## 37.32 Bertini theorems

We continue the discussion started in Varieties, Section 33.47. In this section we prove that general hyperplane sections of geometrically irreducible varieties are geometrically irreducible following the remarkable argument given in [Jou].

reference
Lemma 37.32.1. Let $K/k$ be a geometrically irreducible and finitely generated field extension. Let $n \geq 1$. Let $g_1, \ldots , g_ n \in K$ be elements such that there exist $c_1, \ldots , c_ n \in k$ such that the elements

\[ x_1, \ldots , x_ n, \sum g_ ix_ i, \sum c_ ig_ i \in K(x_1, \ldots , x_ n) \]

are algebraically independent over $k$. Then $K(x_1, \ldots , x_ n)$ is geometrically irreducible over $k(x_1, \ldots , x_ n, \sum g_ ix_ i)$.

**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

\[ \sigma _ a(x_ i) = x_ i + a c_ i \]

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

\[ K_0 = k(x_1, \ldots , x_ n) \subset K_1 = K_0(\xi ) \subset K_2 = K_0(\xi , \delta ) \subset K(x_1, \ldots , x_ n) = \Omega \]

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 infinite^{1}, then we can find distinct elements $a_1, a_2$ of $k$ such that

\[ K_2(\sigma _{a_1}(\theta )) = K_2(\sigma _{a_2}(\theta )) \]

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

\[ K_2 = K_0(\xi _1, \xi _2) \]

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 independent 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

\[ K_0(\xi _1, \xi _2, \theta _1) = K_2(\sigma _{a_1}(\theta )) = K_2(\sigma _{a_2}(\theta )) = K_0(\xi _1, \xi _2, \theta _2) \]

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.
$\square$

Lemma 37.32.2. Let $A$ be a domain of finite type over a field $k$. Let $n \geq 2$. Let $g_1, \ldots , g_ n \in A$ be elements such that $V(g_1, g_2)$ has an irreducible component of dimension $\dim (A) - 2$. Then there exist $c_1, \ldots , c_ n \in k$ such that the elements

\[ x_1, \ldots , x_ n, \sum g_ ix_ i, \sum c_ ig_ i \in \text{Frac}(A)(x_1, \ldots , x_ n) \]

are algebraically independent over $k$.

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

\[ T = \mathop{\mathrm{Spec}}(A[x_1, \ldots , x_ n]) \longrightarrow \mathop{\mathrm{Spec}}(k[x_1, \ldots , x_ n, y, z]) = S \]

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

\[ x_1 - 1, x_2, \ldots , x_ n, \sum x_ ig_ i, g_2 \]

and hence equivalently is cut out by

\[ x_1 - 1, x_2, \ldots , x_ n, g_1, g_2 \]

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

reference
Lemma 37.32.3. In Varieties, Situation 33.47.2 assume

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

$X$ is geometrically irreducible over $k$,

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

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 37.27.5 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

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

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 37.32.1 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 37.32.2 and the fact that $Z \cap U$ is an irreducible component of $V(f_1^{-1}f_2, f_1^{-1}f_3) \subset U$.
$\square$

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