[Theorem 6.3 part 4), Jou]

Lemma 37.31.3. In Varieties, Situation 33.47.2 assume

1. $X$ is of finite type over $k$,

2. $X$ is geometrically irreducible over $k$,

3. there exist $v_1, v_2, v_3 \in V$ and an irreducible component $Z$ of $H_{v_2} \cap H_{v_3}$ such that $Z \not\subset H_{v_1}$ and $\text{codim}(Z, X) = 2$, and

4. every irreducible component $Y$ of $\bigcap _{v \in V} H_ v$ has $\text{codim}(Y, X) \geq 2$.

Then for general $v \in V \otimes _ k k'$ the scheme $H_ v$ is geometrically irreducible over $k'$.

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 $V(f_1^{-1}f_2, f_1^{-1}f_3) \subset U$. $\square$

Comment #5981 by Olivier Benoist on

The numbering of the hypotheses of the lemma is not used correctly in the proof. Namely, when (2) (resp. (3)) is used in the proof, it is (3) (resp. (4)) that is meant.

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