Let $X$ be a scheme with a point $x \in X$ and $\mathcal E$ a locally free sheaf of finite rank on $X$ with a global section $s \in \mathcal E(X)$.

If $\varphi \in Hom_{\mathcal O_X}(\mathcal E, \mathcal O_X)$, then $\varphi(X)(s) \in \mathcal O_X(X)$.

Suppose that the class of $\varphi(X)(s)$ in $\mathcal O_{X,x}/\mathcal m_x$ is zero for all $\varphi \in Hom_{\mathcal O_X}(\mathcal E, \mathcal O_X)$.

Does this imply that the class of $s$ in $\mathcal E_x /\mathcal m_x \cdot \mathcal E_x$ is zero?

  • $\begingroup$ With no additional conditions on $\varphi$ I guess the answer is no. Take $\varphi$ to be the zero-morphism and $s$ a non-zero section of $\mathcal{E}$. $\endgroup$ – Giovanni De Gaetano May 14 '12 at 10:43

No. The most brutal way to see it is to take an $\mathcal E$ which has a non-zero global section $0\neq s\in \Gamma (X, {\mathcal E})$ but such that $\Gamma (X, \check {\mathcal E})=0.$
If $X$ is reduced there exists an $x\in X$ such that $s(x)\neq 0\in \mathcal E_x /\mathcal m_x \cdot \mathcal E_x$ and this gives a negative answer to your question, since obviously $\phi(X)(s)=0$ for all $\phi \in \operatorname {Hom} _{\mathcal O_X}(\mathcal E, \mathcal O_X)=\Gamma (X, \check {\mathcal E})=0$.

Examples of the above set-up abound: the first that comes to mind is $X=\mathbb P_k^n$, $\; \mathcal E=\mathcal O(1)$ and $x\in \mathbb P_k^n$ arbitrary.


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