Lemma 50.23.11. Let $k$ be a field. Let $X$ be an irreducible smooth proper scheme over $k$ of dimension $d$. Let $Z \subset X$ be the reduced closed subscheme consisting of a single $k$-rational point $x$. Then the image of $1 \in k = H^0(Z, \mathcal{O}_ Z) = H^0(Z, \Omega ^0_{Z/k})$ by the map $H^0(Z, \Omega ^0_{Z/k}) \to H^ d(X, \Omega ^ d_{X/k})$ of Remark 50.23.7 is nonzero.

Proof. The map $\gamma ^0 : \mathcal{O}_ Z \to \mathcal{H}^ d_ Z(\Omega ^ d_{X/k}) = R\mathcal{H}_ Z(\Omega ^ d_{X/k})[d]$ is adjoint to a map

$g^0 : i_*\mathcal{O}_ Z \longrightarrow \Omega ^ d_{X/k}[d]$

in $D(\mathcal{O}_ X)$. Recall that $\Omega ^ d_{X/k} = \omega _ X$ is a dualizing sheaf for $X/k$, see Duality for Schemes, Lemma 48.27.1. Hence the $k$-linear dual of the map in the statement of the lemma is the map

$H^0(X, \mathcal{O}_ X) \to \mathop{\mathrm{Ext}}\nolimits ^ d_ X(i_*\mathcal{O}_ Z, \omega _ X)$

which sends $1$ to $g^0$. Thus it suffices to show that $g^0$ is nonzero. This we may do in any neighbourhood $U$ of the point $x$. Choose $U$ such that there exist $f_1, \ldots , f_ d \in \mathcal{O}_ X(U)$ vanishing only at $x$ and generating the maximal ideal $\mathfrak m_ x \subset \mathcal{O}_{X, x}$. We may assume assume $U = \mathop{\mathrm{Spec}}(R)$ is affine. Looking over the construction of $\gamma ^0$ we find that our extension is given by

$k \to (R \to \bigoplus \nolimits _{i_0} R_{f_{i_0}} \to \bigoplus \nolimits _{i_0 < i_1} R_{f_{i_0}f_{i_1}} \to \ldots \to R_{f_1\ldots f_ r})[d] \to R[d]$

where $1$ maps to $1/f_1 \ldots f_ c$ under the first map. This is nonzero because $1/f_1 \ldots f_ c$ is a nonzero element of local cohomology group $H^ d_{(f_1, \ldots , f_ d)}(R)$ in this case, $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).