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Let $X:=V(x^m-y^n)$ be a subspace of $\mathbb{A}^2$. How can I prove that if $(n,m)=1$ then $X$ is irreducible?

I think that it is isomorphic to $\mathbb{P}^1$ but I can't prove that.

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It's not isomorphic to $\mathbb{P}^1$; if $m, n \ge 2$ it has a singularity at the origin. –  Qiaochu Yuan Jun 11 '11 at 23:01
It cannot be isomorphic to $P^1$, because there are non-constant regular functions on it (for example, the restrictions to it of the coordinate functions) (More generally, no algebraic subset of an affine space is projective) –  Mariano Suárez-Alvarez Jun 11 '11 at 23:02
Ha. I meant it's not isomorphic to $\mathbb{A}^1$, which I assume is what Jacob meant. –  Qiaochu Yuan Jun 11 '11 at 23:14
@Qiaochu: Yup, that's a more likely conjecture. A simple way to see it is not an $A^1$, avoiding determining singularities, is to see that the coordinate ring of $X$ is not integrally closed (this is almost the same thing, really, but proving so requires a bit of work) –  Mariano Suárez-Alvarez Jun 11 '11 at 23:27
@Mariano: I think you meant to say "no non-finite subset of affine space is projectve" (as a special case of the general fact that finite = projective (or even proper) + affine). –  Akhil Mathew Jun 11 '11 at 23:34

1 Answer 1

up vote 5 down vote accepted

To prove that it is irreducible, show that it is the image of an irreducible space under a continuous map.

For example, you should have no trouble finding a map $\mathbb A^1\to\mathbb A^2$ whose image is your $X$.

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@Jacob Fox: Dear Jacob, If you complete Mariano's exercise, you will also see that it is birational to $\mathbb P^1$, although is is not actually isomorphic to (any open subset of) $\mathbb P^1$, as noted in the comments to the question. Regards, –  Matt E Jun 12 '11 at 0:11

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