The Stacks project

Situation 33.47.2. Let $k$ be a field, let $X$ be a scheme over $k$, let $\mathcal{L}$ be an invertible $\mathcal{O}_ X$-module, let $V$ be a finite dimensional $k$-vector space, and let $\psi : V \to \Gamma (X, \mathcal{L})$ be a $k$-linear map. Say $\dim (V) = r$ and we have a basis $v_1, \ldots , v_ r$ of $V$. Then we obtain a “universal divisor”

\[ H_{univ} = Z(s_{univ}) \subset \mathbf{A}^ r \times _ k X \]

as the zero scheme (Divisors, Definition 31.14.8) of the section

\[ s_{univ} = \sum \nolimits _{i = 1, \ldots , r} x_ i \psi (v_ i) \in \Gamma (\mathbf{A}^ r \times _ k X, \text{pr}_2^*\mathcal{L}) \]

For a field extension $k'/k$ the $k'$-points $v \in \mathbf{A}^ r_ k(k')$ correspond to vectors $(a_1, \ldots , a_ r)$ of elements of $k'$. Thus we may on the one hand think of $v$ as the element $v = \sum _{i = 1, \ldots , r} a_ i v_ i \in V \otimes _ k k'$ and on the other hand we may assign to $v$ the section

\[ \psi (v) = \sum \nolimits _{i = 1, \ldots , r} a_ i \psi (v_ i) \in \Gamma (X_{k'}, \mathcal{L}|_{X_{k'}}) \]

With this notation it is clear that the fibre of $H_{univ}$ over $v \in V \otimes k'$ is the zero scheme of $\psi (v)$. In a formula:

\[ H_ v = H_{univ, v} = Z(\psi (v)) \]

We will denote this common value by $H_ v$ as indicated. Finally, in this situation let $P$ be a property of vectors $v \in V \otimes _ k k'$ for $k'/k$ an arbitrary field extension1. We say $P$ holds for general $v \in V \otimes _ k k'$ if there exists a nonempty Zariski open $U \subset \mathbf{A}^ r_ k$ such that if $v$ corresponds to a $k'$-point of $U$ for any $k'/k$ then $P(v)$ holds.

[1] For example we could consider the condition that $H_ v$ is smooth over $k'$, or geometrically irreducible over $k'$.

Comments (0)

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Your email address will not be published.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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0G47. Beware of the difference between the letter 'O' and the digit '0'.