An ideal maximal w.r.t. the property that $R/I$ does not have a composition series is a prime ideal 
Let $I$ be an ideal of a Noetherian ring $R$. Suppose that $I$ is maximal with respect to the property that $R/I$ does not have a composition series, i.e., if $I\subset J$ is an ideal, then $R/J$ has a composition series. Prove that $I$ is a prime ideal.

Here's what I did so far: suppose that $ab\in I$ and $a\notin I$. Then by assumption $I\subset I+(a)$ and $R/(I+(a))$ has a composition series. That is the series
$$0=I_0 \subset I_1 \cdots \subset I_n\subset R/(I+(a))$$
satisfies $I_{i+1}/I_i$ is simple. But I don't know how to proceed to show that $a\in I$. Any ideas?
 A: Assume $a,b \notin I$ and look at the exact sequence
$$0 \to (I,a)/I \to  R/I \to R/(I,a) \to 0$$
We want to show that the middle term has a composition series, which would be a contradiction and thus finishes the proof. The right term certainly has a composition series by assumption, so we are left to show that the left term has a composition series.
We have $(I,a)/I \cong (a)/(I \cap (a))$. Now look at the surjection
$$R \twoheadrightarrow (a)/(I \cap (a)),$$
given by multiplication with $a$.
We have $aI \subset I \cap (a)$ and $ab \in I \cap (a)$, hence $(I,b)$ is contained in the kernel. Thus the surjection factors over a surjection
$$R/(I,b) \twoheadrightarrow (a)/(I \cap (a)).$$
If $b \notin I$, $R/(I,b)$ has a composition series, hence $(a)/(I \cap (a))$ has a composition series as a homomorphic image. This completes the proof.

One could also note that $(I,a)/I$ is isomorphic to $R/(I:a)$ (the isomorphism is multiplication with $a$) and then finish the proof via $(I:a) \supset (I,b)$.
