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Let $k$ be an algebraically closed field, let $X$ be a projective scheme over $k$ and let $f:X\rightarrow \mathbb{P}^n_k$ be a morphism over $k$. Let $\mathcal{L}$ be an invertible sheaf generated global sections $s_0, \dotsc s_n $ such that $\mathcal{L} \cong f^* \mathcal{O}(1)$ and $f^*(x_i)=s_i$ for $i=0, \dotsc, n$. Let $V \subseteq \Gamma(X, \mathcal{L})$ be the subspace spanned by the $s_i$ and suppose that for any two distinct closed points $P,Q \in X$ there is an $s\in V$ such that $s_P \in \mathfrak{m}_P\mathcal{L}_P$ but $s_Q \notin \mathfrak{m}_Q\mathcal{L}_Q$ or vice versa.

Then, in Hartshorne's book it is claimed that $f$ is clearly injective, but I don't know how to prove this fact.... Help me....

Martin Brandenburg
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Sang Cheol Lee
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1 Answers1

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Since $s\in V$, we have $s = \sum_i c_i s_i$ for some coefficients $c_i\in k$, and the condition says that $s(P) = \sum_i c_i s_i(P) = 0$ and $s(Q) = \sum_i c_i s_i(Q)\neq 0$. If it were true that $f(P) = f(Q)$, i.e., $[s_1(P):\cdots:s_n(P)] = [s_1(Q):\cdots:s_n(Q)]$, then we would have $\lambda s_i(P) = s_i(Q)$ for all $i$ and some $\lambda\in k^\times$. In particular, this gives $s(Q) = \sum_i c_i s_i(Q) = \sum_ic_i(\lambda s_i(P)) = \lambda s(P) = 0$, which contradicts the assumption on the linear series $V$.

Andrew
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  • Thanks Andrew... But I don't understand that $f(P)=[s_1(P): \cdots s_n(P)]$... $f$ be a morphsim of schmes.... i.e.$\mathbb{P}^n_k=\operatorname{Proj}k[x_0,\cdots,x_n]$.... – Sang Cheol Lee Aug 17 '13 at 00:40
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    Dear @SangCheolLee, in scheme notation, maybe it is better to do as follows: if $f(P) = f(Q)$, then for all $t\in\Gamma(\mathbb P^n,\mathcal O(1))$ we have $t_{f(P)} = t_{f(Q)}$. Given $s$ as above, we can find $t$ such that $s = f^\ast t$, and we can show $s_P = (f^\ast t)P = t{f(P)}$, showing that $s_P = s_Q$, a contradiction. – Andrew Aug 17 '13 at 19:11