Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am just learning the basics of algebraic geometry, mainly by reading Hulek's book "Elementary algebraic geometry". I found there a problem concerning proving that the punctured affine plane $A:=\mathbb{A}^2_\mathbb{C}\setminus\{(0,0)\}$ is not isomorphic to either an affine or projective variety.

In many places I found the hint to consider the regular functions ring of $A$, which is simply $\mathbb{C}[x,y]$. And here my main problem starts -- I have a problem to understand properly what regular functions on an affine variety are. I know the definition of regularity at a point $P$: $f\in K(V)$ is regular at $P$ if there is a representation of $f$, say $\frac{g}{h}$, st. $h(P)\neq0$. (By $K(V)$ I denote the field of functions on $V$; $V$ is some affine variety). And now the function $f$ is regular on $A$ if it is regular at each point of $A$.

How can I deduce from these definitions of regularity, that every regular function on $A$ is just a polynomial from $\mathbb{C}[x,y]$. I saw answers using dimension theory or sheaf theory, but I am not familiar with it (and Hulek says nothing about it before giving the exercise). I cannot understand how can we get known something about a rational function, if the regularity is strictly local property. My intuitive point here is: the regular function $f$ can be represented by many functions $\frac{g}{h}$ which domains are not the entire domain of $f$ (but only agreeing on intersections of their domains).

If I understand this, I think I will understand the last part of proof of the affine case, which is as follows. Assume $A$ is affine. Because $\mathcal{O}(A)=\mathbb{C}[x,y]$, $\mathcal{O}(A)=\mathcal{O}(\mathbb{A}^2_\mathbb{C})$, so $K[A]\cong K[\mathbb{A}^2_\mathbb{C}]$. Consider now the rational natural embedding $i:A\to\mathbb{A}^2_\mathbb{C}$ -- it induces the isomorphism $i^*$ of $K[A]$ and $K[\mathbb{A}^2_\mathbb{C}]$, which is simply the identity function (so bijective). But $i$ is not bijective. A contradiction.

Thank you very much in advance for your hints and explanations. I would be also glad if you help with the projective case (ie. that $A$ is not isomorphic to a projective variety).

share|cite|improve this question
Hints: it is true that $f$ can be represented by many functions $g/h$. But as $\mathbb C[x,y]$ is a UFD, there is a more or less unique representation with $g, h$ coprime. – user18119 Nov 6 '11 at 22:23
To prove $A$ is not a projective variety, it depends on what you know (or is written in Hulek's boo) about projective varieties. Do you know something about the ring of regular functions on a projective variety ? And the compactness ? – user18119 Nov 6 '11 at 22:28
@QiL: I know that the ring of regular functions on a projective variety is isomorphic to the ground field, here $\mathbb{C}$. OK, so it is the answer for the projective case, too. – Damian Sobota Nov 6 '11 at 22:34

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.