If \(X\) is a birationally ruled surface, show that the curve \(C\), such that \(X\) is birationally equivalent to \(C \times \PP^1\), is unique (up to isomorphism).
Solution
Suppose \[ C\times\PP^1\dashrightarrow C'\times\PP^1 \] is a birational map \(\Phi\), with inverse \(\Psi\). Let \[ p:C\times\PP^1\to C, \qquad p':C'\times\PP^1\to C' \] be the projections.
1. If \(B\) is a nonsingular projective curve not isomorphic to \(\PP^1\), thenevery rational map\[ \PP^1\dashrightarrow B \]is constant.
A rational map from the nonsingular curve \(\PP^1\) to the projective curve \(B\) extends to a morphism. If it were nonconstant, it would be dominant and would give an inclusion of function fields \[ k(B)\hookrightarrow k(t). \] By Lüroth’s theorem, every intermediate field of transcendence degree one in \(k(t)/k\) is rational. Thus \(k(B)\cong k(u)\), so \(B\) is birational to \(\PP^1\). Since both curves are nonsingular and projective, the uniqueness of smooth projective models T-CRVMINMOD would give \[ B\cong\PP^1, \] contrary to the hypothesis. Hence the map is constant.
2. If \(C'\not\cong\PP^1\), then the rational map\[ p'\circ\Phi:C\times\PP^1\dashrightarrow C' \]factors uniquely as\[ C\times\PP^1\dashrightarrow C\dashrightarrow C', \]that is,\[ p'\circ\Phi=f\circ p \]for a dominant rational map \(f:C\dashrightarrow C'\).
Restrict \(p'\circ\Phi\) to the generic fibre of \(p\). After extending the ground field from \(k\) to \(k(C)\), this is a rational map \[ \PP^1_{k(C)}\dashrightarrow C'_{k(C)}. \] If it were nonconstant, after passage to an algebraic closure of \(k(C)\) it would give a nonconstant rational map from \(\PP^1\) to the base change of \(C'\). Step 1 would force that base-changed curve, and hence \(C'\), to have genus zero, so \(C'\cong\PP^1\). This contradicts the hypothesis. Thus the map is constant on the generic \(p\)-fibre.
Equivalently on function fields, the pullback \[ k(C')\longrightarrow k(C)(t) \] has image in the subfield \(k(C)\). It therefore defines a rational map \[ f:C\dashrightarrow C' \] with \(p'\circ\Phi=f\circ p\). Since both \(\Phi\) and \(p'\) are dominant, so is \(f\).
3. If neither \(C\) nor \(C'\) is isomorphic to \(\PP^1\), then \(C\) and \(C'\) arebirational.
By step 2, \[ p'\circ\Phi=f\circ p \] for a dominant rational map \(f:C\dashrightarrow C'\). Applying the same argument to the inverse birational map \(\Psi\) gives a dominant rational map \(g:C'\dashrightarrow C\) with \[ p\circ\Psi=g\circ p'. \] On the common dense open set where the birational compositions are defined, \[ \begin{aligned} p &=p\circ\Psi\circ\Phi\\ &=g\circ p'\circ\Phi\\ &=g\circ f\circ p. \end{aligned} \] Since \(p\) is dominant, \[ g\circ f=\id_C \] as rational maps. Similarly, \[ f\circ g=\id_{C'}. \] Thus \(f\) and \(g\) are inverse birational maps.
4. If one of \(C,C'\) is isomorphic to \(\PP^1\), then so is the other.
1. Hence \(C\cong\PP^1\). The otherdirection is symmetric.
5. The base curve of a birationally ruled surface is unique up toisomorphism:\[ \boxed{C\cong C'.} \]
If one curve is rational, step 4 shows both are \(\PP^1\). Otherwise step 3 shows that \(C\) and \(C'\) are birational. Nonsingular projective curves have unique smooth projective models by T-CRVMINMOD, so birationality implies isomorphism.
6. Q.E.D.
Steps <1>1–<1>5 prove the required uniqueness of the base curve.