Does every integral domain come from a quotient?

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

• Given any ring R we can extend it to a bigger ring of which R is a quotient; e.g. the ring of polynomials in a variable. – R.C.Cowsik Dec 8 '18 at 17:29

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 $$I/J$$).