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If $R$ is a PID, $S$ an integral domain and $f: R \to S$ is an epimorphism, why is it that either $f$ is an isomorphism or $S$ is a field?

PID - Principal Ideal Domain

What I know:

If $S$ is not a field then we have to show that the function $f$ one to one since it's already an epimorphism (to show that $f$ is an isomorphism). If $f$ is not an isomorphism then we have to show that $S$ is a finite integral domain (= field).

However, I don't know how to use the facts If $R$ is a PID, $S$ an integral domain and $f: R \to S$ is an epimorphism, to prove the latter statement. Please I appreciate clear explanation. Thank you.

Ka Em
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    That statement is not true for the standard meaning of epimorphism (in the category of commutative rings, say). Consider any localisation, e.g. $\mathbb Z \rightarrow \mathbb Z[1/2]$. Cf. https://mathoverflow.net/a/110443/27465 – Torsten Schoeneberg Apr 02 '20 at 03:10

2 Answers2

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The correct statement is: Let $f: R \rightarrow S$ be a surjective ring homomorphism, where $S$ is an integral domain and $R$ a PID. Then either $f$ is an isomorphism or $S$ is a field.

Proof: Let $I := ker(f)$, then via the first isomorphism theorem $f$ induces an isomorphism $R/I \simeq S$. Since $S$ is an integral domain, $I$ is prime.

Case 1: $I=0$. Then $f$, being injective and surjective, is an isomorphism.

Case 2: $I \neq 0$. Since nonzero prime ideals in PIDs are maximal (see link in other answer), $I$ is maximal, so $R/I$ and a fortiori $S$ is a field.

Torsten Schoeneberg
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Let $I$ be the kernel of $f$, if $I$ is not trivial, it is an ideal of $R$ which is prime since $S$ is integral, thus maximal and $S$ is a field.

Proving a prime ideal is maximal in a PID

Tsemo Aristide
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