Let \(k\) be an algebraically closed field of characteristic zero. Show that if \(X\) is a connected complete \(k\)-variety, then \[ \OO_X(X)=k, \] i.e. every global regular function is constant.
Solution
Let \[ f\in\OO_X(X). \] Regard \(f\) as a morphism \[ f:X\longrightarrow\AA^1_k, \] and let \[ j:\AA^1_k\hookrightarrow\PP^1_k \] be the standard open immersion.
1. The image\[ (j\circ f)(X) \]is closed in \(\PP^1_k\).
Because \(X\) is complete, for every \(k\)-variety \(Y\) the projection \[ \operatorname{pr}_2:X\times_k Y\longrightarrow Y \] is a closed map.
Take \(Y=\PP^1_k\). Since \(\PP^1_k\) is separated, the graph \[ \Gamma_{j\circ f} \subseteq X\times_k\PP^1_k \] is closed. Its image under \(\operatorname{pr}_2\) is exactly \[ \operatorname{pr}_2(\Gamma_{j\circ f}) =(j\circ f)(X). \] The projection is closed by completeness of \(X\), so this image is closed in \(\PP^1_k\).
2. The image \((j\circ f)(X)\) is finite.
By construction, \[ (j\circ f)(X)\subseteq j(\AA^1_k) = \PP^1_k\setminus\{\infty\}. \] Hence the closed subset \((j\circ f)(X)\) from step 1 is a proper closed subset of \(\PP^1_k\).
Every proper closed subset of \(\PP^1_k\) is finite: it is contained in the zero locus of a nonzero homogeneous polynomial in two variables, and such a zero locus has only finitely many points on \(\PP^1_k\). Therefore \((j\circ f)(X)\) is finite.
3. The image \((j\circ f)(X)\) consists of one point.
A morphism is continuous in the Zariski topology, so the image of the connected space \(X\) is connected.
By step 2, \((j\circ f)(X)\) is a proper closed subset of \(\PP^1_k\), hence a finite union of closed points. Since \(k\) is algebraically closed, those closed points are \(k\)-rational. A finite union of closed points has the discrete induced topology, and a finite discrete space is connected only when it has one point. Hence \[ (j\circ f)(X)=\{a\} \] for some \[ a\in\AA^1(k)=k. \]
4. Every \(f\in\OO_X(X)\) is the constant function \(a\) from step
3.
The open immersion \(j\) is injective on points. Since \[ (j\circ f)(X)=\{a\}, \] one has \[ f(X)=\{a\}\subseteq\AA^1_k. \] Therefore \[ f=a \] as a global regular function.
5. Consequently,\[ \boxed{\OO_X(X)=k.} \]
Constant functions give the canonical inclusion \[ k\hookrightarrow\OO_X(X). \] Step 4 shows that every global regular function lies in its image. Therefore the inclusion is an equality.
6. Q.E.D.
Step 5 is the desired conclusion.