$R$ a commutative ring. If $S\subset R$ is a multiplicatively and additively closed subset (not necessarily containing 1) and $U_1,...,U_n$ are prime ideals in $R$ and $I$ is an ideal properly contained in $S$ such that $S$ \ $I \subset U_1\cup...\cup U_n$ how can I show that $S \subset U_k$ $\exists k\leq n$
This problem seems similar to the prime avoidance lemma but I can't seem to figure out how to implement that in my proof. I have tried working with induction but I can't seem to figure out even the base case. Any tips on how to proceed?
Edit: I figured it out, for posterity here is the proof
Note that $S\subset U_1 \cup...\cup U_n \cup I$ where at least all but one ideal is prime. Apply prime avoidance lemma and note that $I \subsetneq S$ meaning $S\not\subset I$ thus $S\subset U_k \exists k\leq n$