Let \(X\) be an affine variety. Show that the Zariski topology on \(X\) is Hausdorff if and only if \(X\) is a singleton.
Solution
1. Every two nonempty Zariski-open subsets of \(X\) intersect.
An affine variety is irreducible by definition. Let \[ U,V\subseteq X \] be nonempty open subsets. If \[ U\intersect V=\varnothing, \] then \[ X = (X\sm U) \union (X\sm V). \] Both complements are proper closed subsets because \(U\) and \(V\) are nonempty. This writes the irreducible space \(X\) as the union of two proper closed subsets, a contradiction.
Therefore \[ U\intersect V\ne\varnothing. \]
2. If \(X\) contains two distinct points, then its Zariski topology is notHausdorff.
1. Hence a variety withat least two points is not Hausdorff.
3. A singleton variety is Hausdorff.
Every singleton topological space is Hausdorff: there are no two distinct points whose neighborhoods must be separated.
4. Therefore\[ \boxed{ X\text{ is Hausdorff in its Zariski topology} \quad\Longleftrightarrow\quad X\text{ is a point}. } \]
Step 2 shows that Hausdorffness forces \(X\) to contain at most one point. A variety is nonempty, so it must then be a singleton. Step 3 proves the converse.
5. Q.E.D.
Step 4 is the required conclusion.