Let \(k\) be an algebraically closed field of characteristic zero, and let \(C\) be a curve over \(k\). Show that the Zariski topology on \(C\) is the cofinite topology.
Solution
1. Every proper Zariski closed subset of \(C\) is finite.
Let \[ Z\subsetneq C \] be Zariski closed. A variety is Noetherian, so \(Z\) has only finitely many irreducible components: \[ Z=Z_1\cup\cdots\cup Z_r. \]
The curve \(C\) is irreducible of dimension \(1\). Each \(Z_i\) is an irreducible proper closed subset of \(C\), so \[ \dim Z_i<\dim C=1. \] Hence \[ \dim Z_i=0. \] Choose \(p\in Z_i\). Since \(k\) is algebraically closed, \(\{p\}\) is a closed irreducible subset of \(Z_i\). If \[ \{p\}\subsetneq Z_i, \] this would be a strict chain of nonempty irreducible closed subsets of length \(1\), contradicting \(\dim Z_i=0\). Thus \(Z_i=\{p\}\). Therefore every \(Z_i\) is a point, and \(Z\) is finite.
2. Every finite subset of \(C\) is Zariski closed.
Since \(k\) is algebraically closed, every point of a \(k\)-variety is a closed point. Thus every singleton \[ \{p\}\subseteq C \] is Zariski closed. A finite union of closed subsets is closed, so every finite subset of \(C\) is Zariski closed.
3. The Zariski closed subsets of \(C\) are exactly \(C\) and its finitesubsets.
Step 1 shows that every proper closed subset is finite. Step 2 shows that every finite subset is closed. Together with \(C\) itself, these are precisely the closed subsets.
4. Therefore the Zariski topology on \(C\) is the cofinite topology.
By definition, the cofinite topology is the topology whose closed subsets are the whole space and the finite subsets. Step 3 gives exactly this description for the Zariski topology on \(C\).
5. Q.E.D.
Step 4 is the required conclusion.