If $I$ is not finitely generated and every ideal properly containing $I$ is finitely generated, then $I$ is prime? If $I$ is an ideal in a ring $R$ such that $I$ is not finitely generated and every ideal properly containing $I$ is finitely generated, then $I$ is prime. 
Let $I$ be a non-finitely generated ideal and let $a,b\in{R}$ such that $a,b\notin{I}$. Then $a\in{I'}$ such that $I'$ is finitely generated and $I'$ properly contains $I$; $b\in{I''}$ such that $I''$ is finitely generated and $I''$ properly contains $I$. I am having difficulties in showing that $ab\notin{I}$. 
Would you help me, please?
Thank you in advance.
 A: Let $I$ be such a non-finitely generated ideal so that  all ideals properly containing $I$ are finitely generated. If $I$ is not a prime ideal, we can find $a,b\not\in I$ with $ab\in I$. Let $I'=I+(a)$ and let $I''=\{x\in R| xI'\subset I\}=\{x\in R|xa\in I\}$. Then, since $b\in I''$, both $I',I''$ are finitely generated. We can choose a set of generators for $I'$ of the form $\{z_1,\ldots,z_n,a\}$ with all $z_i\in I$. We also choose a set of generators $\{u_1,\ldots,u_m\}$ for $I''$. Then, I claim $I$ is generated by $\{z_1,\ldots,z_n, u_1a,\ldots,u_ma\}$ which will contradict our assumption on $I$. Notice that all these elements belong to $I$, so sufficient to prove that any element in $I$ can be written as a linear combination of these. Let $v\in I$. Then, since $v\in I'$, we can write $v=\sum r_iz_i+sa$, where $r_i, s\in R$. This says $sa\in I$ and then $s\in I''$ by definition of $I''$. So, we can write $s=\sum p_iu_i$, with $p_i\in R$ and thus, $v=\sum r_iz_i+\sum p_i(u_ia)$, proving what I claimed.
