Let \(X\) and \(Y\) be smooth projective curves over \(\CC\). Show that every birational morphism \[ f:X\longrightarrow Y \] is an isomorphism.
Solution
1. Both \(X\) and \(Y\) are normal.
A smooth variety is regular, and every regular local ring is integrally closed. Hence smooth varieties are normal.
2. Every fiber of \(f\) consists of a single point.
By step 1, \(Y\) is normal, so Zariski’s Main Theorem applies to the birational morphism \[ f:X\longrightarrow Y: \] every fiber \(f^{-1}(y)\) is connected, and a fiber which is not a singleton has positive dimension.
Suppose some fiber had positive dimension. It is a closed subset of the irreducible curve \(X\). Every proper closed subset of a curve has dimension \(0\), so a positive-dimensional fiber must equal \(X\). Then \(f\) would be constant, contradicting birationality. Thus no fiber has positive dimension. Zariski’s Main Theorem therefore forces every fiber to be a singleton.
3. The morphism \(f\) is finite.
Since \(X\) and \(Y\) are projective, \(f\) is proper and of finite type. By step 2 all fibers are finite, so \(f\) is quasi-finite. A proper quasi-finite morphism is finite. Hence \(f\) is finite.
4. A finite birational morphism onto the normal variety \(Y\) is anisomorphism.
Let \[ V=\Spec A\subseteq Y \] be affine. Since \(f\) is finite, \[ f^{-1}(V)=\Spec B \] for a finite \(A\)-algebra \(B\).
Birationality identifies the function fields \[ K(X)\cong K(Y)=K, \] and under this identification the inclusion induced by \(f\) places \[ A\subseteq B\subseteq K. \] Because \(B\) is finite over \(A\), every element of \(B\) is integral over \(A\). By step 1, \(Y\) is normal, so \(A\) is integrally closed in its fraction field \(K\). Therefore \[ B\subseteq A. \] Together with \(A\subseteq B\), this gives \[ A=B. \] Thus \(f^{-1}(V)\to V\) is an isomorphism for every affine open \(V\subseteq Y\). These local isomorphisms show that \(f\) is an isomorphism.
5. Q.E.D.
Step 3 makes \(f\) finite, and step 4 uses birationality and normality of \(Y\) to conclude that \(f\) is an isomorphism.