**Proof.**
In order for assumption (3) to hold, the elements $v_1, v_2, v_3$ must be $k$-linearly independent in $V$ (small detail omitted). Thus we may choose a basis $v_1, \ldots , v_ r$ of $V$ incorporating these elements as the first $3$. Recall that $H_{univ} \subset \mathbf{A}^ r_ k \times _ k X$ is the “universal divisor”. Consider the projection $q : H_{univ} \to \mathbf{A}^ r_ k$ whose scheme theoretic fibres are the divisors $H_ v$. By Lemma 37.26.5 it suffices to show that the generic fibre of $q$ is geometrically irreducible. To prove this we may replace $X$ by its reduction, hence we may assume $X$ is an integral scheme of finite type over $k$.

Let $U \subset X$ be a nonempty affine open such that $\mathcal{L}|_ U \cong \mathcal{O}_ U$. Write $U = \mathop{\mathrm{Spec}}(A)$. Denote $f_ i \in A$ the element corresponding to section $\psi (v_ i)|_ U$ via the isomorphism $\mathcal{L}|_ U \cong \mathcal{O}_ U$. Then $H_{univ} \cap (\mathbf{A}^ r_ k \times _ k U)$ is given by

\[ H_ U = \mathop{\mathrm{Spec}}(A[x_1, \ldots , x_ r]/(x_1f_1 + \ldots + x_ rf_ r)) \]

By our choice of basis we see that $f_1$ cannot be zero because this would mean $v_1 = 0$ and hence $H_{v_1} = X$ which contradicts assumption (3). Hence $\sum x_ if_ i$ is a nonzerodivisor in $A[x_1, \ldots , x_ r]$. It follows that every irreducible component of $H_ U$ has dimension $d + r - 1$ where $d = \dim (X) = \dim (A)$. If $U' = U \cap D(f_1)$ then we see that

\[ H_{U'} = \mathop{\mathrm{Spec}}(A_{f_1}[x_1, \ldots , x_ r]/(x_1f_1 + \ldots + x_ rf_ r)) \cong \mathop{\mathrm{Spec}}(A_{f_1}[x_2, \ldots x_ r]) = \mathbf{A}^{r - 1}_ k \times _ k U' \]

is irreducible. On the other hand, we have

\[ H_ U \setminus H_{U'} = \mathop{\mathrm{Spec}}(A/(f_1)[x_1, \ldots , x_ r]/(x_2f_2 + \ldots + x_ rf_ r)) \]

which has dimension at most $d + r - 2$. Namely, for $i \not= 1$ the scheme $(H_ U \setminus H_{U'}) \times _ U D(f_ i)$ is either empty (if $f_ i = 0$) or by the same argument as above isomorphic to an $r - 1$ dimensional affine space over an open of $\mathop{\mathrm{Spec}}(A/(f_1))$ and hence has dimension at most $d + r - 2$. On the other hand, $(H_ U \setminus H_{U'}) \times _ U V(f_2, \ldots , f_ r)$ is an $r$ dimensional affine space over $\mathop{\mathrm{Spec}}(A/(f_1, \ldots , f_ r))$ and hence assumption (4) tells us this has dimension at most $d + r - 2$. We conclude that $H_ U$ is irreducible for every $U$ as above. It follows that $H_{univ}$ is irreducible.

Thus it suffices to show that the generic point of $H_{univ}$ is geometrically irreducible over the generic point of $\mathbf{A}^ r_ k$, see Varieties, Lemma 33.8.6. Choose a nonempty affine open $U = \mathop{\mathrm{Spec}}(A)$ of $X$ contained in $X \setminus H_{v_1}$ which meets the irreducible component $Z$ of $H_{v_2} \cap H_{v_3}$ whose existence is asserted in assumption (3). With notation as above we have to prove that the field extension

\[ \text{Frac}(A[x_1, \ldots , x_ r]/(x_1f_1 + \ldots + x_ rf_ r))/ k(x_1, \ldots , x_ r) \]

is geometrically irreducible. Observe that $f_1$ is invertible in $A$ by our choice of $U$. Set $K = \text{Frac}(A)$ equal to the fraction field of $A$. Eliminating the variable $x_1$ as above, we find that we have to show that the field extension

\[ K(x_2, \ldots , x_ r)/ k(x_2, \ldots , x_ r, -\sum \nolimits _{i = 2, \ldots , r} f_1^{-1}f_ i x_ i) \]

is geometrically irreducible. By Lemma 37.31.1 it suffices to show that for some $c_2, \ldots , c_ r \in k$ the elements

\[ x_2, \ldots , x_ r, \sum \nolimits _{i = 2, \ldots , r} f_1^{-1}f_ i x_ i, \sum \nolimits _{i = 2, \ldots , r} c_ if_1^{-1}f_ i \]

are algebraically independent over $k$ in the fraction field of $A[x_2, \ldots , x_ r]$. This follows from Lemma 37.31.2 and the fact that $Z \cap U$ is an irreducible component of $V(f_1^{-1}f_2, f_1^{-1}f_3) \subset U$.
$\square$

## Comments (2)

Comment #5981 by Olivier Benoist on

Comment #6155 by Johan on