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Can any one help me to understand why all the proper Zariski closed sets of the parabola $V=\mathbb{V}(y-x^2)\subseteq\mathbb{A}^2$ are finite?Indeed, the Zariski topology on any plane curve is the co-finite topology, provided the curve is not a union of two other curves.

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Try to prove that $V$ is isomorphic, hence homeomorphic, to $\mathbb A^1$. – Georges Elencwajg Feb 9 '12 at 15:29
The parabola is homeomorphic to line under the morphism $\mathbb{A}^1 ---->\mathbb{C}$,( t----->(t,t^2)), so? – Un Chien Andalou Feb 9 '12 at 15:32
$k[x]$ is a principal ideal domain, so any closed subset of $\mathbb A^1$ is the set of zeros of a single polynomial. So the proper Zariski closed sets are finite subsets of $\mathbb A^1$ – Matt Feb 9 '12 at 15:58
@puchkusona: you are reduced to the same question for $\mathbb A^1$. And there a closed set is given by the zero set of a polynomial. – Georges Elencwajg Feb 9 '12 at 15:58
up vote 1 down vote accepted

The key of understanding what you said is that for a curve $f(x,y)=0$ in $\Bbb A(K)^2$ the closed Zariski subsets correspond (roughly) to ideals $I\subseteq A:=K[x,y]/(f(x,y))$. The geometrical points in the closed set corresponding to the ideal $I$ are given by the maximal ideals containing $I$ (to avoid headaches one may assume that $K$ is algebraically closed, e.g. $K=\Bbb C$).

One knows that:

  • There is a 1-1 correspondence between ideals in $A$ and ideals in $K[x,y]$ containing $f(x,y)$

  • If the curve (equivalently: the polynomial $f(x,y)$) is irreducible, every non-zero prime ideal in $A$ is maximal.

Thus, either $I$ in $A$ is the zero ideal and the whole curve is the closed set corresponding to it, or $I$ is not zero and so is contained in only finitely many maximal ideals.

When the curve is not irreducible, say $f(x,y)=g_1(x,y)g_2(x,y)$ as product of irreducibles, then the ideals $I_1=(g_1)$ and $I_2=(g_2)$ are prime in $A$ but not maximal and correspond to infinite Zariski closed sets contained in the curve (namely, the two components).

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