Let $A$ be conmutative ring with identity and $\mathfrak{p}, \mathfrak{m}$ ideals. Then
$$\begin{array}{ll}
\mathfrak{p}\text{ is a prime}\iff A/\mathfrak{p}\text{ is an integral domain}\\
\mathfrak{m}\text{ is maximal}\iff A/\mathfrak{m}\text{ is a field}
\end{array}$$
Does every integral domain/field come from some other ring and prime/maximal ideal? An equivalent question is can we un-quotient every ring? That is, given a ring $A$ can we find another ring $B$ and some ideal $I$ such that $B/I\cong A$? If there is a way to do so is it unique (up to isomorphism)?
I want to answer in a not trivial way: every integral domain $A$ is isomorphic to $A/(0)$ and $(0)$ is prime in $A$.
Note: this is not an exercise and I will try to answer the question myself but don't have time now
Best Answer
Any ring $A$ can be written as a quotient of another ring in many many different ways. For instance, you could take any ring $B$ and consider the product ring $A\times B$. The first projection $A\times B\to A$ is a surjective ring-homomorphism and thus exhibits $A$ as a quotient of $A\times B$ (by the ideal $0\times B$).
This is far from the only possibility. For instance, you could take a polynomial ring $B$ over $A$ in any number of variables, and get a surjective ring homomorphism $B\to A$ by mapping all the variables to $0$. Or, you could map the variables to any elements of $A$ instead of $0$ (assuming $A$ is commutative; if $A$ is noncommutative you would need to instead use a ring of noncommuting polynomials). Or, you could take a polynomial ring $B$ over $\mathbb{Z}$ in lots of variables, and map those variables to elements of $A$. As long as the chosen elements of $A$ generate $A$ as a ring, the homomorphism $B\to A$ will be surjective.
Another way to get more examples: given any ring $B$ with an ideal $I$ such that $B/I\cong A$, you can take any ideal $J\subseteq B$ such that $J\subseteq I$ and then $A$ is also a quotient of $B/J$ (by the ideal $I/J$).