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

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.

• 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 $1$. – Alex Youcis Sep 23 '13 at 4:05
• 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? – Rankeya Sep 24 '13 at 2:31

## 2 Answers

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?

• 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. – Shine On You Crazy Diamond Sep 23 '13 at 4:04
• @EnjoysMath: Let $x$ be irreducible and suppose $I\supset(x)$. Write $I=(a)$; express the relationship $(a)\supset(x)$ in a different way. – Karl Kronenfeld Sep 23 '13 at 4:06
• If you mean irreducible as a polynomial, I would like to prove it in the general setting of PID. – Shine On You Crazy Diamond Sep 23 '13 at 4:13
• @EnjoysMath Yes, a nonzero nonunit which is not the product of two nonunits. – Karl Kronenfeld Sep 23 '13 at 4:18
• @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. – Karl Kronenfeld Sep 23 '13 at 4:48

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.

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