$\varphi: F[X] \rightarrow R'$ an integral domain, then $\ker \varphi$ is maximal or $(0)$

Question

Let $\varphi : F[X] \rightarrow R'$ be a ring homomorphism where $F$ is a field and $R'$ is an integral domain. $P = \ker \varphi$ is either maximal or $(0)$.

I know that the maximal ideals of $F[X]$ correspond to the principal ideals generated by irreducible monic polynomials, and that $P$ is maximal iff $F[X]/P$ is a a field. Please only give a hint. Thanks.

Answer 1

Hint: $(0)$ is prime in $R'$, and the inverse image of a prime ideal under a ring map is prime. Is $ker(\varphi)$ prime then?

What do you know about the prime ideals of a PID?

Answer 2

Hint: The ring $F[X]$ is a principal ideal domain. In any PID, the prime ideals are precisely

• the zero ideal, and
• the maximal ideals.

Thus, for this ring homomorphism $\varphi$, saying that $\ker(\varphi)$ is either $(0)$ or maximal is just the same thing as saying that $\ker(\varphi)=\varphi^{-1}(0)$ is prime.