirected from Prime avoidance)
Jump to navigation Jump to search

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 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.

Theorem — [2] Let A be a ring, prime ideals, x an element of A and J an ideal. For the ideal , if for each i, then there exists some y in J such that for each i.

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.

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.

Application

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.

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.
  2. ^ Matsumura, Exercise 16.8.
  3. ^ Adapted from the solution to Matsumura, Exercise 1.6.

References

  • Mel Hochster, Dimension theory and systems of parameters, a supplementary note
  • Matsumura, Hideyuki (1986). Commutative ring theory. Cambridge Studies in Advanced Mathematics. Vol. 8. Cambridge University Press. ISBN 0-521-36764-6. MR 0879273. Zbl 0603.13001.