The Stacks project

Proof. Let $c_1, \ldots , c_ n \in k$ be as in the statement of the lemma. Write $\xi = \sum g_ ix_ i$ and $\delta = \sum c_ ig_ i$. For $a \in k$ consider the automorphism $\sigma _ a$ of $K(x_1, \ldots , x_ n)$ given by the identity on $K$ and the rules

Observe that $\sigma _ a(\xi ) = \xi + a \delta $ and $\sigma _ a(\delta ) = \delta $. Consider the tower of fields

Observe that $\sigma _ a(K_0) = K_0$ and $\sigma _ a(K_2) = K_2$. Let $\theta \in \Omega $ be separable algebraic over $K_1$. We have to show $\theta \in K_1$, see Algebra, Lemma 10.47.12.

Denote $K'_2$ the separable algebraic closure of $K_2$ in $\Omega $. Since $K'_2/K_2$ is finite (Algebra, Lemma 10.47.13) and separable there are only a finite number of fields in between $K'_2$ and $K_2$ (Fields, Lemma 9.19.1). If $k$ is infinite¹, then we can find distinct elements $a_1, a_2$ of $k$ such that

as subfields of $\Omega $. Write $\theta _ i = \sigma _{a_ i}(\theta )$ and $\xi _ i = \sigma _{a_ i}(\xi ) = \xi + a_ i \delta $. Observe that

as we have $\xi _ i = \xi + a_ i \delta $, $\xi = (a_2 \xi _1 - a_1 \xi _2)/(a_2 - a_1)$, and $\delta = (\xi _1 - \xi _2)/(a_1 - a_2)$. Since $K_2/K_0$ is purely transcendental of degree $2$ we conclude that $\xi _1$ and $\xi _2$ are algebraically indepedent over $K_0$. Since $\theta _1$ is algebraic over $K_0(\xi _1)$ we conclude that $\xi _2$ is transcendental over $K_0(\xi _1, \theta _1)$.

By assumption $K/k$ is geometrically irreducible. This implies that $K(x_1, \ldots , x_ n)/K_0$ is geometrically irreducible (Algebra, Lemma 10.47.10). This in turn implies that $K_0(\xi _1, \theta _1)/K_0$ is geometrically irreducible as a subextension (Algebra, Lemma 10.47.6). Since $\xi _2$ is transcendental over $K_0(\xi _1, \theta _1)$ we conclude that $K_0(\xi _1, \xi _2, \theta _1)/K_0(\xi _2)$ is geometrically irreducible (Algebra, Lemma 10.47.11). By our choice of $a_1, a_2$ above we have

Since $\theta _2$ is separably algebraic over $K_0(\xi _2)$ we conclude by Algebra, Lemma 10.47.12 again that $\theta _2 \in K_0(\xi _2)$. Taking $\sigma _{a_2}^{-1}$ of this relation givens $\theta \in K_0(\xi ) = K_1$ as desired.

This finishes the proof in case $k$ is infinite. If $k$ is finite, then we can choose a variable $t$ and consider the extension $K(t)/k(t)$ which is geometrically irreducible by Algebra, Lemma 10.47.10. Since it is still be true that $x_1, \ldots , x_ n, \sum g_ ix_ i, \sum c_ ig_ i$ in $K(t, x_1, \ldots , x_ n)$ are algebraically independent over $k(t)$ we conclude that $K(t, x_1, \ldots , x_ n)$ is geometrically irreducible over $k(t, x_1, \ldots , x_ n, \sum g_ ix_ i)$ by the argument already given. Then using Algebra, Lemma 10.47.10 once more finishes the job. $\square$

Proof. The algebraic independence over $k$ means that the morphism

given by $y = \sum g_ ix_ i$ and $z = \sum c_ ig_ i$ is dominant. Set $d = \dim (A)$. If $T \to S$ is not dominant, then the image has dimension $< n + 2$ and hence every irreducible component of every fibre has dimension $> d + n - (n + 2) = d - 2$, see Varieties, Lemma 33.20.4. Choose a closed point $u \in V(g_1, g_2)$ contained in an irreducible component of dimension $d - 2$ and in no other component of $V(g_1, g_2)$. Consider the closed point $t = (u, 1, 0, \ldots 0)$ of $T$ lying over $u$. Set $(c_1, \ldots , c_ n) = (0, 1, 0, \ldots , 0)$. Then $t$ maps to the point $s = (1, 0, \ldots , 0)$ of $S$. The fibre of $T \to S$ over $s$ is cut out by

and hence equivalently is cut out by

By our condition on $g_1, g_2$ this subscheme has an irreducible component of dimension $d - 2$. $\square$

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 37.27.5 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

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

is irreducible. On the other hand, we have

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

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

is geometrically irreducible. By Lemma 37.32.1 it suffices to show that for some $c_2, \ldots , c_ r \in k$ the elements

are algebraically independent over $k$ in the fraction field of $A[x_2, \ldots , x_ r]$. This follows from Lemma 37.32.2 and the fact that $Z \cap U$ is an irreducible component of $V(f_1^{-1}f_2, f_1^{-1}f_3) \subset U$. $\square$