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. 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. There is the following variant of prime avoidance due to E. Davis. Proof: 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. Let A be a Noetherian ring, I an ideal generated by n elements and M a finite A-module such that .