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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.

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  • $\begingroup$ TO add words to the answers of people, the ring $F[x]$ is dimension $1$ if $F$ is a field. That means that all non-zero primes are maximal. This follows since $F$ is dimension $0$, and adjoining a variable raises dimension by $1$. Or, because $F[x]$ is a PID, and all PIDs are dimension $1$. $\endgroup$ – Alex Youcis Sep 23 '13 at 4:05
  • $\begingroup$ Dear @AlexYoucis: It is my impression that the OP probably doesn't know what dimension of a ring means. In my humble opinion your explanation is over-complicating the matter don't you think? $\endgroup$ – Rankeya Sep 24 '13 at 2:31
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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?

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  • $\begingroup$ Do you think I could figure out the proof that every prime ideal in a PID is maximal or $(0)$ - it's not in my Michael Artin Algebra book and I googled. $\endgroup$ – BananaCats Category Theory App Sep 23 '13 at 4:04
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    $\begingroup$ @EnjoysMath: Let $x$ be irreducible and suppose $I\supset(x)$. Write $I=(a)$; express the relationship $(a)\supset(x)$ in a different way. $\endgroup$ – Karl Kronenfeld Sep 23 '13 at 4:06
  • $\begingroup$ If you mean irreducible as a polynomial, I would like to prove it in the general setting of PID. $\endgroup$ – BananaCats Category Theory App Sep 23 '13 at 4:13
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    $\begingroup$ @EnjoysMath Yes, a nonzero nonunit which is not the product of two nonunits. $\endgroup$ – Karl Kronenfeld Sep 23 '13 at 4:18
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    $\begingroup$ @EnjoysMath: Split into two cases, in both you will be able to determine $(a)$. First, assume $b$ is unit. Second, assume $b$ is a nonunit. $\endgroup$ – Karl Kronenfeld Sep 23 '13 at 4:48
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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.

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  • $\begingroup$ Why the downvote? $\endgroup$ – Zev Chonoles Dec 9 '13 at 0:48

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