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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?

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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.

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That is simple indeed! Thanks. –  solbap Nov 24 '10 at 22:53
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