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. 
 A: 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.
A: 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.
