Ring homomorphism with field as image, is the pre-image also a field?

Question

Let $f:R\rightarrow S$ be a surjective ring homomorphism. $R,S$ are both integral domains. Suppose $S$ is a field, then is $R$ also a field?

A possible useful fact: A finite integral domain is a field.

Answer 1

No, $R$ need not be a field. For instance, if $R$ is the polynomial ring $S[x]$, then the map $R\to S$ given by $f(x)\mapsto f(0)$ is a surjective ring homomorphism.

Answer 2

In general, no. Simple family of counterexamples: let $R = \Bbb Z$, the integral domain of integers. Let $S = \Bbb Z_p$ for any prime $p$, and let $f:\Bbb Z \to \Bbb Z_p$ be the usual projection given by $f(z) = z (\mod p)$. All the hypotheses of the question are satisfied, but $R = \Bbb Z$ is not a field. Generalization: let $R$ be any non-field integral domain with a proper maximal ideal $M$, that is, $\{0\} \ne M \ne R$. (This may be a bit redundant if one defines a maximal ideal to be a proper subset of $R$, but bear with me, if you please.) Then $R/M$ is a field and the natural map $f:R \to R/M$ is surjective. So we see there are very many counterexamples.

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