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Let $F$ be a field. Find all prime and maximal ideals of $F[x]$.

I have no idea on how to start this question. I try to use the theorem that '$F[x]/I$ is a field $\Leftrightarrow$ $I$ is a maximal ideal $\Rightarrow$ I is a prime ideal'. But I realised that $F[x]/I$ is not unique, that is different $I$ may yield different field.

Can anyone guide me?

Idonknow
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  • Realizing that the quotient is a field iff the ideal is prime is a good start. But you should look closer at why this is the case. The reason is that ideals of this ring have a very special form, which should help you determine what the prime ideals look like. – Tobias Kildetoft Apr 03 '13 at 12:48
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    Perhaps you can use the fact that $F[x]$ is a PID and that prime elements and irreducible elements in this ring coincide. – sxd Apr 03 '13 at 12:49

1 Answers1

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You are correct that for any ring $R$, $$R/I\text{ is a field} \iff I \text{ is maximal}\implies I\text{ is prime}.$$ It is not in general true that $$I\text{ is maximal}\impliedby I\text{ is prime}.$$ The corresponding characterization of prime ideals is $$R/I\text{ is a domain}\iff I\text{ is prime}.$$

Hint for the problem: The ring $F[x]$ is a principal ideal domain. That is,

  • every ideal of $F[x]$ is principal; in other words, for any ideal $I$, there is an $f$ such that $$I=(f)=\{f\cdot g\mid g\in F[x]\}.$$

  • $F[x]$ is a domain; in other words, if neither of $f,g\in F[x]$ are the zero polynomial, then $fg\neq 0$.

Now characterize when $I$ is prime in terms of properties of how $f$ factors (remember, every polynomial in $F[x]$ factors into irreducibles). In any ring every maximal ideal is prime, but the converse is not true. In the case of $F[x]$, there will be exactly one prime ideal that is not maximal. Here is a hint for what it is: the ring $F[x]$ is not a field.

Zev Chonoles
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