Prime avoidance lemma

Question

Let $R$ be a commutative ring. $A$ is an ideal of $R$, and $P_1,P_2,...,P_s$ are prime ideals of $R$ such that $$A \subseteq \bigcup_{j=1}^{s} P_j.$$ Prove that there exists some $P_j$ such that $A\subseteq P_j$.

Answer 1

Ideas & hints:

Induction on $\;s\;$: for $\;s=1\;$ it is trivial (for $\;s=2\;$ we don't even need the ideals in the union to be prime! Yet for more than two we do need primality).

Assume for any union of prime ideals up to $\;s-1\;,\;\;s>2\;$ . Now, if $\;A\;$ is already contained in the union of less than $\;s\;$ of those prime ideals the inductive hypothesis kicks in and we're done, so we can go now by reductio ad absurdum and assume the claim is false:

$$\exists\,a_1,a_2,...,a_s\in A\;\;\text{s.t.}\;\;a_i\in P_i\;,\;\;a_i\notin\bigcup_{k=1,\,k\neq i}^s P_k\;,\;\;\forall\;i=1,2,...,s$$

and let us take now for all $\;i\;$ :

$$x_i:=\prod_{k=1,\,k\neq i}^s a_k\implies x_i\notin P_i\;\;\text{(why?)}$$

Reach now a contradiction by considering where does the following element lie

$$r:=\sum_{i=1}^sx_i.$$