I'm reviewing past questions on the algebra prelim at my university, and I came across this question.
"Let $R$ be a UFD and I a proper principal ideal. Prove that $R$ has a proper principal ideal that is maximal with respect to the property of containing I and identify its generator interms of the generator of $I$." This is what I have so far:
Let $I=(x)$. In a UFD, we have the unique decomposition into irreducibles/primes. so $$x=p_1p_2\dots p_n$$ This implies that $I\subset (p_i)$ for $1\leq i \leq n$. Now I must show that all (or just one?) of these prime ideals are maximal. If $R$ was also PID, then prime ideals are maximal, but this is not the case with UFDs in general. I don't know how to show if $(p_i)$ is contained in some ideal $M$, then either $(p_i)=M$ or $M=R$ given that $M$ may not be principal.