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.
Statement and proof The following statement and argument are perhaps the most standard. Theorem (Prime Avoidance Lemma): Let E be a subset of commutative ring R that is an additive subgroup of R and is multiplicatively closed. (In particular, E could be a subring or ideal of R.) Let I 1 , I 2 , … , I n , n ≥ 1 {\displaystyle I_{1},I_{2},\dots ,I_{n},n\geq 1} be ideals such that I i {\displaystyle I_{i}} are prime ideals for i ≥ 3 {\displaystyle i\geq 3} . If E is not contained in any of the I i {\displaystyle I_{i}} , then E is not contained in the union ⋃ I i {\textstyle \bigcup I_{i}} . Proof by induction on n: The idea is to find an element of R that is in E and not in any of the I i {\displaystyle I_{i}} . The base case n = 1 {\displaystyle n=1} is trivial. Next suppose n ≥ 2 {\displaystyle n\geq 2} . For each i, choose
z i ∈ E ∖ ⋃ j ≠ i I j {\displaystyle z_{i}\in E\setminus \bigcup _{j\neq i}I_{j}} , where each of the sets on the right is nonempty by the inductive hypothesis. We can assume z i ∈ I i {\displaystyle z_{i}\in I_{i}} for all i; otherwise, there is some z k {\displaystyle z_{k}} among them that avoids all of the I i {\displaystyle I_{i}} , and we are done. Put
z = z 1 ⋯ z n − 1 + z n {\displaystyle z=z_{1}\cdots z_{n-1}+z_{n}} . Because E is closed under addition and multiplication, z is in E by construction. We claim that z is not in any of the I i {\displaystyle I_{i}} . Indeed, if z ∈ I i {\displaystyle z\in I_{i}} for some i ≤ n − 1 {\displaystyle i\leq n-1} , then z n ∈ I i {\displaystyle z_{n}\in I_{i}} , a contradiction. Next suppose z ∈ I n {\displaystyle z\in I_{n}} . Then z 1 ⋯ z n − 1 ∈ I n {\displaystyle z_{1}\cdots z_{n-1}\in I_{n}} . If n = 2 {\displaystyle n=2} , this is already a contradiction. If n > 2 {\displaystyle n>2} , then, since I n {\displaystyle I_{n}} is a prime ideal, z i ∈ I n {\displaystyle z_{i}\in I_{n}} for some i ≤ n − 1 {\displaystyle i\leq n-1} , again a contradiction. ◻ {\displaystyle \square }
E. Davis' prime avoidance There is the following variant of prime avoidance due to E. Davis.
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