4.4.1 Nagata’s Example
Then take \(C\) to be an smooth elliptic curve over \(\mathbb {C}\) with origin \(P_0\in C\), and embed \(C\) in \(X_0 = \mathbb {P}^2_{\mathbb {C}}\) using the complete linear system \(|3P_0|\), and let \(E_0\) denote the image of \(C\) in \(X_0\). Then, if we identify \(C\) with \(X_0\), the group structure on \(X_0\) is such that, if \(L\) is a line in \(\mathbb {P}^2 = X_0\) and \(L\cap X_0 = Q_1 + Q_2 + Q_3\), with the \(Q_i\) possibly nondistinct points, then \(Q_1 + Q_2 + Q_3 = 0\) on \(E_0\). And, conversely, if \(Q_1\), \(Q_2\) and \(Q_3\) are three possibly nondistinct points on \(X_0\), then \(Q_1 + Q_2 + Q_3 = 0\) on \(X_0\) if and only if there is a line \(L\) with \(L\cap X_0 = Q_1 + Q_2 + Q_3\).
A point \(Q\in E_0\) has finite order if and only if there exists a curve \(D\in \mathbb {P}^2\) such that \(D\cap E_0 \subseteq \{ Q\} \).
(\(\Rightarrow \)) Suppose \(Q\) has order \(m{\lt}\infty \). Then \(m(Q - P_0)\sim 0\). So \(mQ \sim mP_0\), and, therefore, \(3mQ \sim 3mP_0\). So, as \(3P_0\in |\mathcal{O}_{C}(1)|\), \(3mQ\in |\mathcal{O}_C(m)|\).
(\(\Leftarrow \)) Suppose there exists a curve \(D\in \mathbb {P}^2\) such that \(D\cap E_0 = \{ Q\} \). Let \(d = \deg D\) so that \(D\in \mathcal{O}_{\mathbb {P}^2}(d)\). Then \(D\cap E_0 = 3dQ\) as a divisor. So \(3dQ \sim 3dP_0\). So the order of \(Q\) divides \(3d\).
Now, take \(Q_0\in X_0\) to be a point of infinite order, let \(X_1\) be the blowup of \(X_0\) and \(Q_0\) and let \(E_1\) be the proper transform of \(E_0\) in \(X_1\). The point \(Q_0\) has multiplicity \(1\) in \(E_0\) (as \(E_0\) is smooth), and \(E_0^2 = 9\). So \(E_1^2 = 8\). The restriction of the blow up map \(\pi _1:X_1\to X_0\) to \(E_1\) induces an isomorphism \(E_1\to E_0\) (as \(E_1\) is the blowup of \(E_0\) at \(Q\)). So there is a unique point \(Q_1\) in \(E_1\) lying over \(Q_0\).
So let \(X_2\) be the blowup of \(X_1\) and \(Q_1\). Continuing in this way, we get an infinite sequence of blowups \(\pi _{i+1}:X_{i+1}\to X_{i}\) of surfaces \(X_{i}\) at points \(Q_{i}\) in elliptic curves \(E_i\) with \(E_i^2 = 9 - i\). In particular, \(E_{10}\) is an elliptic curve on \(X_{10}\) with \(E_{10}^2 = -1\).
Set \(X=X_{10}, E=E_{10}\) and \(Q=Q_{10}\). There is no morphism \(f:X\to Y\) to a scheme \(Y\) with the following properties:
\(f\) is proper.
\(f(E)\) is a normal point \(y\in Y\).
The restriction of \(f\) to \(X\setminus E\) induces an isomorphism \(X\setminus E\xrightarrow {\sim } Y\setminus \{ y\} \).
Suppose such a morphism exists. By b and c, \(f\) is onto. Since \(f\) is proper and connected, it follows that \(Y\) is also proper and connected. Moreover, by c, \(\dim Y = \dim X = 2\). By taking the complement of a Zariski open neighborhood of \(Y\), we can find a curve \(D\subseteq Y\) such that \(y\not\in D\). The, using c again, we can regard \(D\) as a curve in \(X\).
Write \(\pi :X\to X_0 = \mathbb {P}^2\) for the map down to \(\mathbb {P}^2\). Then it follows that \(F = \pi (D)\) is a curve in \(\mathbb {P}^2\) containing no points of \(E_0\) other than (possibly) \(Q_0\). Since \(\deg F\cap E_0 = 3\deg F {\gt} 0\) (by Bezout), we must have \(F\cap E_0 = 3(\deg F) Q_0\). But this contradicts the assumption that \(Q_0\) had infinite order.