Chapter 6: Problem 43
Let \(C\) be a projective curve, \(P \in C .\) Then there is a birational morphism \(f: C \rightarrow\) \(C^{\prime}, C^{\prime}\) a projective plane curve, such that \(f^{-1}(f(P))=\\{P\\} .\) We outline a proof: (a) We can assume: \(C \subset \mathbb{P}^{n+1}\) Let \(T, X_{1}, \ldots, X_{n}, Z\) be coordinates for \(\mathbb{P}^{n+1}\); Then \(C \cap V(T)\) is finite; \(C \cap V(T, Z)=\varnothing ; P=[0: \ldots: 0: 1] ;\) and \(k(C)\) is algebraic over \(k(u)\) where \(u=\bar{T} / \bar{Z} \in k(C)\). (b) For each \(\lambda=\left(\lambda_{1}, \ldots, \lambda_{n}\right) \in k^{n}\), let \(\varphi_{\lambda}: C \rightarrow \mathbb{P}^{2}\) be defined by the formula \(\varphi([t:\) \(\left.\left.x_{1}: \ldots: x_{n}: z\right]\right)=\left[t: \sum \lambda_{i} x_{i}: z\right] .\) Then \(\varphi_{\lambda}\) is a well-defined morphism, and \(\varphi_{\lambda}(P)=\) \([0: 0: 1] .\) Let \(C^{\prime}\) be the closure of \(\varphi_{\lambda}(C)\) (c) The variable \(\lambda\) can be chosen so \(\varphi_{\lambda}\) is a birational morphism from \(C\) to \(C^{\prime}\), and \(\varphi_{\lambda}^{-1}([0: 0: 1])=\\{P\\} .\) (Use Problem \(6.32\) and the fact that \(C \cap V(T)\) is finite).
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