Why are affine varieties except points not compact in the standard topology on $C^n$ ?

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Solution 1

They are never bounded. ${}{}{}{}{}{}{}$

Solution 2

Noether's normalization theorem (Mumford, Red book, page 42 ) says that if $X$ is a variety of dimension $n$ , there exists a finite surjective morphism $X\to \mathbb A^n$.
Since, in the transcendental topology over $\mathbb C$, affine space $\mathbb A^n$ is not compact for $n\geq 1$ , $X$ is not compact either.

A comment
Exciting as it definitely is, algebraic geometry has the drawback that many very intuitive facts, like the above, are difficult to justify without some fairly technical tools.
I think it is the duty of a teacher to acknowledge this explicitly in an introductory course (and maybe give a reference for the student to come back to later), rather than throw offhand remarks which might discourage a student and make him feel it is his fault that he can't find the (actually quite hard) rigorous proof.

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Updated on August 09, 2022

Comments

  • readingframe
    readingframe over 1 year

    I am starting to learn algebraic geometry and in the notes I am reading there is the following remark:

    " Over the complex numbers and with the strong topology we see that $A^n$ and affine varieties (except for points) are not compact. " (The strong topology seems to be the standard topology in this text.)

    Why is that true ?

    • Georges Elencwajg
      Georges Elencwajg over 11 years
      Dear readingframe, this is an excellent question and you are quite right to ask about this non trivial assertion: +1.
  • readingframe
    readingframe over 11 years
    Thank you for your answer. I can see that unboundedness implies non-compactness. But since i know essentially nothing about affine varieties, i am not familiar with the fact that they are never bounded. Could you please explain why ?
  • Ben Blum-Smith
    Ben Blum-Smith over 11 years
    Morally it's because $\mathbb{C}$ is algebraically closed. For example, take $x^2+y^2=1$ (whose solution set is bounded over $\mathbb{R}$). You can pick unboundedly large values for $x$ and due to algebraic closure there will always be a solution for $y$.
  • readingframe
    readingframe over 11 years
    That is nice. Thank you a lot.
  • readingframe
    readingframe over 11 years
    Thank you for this answer. Right now it is beyond my current reach to really understand it. But hopefully i will understand it as i learn a little more about algebraic geometry in the weeks to come.
  • Admin
    Admin over 11 years
    Excellent comment.
  • Admin
    Admin over 11 years
    yes, my comment was on the remark called "comment":)