Let \(X\) be a nonsingular projective curve. Suppose that \(X\) is a locally closed subvariety of a variety \(Y\). Show that \(X\) is in fact a closed subset of \(Y\).
Solution
1. The locally closed inclusion of \(X\) in \(Y\) is a morphism of varieties.
By hypothesis there is an open subset \(U\subseteq Y\) such that \(X\) is closed in \(U\). The inclusion \(X\hookrightarrow U\) is a closed immersion of varieties and the inclusion \(U\hookrightarrow Y\) is an open immersion. Their composite \[ i:X\longrightarrow Y \] is therefore a morphism whose image is the given subset \(X\subseteq Y\).
2. The variety \(X\) is complete.
The curve \(X\) is projective over \(k\) by hypothesis. Projective varieties are complete, so \(X\) is complete in the sense of Chapter I.
3. The image \(i(X)\) is closed in \(Y\).
1.Since \(X\) is complete by step 2, the subset\[ i(X)=X \]is closed in \(Y\).
4. Q.E.D.
Step 3 is exactly the required conclusion.