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I am looking for a completely elementary proof of the fact that the maximal ideals of $\mathbb{C}[X,Y]$ are of the form $(X-a,Y-b)$ for $a,b \in \mathbb{C}$.

(The proof should not use that $(0),(X-a,Y-b),(f), f$ irred. are the prime ideals as I want to use the fact about maximal ideals to prove this).

Elementary means: No use of the Nullstellensatz, or Krull dimension.

fish_monster
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    You can base a proof on the fact that $\Bbb{C}$ is algebraically closed. If you disallow that piece of information then you absolutely need to prove it first. If $R$ is a commutative ring, then an ideal $M\subset R$ is maximal if and only if $R/M$ is a field. Combine that with the fact that an algebraically closed field has no algebraic extensions. – Jyrki Lahtonen Dec 08 '21 at 07:08
  • Hm. OK, assuming $\mathbb{C}$ is algebraically closed, I unfortunately still don't see how to conclude from what you say. – fish_monster Dec 08 '21 at 07:11
  • Actually the details are probably a bit hairier than I initially thought. $\Bbb{C}[X,Y]/M$ is a field, and I want to conclude that $X+M$ and $Y+M$ are algebraic over $\Bbb{C}$, implying the claim. I think I have seen it here, but don't remember right away. Sorry to leave it at this :-( – Jyrki Lahtonen Dec 08 '21 at 07:14
  • If the extension were transcendental it would have uncountable dimension, I think... More later (unless somebody else clears up this mess). – Jyrki Lahtonen Dec 08 '21 at 07:16
  • I found a copy/generalization of the argument I had in mind (and in my deleted answer). I vote to close this as a duplicate of that. Because I have the dupehammer privilege in [tag:abstract-algebra], my vote took immediate effect. If somebody disagrees, or notices that I overlooked something, feel free to @-ping me and explain the problem. – Jyrki Lahtonen Dec 08 '21 at 08:54

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