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”

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

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

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:

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 extension^{1}. 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.

## Comments (0)