Let $R$ be a ring, $I$ a finitely generated ideal of $R$ and $J$ an ideal of $R$ with $J \subset I$. Show that $R$ has an ideal $M$ so that $J \subset M \subset I$ and for every ideal $N$ with $M \subset N \subset I$ $\Rightarrow$ $N=M$ or $N=I$.
If I choose $M:=I+J=\{a+b,a \in I, b \in J\}$ then the first part should be proved?
For the second part I have to assume that if $N \neq M$ then it has to be equal to $I$ meaning there is at least one element in $N$ that is not in $I$?
Thanks for any help!
There is a finite sequence $x_1,\cdots,x_n\in I$ such that $$I=J+(x_1)+\cdots+(x_n).$$
So among all such sequences there is one of minimial length, and so we may assume $x_1,\cdots,x_n$ has a minimal length.Then $$L=J+(x_2)+\cdots+(x_n)$$ is a proper ideal of $I$.
Let $\mathscr{F} $ be the set of all proper ideal of $I$ that contanin $L$. Clearly, $\mathscr{F} $ is a non-empty. Now an ideal $N$ that contains $L$ is in $\mathscr{F} $ if and only if $x_1\notin N$.
By zorn's lemma, $\mathscr{F} $ has a maximal element, say $M$. $M$ is the ideal satisfying the condition.