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.

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

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