$R/I$ is not Noetherian. Prove that $I$ is a prime ideal.

Let $R$ be a commutative ring with $1$ and let $I$ be an ideal of $R$, maximal with respect to the property that $R/I$ is not Noetherian. Prove that $I$ is a prime ideal.

I need some hints to start.

Solution 1:

Let $a,b\in R$ such that $ab\in I$. Suppose that $a\notin I$ and $b\notin I$. Then $I\subsetneq (I:a)$ and $I\subsetneq I+(a)$. We have $R/(I:a)\cong (I+(a))/I$, so $R/(I:a)$ is a noetherian ring, or equivalently, a noetherian $R$-module. On the other side, $\frac{R/I}{(I+(a))/I}\cong R/(I+(a))$ is also noetherian, and thus you have got an $R$-submodule and a factor module of $R/I$ which are both noetherian, so $R/I$ is a noetherian $R$-module, that is, a noetherian ring, a contradiction.

Solution 2:

Hint: If $AB \subseteq I$, study the $R$-module $(A+I)/I$