Bijection between sets of ideals

Solution 1:

In order to better distinguish between elements and ideals I'll use Fraktur letters for the latter: $\mathfrak{b}$ and $\mathfrak{c}$.

With $\mathfrak{c}/\mathfrak{b}$ the ideal $$ \{x+\mathfrak{b}:x\in \mathfrak{c}\} $$ is meant.

You have to prove that

1. $\mathfrak{c}/\mathfrak{b}$ is an ideal of $A/\mathfrak{b}$
2. Every ideal of $A/\mathfrak{b}$ is of this form for a unique ideal $\mathfrak{c}$ of $A$ such that $\mathfrak{c}\supseteq\mathfrak{b}$.

The proof of 1 is easy, just a verification. For 2, the hint is

given an ideal $\mathfrak{d}$ of $A/\mathfrak{b}$, consider $\mathfrak{c}=\{x\in A:x+\mathfrak{b}\in\mathfrak{d}\}$

Solution 2:

Hint: $b$ induces a congruence on $A$, and $c/b$ is a set of certain congruence classes modulo $b$. How do you reconstruct $c$ from $c/b$?