The following statement holds: if \(X\) is a nonsingular curve and \(Y\) is a projective variety, then every rational map \(\varphi: X \dashrightarrow Y\) extends to a morphism on all of \(X\).
Show by example that this result is false if either
\(\dim X \geq 2\), or
\(Y\) is not projective.
Solution
1. The extension theorem fails in dimension two even when the target is projective.
Take \[ X=\AA_k^2=\Spec k[x,y],\qquad Y=\PP_k^1, \] and define on \(X\setminus\{(0,0)\}\) the morphism \[ \varphi(x,y)=[x:y]. \] This represents a rational map \(X\dashrightarrow\PP^1\).
Suppose it extended to a morphism \[ \overline\varphi:\AA^2\longrightarrow\PP^1. \] On the punctured \(x\)-axis, \(\varphi(t,0)=[1:0]\). The restriction of \(\overline\varphi\) to the whole \(x\)-axis is a morphism \(\AA^1\to\PP^1\) which agrees with the constant map \([1:0]\) on the dense open subset \(\AA^1\setminus\{0\}\). Since \(\PP^1\) is separated, two morphisms agreeing on a dense open subset of the integral scheme \(\AA^1\) agree everywhere. Hence \[ \overline\varphi(0,0)=[1:0]. \]
Applying the same argument to the \(y\)-axis, where \[ \varphi(0,t)=[0:1]\qquad(t\ne0), \] gives \[ \overline\varphi(0,0)=[0:1], \] a contradiction. Thus the rational map does not extend across the origin. This disproves the dimension-\(\ge2\) analogue while retaining a projective target.
2. The extension theorem fails for a nonprojective target even when the source is a nonsingular curve.
Take \[ X=\AA_k^1=\Spec k[t],\qquad Y=\AA_k^1, \] and define on \(D(t)=\AA^1\setminus\{0\}\) the morphism \[ \psi(t)=t^{-1}. \] This is a rational map \(X\dashrightarrow Y\).
If it extended to a morphism \(\overline\psi:\AA^1\to\AA^1\), then \(\overline\psi\) would correspond to a polynomial \(p(t)\in k[t]\). On the dense open set \(D(t)\) we would have \[ p(t)=t^{-1}, \] so \(tp(t)=1\) in the function field \(k(t)\). Since both sides lie in \(k[t]\), this would be a polynomial identity, impossible because \(t\) is not a unit in \(k[t]\). Thus \(t^{-1}\) has no extension over \(0\) with values in \(\AA^1\).
The same rational function does extend to \(\PP^1\) by sending \(0\) to \(\infty\), which isolates precisely the missing projectivity of the target in this example.
3. Q.E.D.
Step 1 proves failure when the source has dimension at least two, and step 2 proves failure when the target is not projective.