Prime avoidance lemma

Remarks

The proof is by induction on $n$. If $I$ is contained in no $P_i$, the induction hypothesis gives, for each $i$, an element $x_i \in I$ that lies in no $P_j$ with $j \ne i$; one may assume $x_i \in P_i$. Then $y = x_1 + x_2 x_3 \cdots x_n$ lies in $I$ but in none of the $P_i$, using that $P_1$ is prime for the term $x_2 \cdots x_n$. The same proof works when at most two of the $P_i$ are not prime. Primality is needed: in $R = \F_2[x, y]/(x, y)^2$, the maximal ideal $(x, y) = \set{0, x, y, x + y}$ is the union of the three ideals $(x)$, $(y)$, $(x + y)$, but it is contained in none of them. Finiteness is needed: in $k[x, y]$ for a field $k$, every nonzero element of $(x, y)$ has an irreducible factor vanishing at the origin. So $(x, y)$ is the union of infinitely many principal prime ideals, none of which contains it.

Going deeper

For a student

Prime avoidance is a tool for choosing good elements. For example, let $M$ be a finitely generated module over a Noetherian ring, and let $I$ be an ideal whose elements are all zero divisors on $M$. Since zero divisors are the union of associated primes, $I$ lies in the union of the finitely many associated primes of $M$. So $I$ lies in one associated prime. In dimension theory the lemma lets one choose elements outside all minimal primes, which is how regular sequences and systems of parameters are built one element at a time. If $R$ contains an infinite field, the conclusion holds even when none of the ideals $P_i$ is prime, by a vector space argument.