$R$ is PID, so $R/I$ is PID, and application on $\mathbb{Z}$ and $\mathbb{N}$ I'm supposed to show in a part of an exercise that if we have a ring $R$ that is a principal ideal domain, then for any ideal $I$ in $R$, $R/I$ will also be a PID.
So $I=(i)$ for some $i \in R$, and $R/I$ is $\{rI |r\in R\}$, and we know: 
An ideal $J$ in $R/I$ will be an ideal $J$ in $R$ containing $I$.
So if $I \subset J$ and $I=(i)$, then $J$ should be $(j)$, where $j|i$. but wherefrom do I get the necessity that $J=(j)$, so that $J$ is also principal? and there will be several $j$ that divide $i$, but should it not be only one?
And then I shall use this to show a 1-1 correspondence between ideals of $\mathbb{Z}$ and $\mathbb{N}$.
Can someone support with with the proofs?
 A: The claimed result is false: For $R/I$ to be a principal ideal domain, it must be a domain. But $R/I$ is a domain if and only if $I$ is a prime ideal. In particular, $\mathbb{Z}/6\mathbb{Z}$ is a quotient of a PID that is not a domain, hence not a PID.
What is true is that a quotient of a PID will be a principal ideal ring: every ideal of $R/I$ is principal. Your argument is confused because you are using the same letter to denote the ideal of $R/I$ and the ideal it corresponds to inside $R$. Better to use different letters.
Indeed, let $K$ be an ideal of $R/I$. By the Correspondence Theorem $K$ corresponds to an ideal $J$ of $R$ that contains $I$. Since $R$ is assumed to be a PID, then $J=(j)$ for some $j\in R$. The claim is that $K = (j+I)(R/I)$: let $k+I\in K$. Then $k+I \in J+I$, so there exists $a\in J$ such that  $k+I = a+I$, which means $a-k\in I$; since $I\subseteq J$, we conclude that $a-(a-k) = k\in J$. Therefore, $k=jx$ for some $x\in R$, so $k+I = jx+I = (j+I)(x+I)\in (j+I)(R/I)$. Thus, $K\subseteq (j+I)(R/I)$. And since $j+I\in K$ and $K$ is an ideal, then $(j+I)(R/I)\subseteq K$, giving equality.
