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A hypersurface $C \subset \mathbb{A}^2$ we call a plane curve. Show that any infinite subset of an irreducible plane curve $C \subset \mathbb{A}^2$ is dense in $C.$

Note. We call hypersurface the zero set $Z(f)$ of a non-constant polynomial $f \in k[x_1,\ldots,x_n],$ and we don't know the dimension of a hypersurface, nor that $\dim \mathbb{A}^n=n.$

Edit: $k$ algebraically closed.

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There is a quicker proof than mine given below, which is also more elementary. Namely, use the fact that two non-constant polynomials $f,g \in k[x,y]$ with no common factor have only finitely many common zeros (c.f. Fulton, "Algebaic Curves", p.9). –  Nils Matthes Mar 26 '13 at 17:19
    
Thanks for the hints: using a corollary of your hint, which is proved in the book you mention, I can easily convince myself that $\dim\mathbb{A}^2=2;$ from this it is straightforward to solve the exercise. –  jj_p Mar 27 '13 at 11:16
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up vote 2 down vote accepted

Hint: Every plane curve is homeomorphic (though not isomorphic!!) to $\mathbb{A}^1$ (why?). So it suffices to show that every infinite subset of $\mathbb{A}^1$ is dense in $\mathbb{A}^1$. This comes down (why?) to the fact that the only polynomial in $k[x]$ having infinitely many zeros is the zero polynomial.

Edit: I should add that I have $k$ algebraically closed in mind.

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Hmm, I think it can. It cannot be isomorphic to the affine line though (by homeomorphic I mean homeomorphism of the underlying topological spaces, not isomorphism of the curves as varieties). It that what you meant? –  Nils Matthes Mar 26 '13 at 16:22
    
Actually, I just came back to say something like "Hmm, I think it can"! I might remove my comments to avoid confusion... –  Rhys Mar 26 '13 at 16:26
    
Still thanks for you inital objection, I think I have to make the distinction homeomorphic vs. isomorphic clearer. –  Nils Matthes Mar 26 '13 at 16:28
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