# Zariski topology irreducible affine curve is same as the cofinite topology

Is Zariski topology on any irreducible affine curve is same as the cofinite topology. Try: I proved the satement for irreducible affine plane curve.

1) No, despite a widespread misconception, the Zariski topology for any curve seen as a scheme never coincides with the cofinite topology.
This is because the curve has a generic point if it is irreducible and several generic points if it is reducible and these points are not closed, so that the Zariski topology cannot be the cofinite topology. (In a cofinite topology all points are closed.)

2) However in elementary algebraic geometry over an algebraically closed field $k$ one may consider only the closed points of the curve, which in the affine pieces correspond to maximal ideals of the relevant $k-$algebra. This is, for example, Fulton's point of view in his celebrated (now freely available online) book Algebraic Curves.
In that context it is indeed true that the Zariski topology of the curve coincides with the cofinite topology.

• @ Geoges Elencwajg . Sir, I donot know what are closed points of the curve. I am only doing an introductory course in Algebraic geometry. Commented Apr 13, 2013 at 8:08
• Dear kushal, in the version of algebraic geometry you use, yes, the Zariski topology coincides with the cofinite topology. Do not take point 1) of my answer into account (for the time being: come back to it later, when you learn about schemes!) and only consider point 2) . Commented Apr 13, 2013 at 8:14
• @ Geoges Elencwajg .What is the proof of that? Can it be done using Noether`s normalization as stated below. If yes, then how? Commented Apr 13, 2013 at 8:17
• Dear kushal: no, you don't need anything deep. This is more or less by definition : the curve $X$ has dimension $1$ so any irreducible strict algebraic subset $Y$ is a point, else we would get a chain $\{y\}\subsetneq Y\subsetneq X$, contradicting that $X$ has dimension $1$. Commented Apr 13, 2013 at 8:31
• @GeorgesElencwajg: I have removed the wiki, however, it appears that the wiki was accidentally set when you created the post.
– robjohn
Commented Apr 13, 2013 at 8:44

Let $\mathfrak{p} \subset k[x_1,x_2 \ldots x_n]=A$ be the prime ideal of the irreducible curve. Then $A/\mathfrak{p}$ is a one dimensional domain. So by Noether's normalization its an integral extension of $k[y]$ Now you prove that Zariski topology on an integral extension of $k[y]$ is the cofinite topology.