Prime avoidance lemma explained

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,} ... , I _{n,} n \ge 1 be ideals such that I _{i} are prime ideals for i \ge 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, choosez _{i} \in E - \cup _{j} 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. Putz = z _{1} ... 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 \le n - 1 , then z _{n} is in I _{i} , a contradiction. Suppose z is in I _{n} . Then z _{1} ... 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.

E. Davis' prime avoidance

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

Proof:^[2] We argue by induction on r. Without loss of generality, we can assume there is no inclusion relation between the

ak{p} _{i} 's; since otherwise we can use the inductive hypothesis.

Also, if

x \not\in ak{p} _{i} for each i, then we are done; thus, without loss of generality, we can assume x \in ak{p} _{r} . By inductive hypothesis, we find a y in J such that x + y \in I - \cup ^{r -1} _{1} ak{p} _{i} . If x + y is not in ak{p} _{r} , we are done. Otherwise, note that J \not\subset ak{p} _{r} (since x \in ak{p} _{r} ) and since ak{p} _{r} is a prime ideal, we have:ak{p} _{r} \not\supset J ak{p} _{1} … ak{p} _{r-1} .Hence, we can choose y' in J ak{p} _{1} … ak{p} _{r-1} that is not in ak{p} _{r} . Then, since x + y \in ak{p} _{r} , the element x + y + y' has the required property. \square

Application

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

IM \ne M . Also, let d = \operatorname{depth} _{A(I,} M) = the maximal length of M-regular sequences in I = the length of every maximal M-regular sequence in I. Then d \le n ; this estimate can be shown using the above prime avoidance as follows. We argue by induction on n. Let \{ ak{p} _{1,} ... , ak{p} _{r} \} be the set of associated primes of M. If d > 0 , then I \not\subset ak{p} _{i} for each i. If I = (y _{1,} ... , y _{n)} , then, by prime avoidance, we can choosex _{1} = y _{1} + \sum ^{n} _{i =2} a _{i} y _{i} for some a _{i} in A such that x _{1} \not\in \cup ^{r} _{1} ak{p} _{i} = the set of zero divisors on M. Now, I/(x _{1)} is an ideal of A/(x _{1)} generated by n - 1 elements and so, by inductive hypothesis, \operatorname{depth} _{A/(x _{1)}} (I/(x _{1),} M/x _{1M)} \le n - 1 . The claim now follows.

References

• Mel Hochster, Dimension theory and systems of parameters, a supplementary note
• Book: Matsumura. Hideyuki. 1986. [{{google books|yJwNrABugDEC|Commutative ring theory|plainurl=yes|page=123}} Commutative ring theory ]. Cambridge Studies in Advanced Mathematics. 8. Cambridge University Press. 0-521-36764-6. 0879273. 0603.13001.

Notes and References

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. Adapted from the solution to