# Ideals in a Quotient Ring

Let $R$ be a commutative ring with identity and $I$ na ideal of $R$. It is well known that there the is a one-to-one correspondence between the set of all ideals of $R$ which contain $I$ and the set of all ideals of $R/I$. In this correspondence to a principal ideal in $R/I$ corresponds necessarily a principal ideal in $R$?

Let $R=\mathbb{Z}[x]$, $I=(2)$. Consider $J=(2,x)$ in $R$.
No. For example consider the following type of situation where $(a,b)\ne(a)$:
$$\begin{array}{ccccc} R & \triangleright & (a,b) & \triangleright & (a) \\[5pt] & & \Updownarrow \\[5pt] R/(a) & \triangleright & (\bar{b}) & \triangleright & 0 \end{array}$$
Above $\bar{b}$ is the image of $b$ mod $(a)$, i.e. under the projection $R\to R/(a)$.
The perspective here is that in nicely-behaved rings ideals will have minimal generating sets (this isn't always the case but it occurs in comfy settings), and 'modding out' by a thinner ideal, thereby passing between the ring and a quotient ring, will remove generators from this minimal set. In this way we can artificially create a situation where $(a,b)$ lattice-corresponds to $(\bar{b})$.
A simple example would be let $R=k[x,y]$ where $k$ is a field, say algebraically closed to be nice. Then $k[x,y]/(y)\cong k[x]$ (This is easily shown by using the first isomorphism theorem and the map $f(x,y)\mapsto f(x,0)$.) Though $(x)\in k[x,y]/(y)$ is principal. The corresponding ideal is actually $(x,y)$.