Quotient ring is cyclic group implies every ideal is generated by 2 elements

Question

I'm trying to solve the following exercise:

Let $R$ be a commutative ring with identity. If for every ideal $\mathfrak{a} \neq 0$ of $R$ we have ($R/\mathfrak{a}$,+) is a cyclic group then every ideal of $R$ can be generated by two elements.

I've tried a lot of things (tried to use the correspondence theorem and the isomorphism between $R/\mathfrak{a}$ and $\mathbb{Z}$ or $\mathbb Z_n$, but I don't know how to relate $R/\mathfrak{a}$ as a group and $\mathfrak{a}$ as an ideal.

I'd love any hint, thank you.

Answer 1

Let $\mathfrak{a}$ be an ideal of $R$ and $x\in\mathfrak{a}$ a non-zero element. You have that each ideal of $R/(x)$ is principal (here you use the hypothesis over $R/(x)$). Consider the kernel of the projection $R/(x)\to R/\mathfrak{a}$. This is an ideal generated by the class of some $y\in R$. Try to prove that $\mathfrak{a}=(x,y)$.

Answer 2

If $I\neq 0$ is an ideal, and $x\in I$ is any nonzero element, then consider the surjective map of cyclic groups $R/(x) \to R/I$. If $y$ is an element of $R$ that generates the kernel of this map, what can we say about the ideal $(x,y)$?