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I used the first isomorphism theorem and got $\mathbb{C}[x]/\ker\pi\cong\mathscr{F}$ but was not able to proceed any further for a couple of hours. I suppose it has to do with the fact that $\mathscr{F}$ is a field but so far I haven't found a single clue. Could someone please give me a hint?

user26857
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1798bebe
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So you know $\mathbb{C}[x]/\ker(\pi)\simeq\mathscr{F}$, and thus $\ker(\pi)$ is a maximal ideal since the quotient is a field iff the ideal is maximal. We also know that maximal ideals in $\mathbb{C}[x]$ are of the form $(x-a)$ for $a \in \mathbb{C}$. This gives you $$ \mathscr{F} \simeq \mathbb{C}[x] / \ker(\pi) \simeq \mathbb{C}[x]/(x-a) \simeq \mathbb{C} $$ as desired.

Derek Allums
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  • But how do you know that maximal ideals are of this form? It feels to me like it's hard to avoid circular reasoning with this approach. – Andrew Dudzik Jun 07 '23 at 08:22
  • @AndrewDudzik I don't think you necessarily need the Nullstellensatz (also not 100% sure that it would be circular if you do but maybe I am missing something) - you can do this from first principles with something like this: https://math.stackexchange.com/questions/56916/what-do-prime-ideals-in-kx-y-look-like/56921#56921, or something like Theorem 3.1 here, no: https://kconrad.math.uconn.edu/blurbs/ringtheory/maxideal-polyring.pdf ? – Derek Allums Jun 07 '23 at 08:40
  • To proceed from first principles, we have to prove the Fundamental Theorem of Algebra. We do need some version of Zariski's lemma as well, but this isn't enough as we need to know all the finite extensions of $\mathbb{C}$. Basically, in proving the classification of maximal ideals of $\mathbb{C}[x]$, I would expect to see the OP's question or an equivalent as a fundamental lemma, so I don't agree with using this classification to answer the question. – Andrew Dudzik Jun 07 '23 at 12:22
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Without using stronger facts like the classification of maximal ideals in $\mathbb{C}[x]$, we can proceed as follows.

$\mathscr{F}$ is generated by $\mathbb{C}$ and $\pi(x)$. If $\pi(x)$ is transcendental, we can show that $\pi$ cannot be surjective. Otherwise, $\mathscr{F}$ is a finite, therefore algebraic extension of $\mathbb{C}$.

Now apply the Fundamental Theorem of Algebra.

Andrew Dudzik
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    The fundamental theorem of algebra is equivalent to the classification of maximal ideals in $\Bbb C[x]$, isn't it? Also it would be good to explain, why $\pi$ cannot be surjective (if $\pi(x)$ is transcendental, $\pi$ is injective and surjective by asssumption, so $\pi$ is an isomorphism of a non-field $\Bbb C[x]$ to a field $\cal F$, a contradiction). – Jonas Linssen Jun 07 '23 at 09:44
  • @JonasLinssen No, they aren't quite equivalent; you need an argument like the above (basically a special case of Zariski's lemma that finitely generated field extensions are finite) to bridge the gap, while the reverse direction is trivial. – Andrew Dudzik Jun 07 '23 at 12:05