# 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 *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 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

- 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.

## References

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