Lemma 33.22.2. Let $k$ be an infinite field. Let $X$ be a scheme of finite type over $k$. Let $\mathcal{L}$ be a very ample invertible sheaf on $X$. Let $n \geq 0$ and $x, x_1, \ldots , x_ n \in X$ be points with $x$ a $k$-rational point, i.e., $\kappa (x) = k$, and $x \not= x_ i$ for $i = 1, \ldots , n$. Then there exists an $s \in H^0(X, \mathcal{L})$ which vanishes at $x$ but not at $x_ i$.
Proof. If $n = 0$ the result is trivial, hence we assume $n > 0$. By definition of a very ample invertible sheaf, the lemma immediately reduces to the case where $X = \mathbf{P}^ r_ k$ for some $r > 0$ and $\mathcal{L} = \mathcal{O}_ X(1)$. Write $\mathbf{P}^ r_ k = \text{Proj}(k[T_0, \ldots , T_ r])$. Set $V = H^0(X, \mathcal{L}) = kT_0 \oplus \ldots \oplus kT_ r$. Since $x$ is a $k$-rational point, we see that the set $s \in V$ which vanish at $x$ is a codimension $1$ subspace $W \subset V$ and that $W$ generates the homogeneous prime ideal corresponding to $x$. Since $x_ i \not= x$ the corresponding homogeneous prime $\mathfrak p_ i \subset k[T_0, \ldots , T_ r]$ does not contain $W$. Since $k$ is infinite, we then see that $W \not= \bigcup W \cap \mathfrak q_ i$ and the proof is complete. $\square$
Post a comment
Your email address will not be published. Required fields are marked.
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).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)