Constructing finite fields as quotients of a domain by its ideals

Question

All of the finite fields of prime order can be constructed as quotients of $\mathbb{Z}$ by one of its ideals.

Is there a domain such that every finite field can be constructed as a quotient of that domain by one of its ideals?

What if "finite" is replaced with "of cardinality strictly smaller than a given cardinal"?

Answer 1

A finite field of cardinality $p^n$ is of the form $\mathbb{F}_p[x]/(f(x))$, where $f(x)$ is an irreducible polynomial of degree $n$ over $\mathbb{F}_p$ (the prime field of characteristic $p$). Such a polynomial is the image of $\hat{f}(x)\in\mathbb{Z}[x]$, via reduction modulo $p$.

Prove that $$ \mathbb{F}_p[x]/(f(x))\cong\mathbb{Z}[x]/(p,\hat{f}(x)) $$

Answer 2

Let $\kappa$ be the given (infinite) cardinal and let $R_{\kappa} = \Bbb{Z}[x_{\alpha} \mid \alpha < \kappa]$ be the ring of multivariate polynomials over a set of variables $x_{\alpha}$ of cardinality $\kappa$. Then any ring $S$ of cardinality at most $\kappa$ is a homomorphic image of $R_{\kappa}$ (under the homomorphism induced by any surjection carrying the $x_{\alpha}$ onto $S$). So $R_{\kappa}$ is an integral domain meeting your requirements.