Use the result of (Ex. 9.1) to prove the following geometric result. Let \(k\) be algebraically closed, let \(Y\subseteq X=\PP_k^n\) be a connected nonsingular closed subvariety of positive dimension, and let \(f:X\to Z\) be a morphism of \(k\)-varieties. Suppose that \(f(Y)\) is a single closed point \(P \in Z\). Then \(f(X) = P\) also.
Solution
Let \(\mathcal I\) be the ideal sheaf of \(Y\) in \(X\) and let \(\hat X\) be the formal completion along \(Y\). By P-AGH291FORMALREG, the constants map identifies \(\Gamma(\hat X,\OO_{\hat X})\) with \(k\).
1. For every open neighborhood \(U\) of \(Y\) in \(X\), the restriction homomorphism\[ \Gamma(U,\OO_X)\longrightarrow\Gamma(\hat X,\OO_{\hat X}) \]is injective.
The completion of \(U\) along \(Y\) is canonically \(\hat X\), since the ideal quotients on the common underlying space \(Y\) are unchanged after restricting to a neighborhood of \(Y\). Reduction modulo each \(\mathcal I^r\) therefore defines the stated homomorphism to the inverse-limit section ring.
Suppose \(a\in\Gamma(U,\OO_X)\) maps to zero. Choose a point \(y\in Y\), which is possible by the positive-dimensional hypothesis. The germ \(a_y\) lies in \((\mathcal I_y)^r\) for every \(r\ge1\), since its restriction to every infinitesimal neighborhood is zero. The ring \(\OO_{X,y}\) is noetherian and local, and \(\mathcal I_y\) is a proper ideal. Thus Krull’s intersection theorem gives \[ \bigcap_{r\ge1}(\mathcal I_y)^r=0, \] so \(a_y=0\). The section \(a\) is then zero on some nonempty open neighborhood of \(y\) in \(U\). Since \(U\) is integral, a regular function is determined by its generic value; hence \(a=0\) on \(U\). This proves injectivity without assuming that \(U\) is affine.
2. Every regular function on an open neighborhood \(U\) of \(Y\) is constant.
By step 1 and the formal-function calculation, there is an injective \(k\)-algebra map \(\Gamma(U,\OO_X)\to k\) that is the identity on constants. For any \(a\) in its source, let \(c\in k\) be its image. Then \(a-c\) has image zero, so injectivity gives \(a=c\). Consequently \(\Gamma(U,\OO_X)=k\) via the constants map.
3. The morphism \(f\) is constant with value \(P\) on a dense open neighborhood of \(Y\).
Choose an affine open \(V=\Spec B\subseteq Z\) containing \(P\) and set \(U=f^{-1}(V)\). The hypothesis \(f(Y)=P\) implies \(Y\subseteq U\). Thus \(U\) is nonempty and dense in the integral scheme \(X\). Step 2 identifies the global-functions map of \(f|_U\) with a \(k\)-algebra homomorphism \[ B\longrightarrow\Gamma(U,\OO_U)=k. \] It defines a \(k\)-point \(Q\in V\), and \(f|_U\) is the composite \(U\to\Spec k\xrightarrow{Q}V\). Indeed, this composite and \(f|_U\) have identical maps on coordinate rings on every affine open of \(U\), so they agree as morphisms. As \(U\) contains \(Y\) and \(f(Y)=P\), the point \(Q\) must be \(P\). The closed point \(P\) is \(k\)-rational because \(k\) is algebraically closed.
4. Q.E.D.
3.It therefore has underlying set all of \(X\), proving \(f(X)=\{P\}\).In fact the morphism itself is the constant morphism to \(P\): it agrees with that morphism on \(U\), the source is reduced, and the target variety is separated, so P-AGH242AGREEDENSE applies.