MathOverflow is a question and answer site for professional mathematicians. 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 recently stumbled upon a paper (NON CANCELLATION FOR SMOOTH CONTRACTIBLE AFFINE THREEFOLDS) about the cancellation problem: If $X$ is a variety over $\mathbb{C}$ of dimension $d$ such that $X \times \mathbb{A}^n \cong \mathbb{A}^{n+d}$ when is $X \cong \mathbb{A}^d$?

Apparently when $d = 1,2$ the answer is always. The paper remarked that the case $d = 1$ is trivial, but I'm having trouble coming up with an argument. The paper does reference a paper (ON THE UNIQUENESS OF THE COEFFICIENT RING IN A POLYNOMIAL RING) which proves the more general cancellation problem -when does $X \times \mathbb{A}^n \cong Y \times \mathbb{A}^n$ imply $X \cong Y$- for curves (again the answer for curves is always).

However I wonder if there is a simple argument for the less general question. I can see that $X \times \mathbb{A}^n \cong \mathbb{A}^{n+1}$ implies that $X$ should be irreducible, affine and smooth. Seems like if you could show genus$(X) = 0$ you would be done, but I'm kinda stuck here.

Does someone have a simple argument for this?

share|cite|improve this question
up vote 13 down vote accepted

The only smooth affine curve admitting a non-constant map from an affine space is $\mathbb A^1$. It must have genus 0, because of Lüroth's theorem, so it is $\mathbb P^1$ minus $d$ points for some $d$. But if $d$ were larger than 1 the curve would have a non-constant invertible function, which would pull back to a non-constant invertible function on an affine space, and this is impossible.

share|cite|improve this answer
That is simple indeed! Thanks. – solbap Nov 24 '10 at 22:53

Your Answer


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

Not the answer you're looking for? Browse other questions tagged or ask your own question.