In the proof of theorem 2.14 of David Eisenbud textbook "Commutative algebra with a view towards algebraic gometry", he says the following:
If R is Noetherian, and not of finite length, let $I \subset R $ be an ideal maximal with respect to the property that $R/I$ is not of finite length. We claim that $I$ is prime. Indeed, if $ab \in I$ and $a \notin I$, then we may form an exact sequence $$ 0 \rightarrow R/(I:a)\stackrel{a}{\rightarrow} R/I \rightarrow R/(I + (a)) \rightarrow 0.$$
I don't quite see why this is an exact sequence and why the author needs to say that $I$ is an ideal maximal $\textbf{with respect to the property that $R/I$ is not of finite length}$, could someone explain these two parts to me? Thanks!
The short exact sequence is deduced from the exact sequence: $$R\xrightarrow{\ {}\times a\enspace }R/I\longrightarrow R/(I+(a))\longrightarrow0,$$
observing the kernel of multiplication by $a$ is precisely $(I:a)$, so that the left arrow can be factored as $$\begin{matrix} R&\xrightarrow{\qquad{}\times a\qquad }R/I&\\ &\searrow\phantom{R/(I:a)} \nearrow\qquad&\hspace{-2.5em}\leftarrow\text{injection}\\ &R/(I:a)\qquad \end{matrix}$$ This exact sequence has nothing to do with $R/I$ being of finite length or not. I guess this hypothesis, is related to the rest of the proof.