33.47 Bertini theorems
In this section we prove results of the form: given a smooth projective variety $X$ over a field $k$ there exists an ample divisor $H \subset X$ which is smooth.
Lemma 33.47.1. Let $k$ be a field. Let $X$ be a proper scheme over $k$. Let $\mathcal{L}$ be an ample invertible $\mathcal{O}_ X$-module. Let $Z \subset X$ be a closed subscheme. Then there exists an integer $n_0$ such that for all $n \geq n_0$ the kernel $V_ n$ of $\Gamma (X, \mathcal{L}^{\otimes n}) \to \Gamma (Z, \mathcal{L}^{\otimes n}|_ Z)$ generates $\mathcal{L}^{\otimes n}|_{X \setminus Z}$ and the canonical morphism
\[ X \setminus Z \longrightarrow \mathbf{P}(V_ n) \]
is an immersion of schemes over $k$.
Proof.
Let $\mathcal{I} \subset \mathcal{O}_ X$ be the quasi-coherent ideal sheaf of $Z$. Observe that via the inclusion $\mathcal{I} \otimes _{\mathcal{O}_ X} \mathcal{L}^{\otimes n} \subset \mathcal{L}^{\otimes n}$ we have $V_ n = \Gamma (X, \mathcal{I} \otimes _{\mathcal{O}_ X} \mathcal{L}^{\otimes n})$. Choose $n_1$ such that for $n \geq n_1$ the sheaf $\mathcal{I} \otimes \mathcal{L}^{\otimes n}$ is globally generated, see Properties, Proposition 28.26.13. It follows that $V_ n$ gererates $\mathcal{L}^{\otimes n}|_{X \setminus Z}$ for $n \geq n_1$.
For $n \geq n_1$ denote $\psi _ n : V_ n \to \Gamma (X \setminus Z, \mathcal{L}^{\otimes n}|_{X \setminus Z})$ the restriction map. We get a canonical morphism
\[ \varphi = \varphi _{\mathcal{L}^{\otimes n}|_{X \setminus Z}, \psi _ n} : X \setminus Z \longrightarrow \mathbf{P}(V_ n) \]
by Constructions, Example 27.21.2. Choose $n_2$ such that for all $n \geq n_2$ the invertible sheaf $\mathcal{L}^{\otimes n}$ is very ample on $X$. We claim that $n_0 = n_1 + n_2$ works.
Proof of the claim. Say $n \geq n_0$ and write $n = n_1 + n'$. For $x \in X \setminus Z$ we can choose $s_1 \in V_1$ not vanishing at $x$. Set $V' = \Gamma (X, \mathcal{L}^{\otimes n'})$. By our choice of $n$ and $n'$ we see that the corresponding morphism $\varphi ' : X \to \mathbf{P}(V')$ is a closed immersion. Thus if we choose $s' \in \Gamma (X, \mathcal{L}^{\otimes n'})$ not vanishing at $x$, then $X_{s'} = (\varphi ')^{-1}(D_+(s'))$ (see Constructions, Lemma 27.14.1) is affine and $X_{s'} \to D_+(s')$ is a closed immersion. Then $s = s_1 \otimes s' \in V_ n$ does not vanish at $x$. If $D_+(s) \subset \mathbf{P}(V_ n)$ denotes the corresponding open affine space of our projective space, then $\varphi ^{-1}(D_+(s)) = X_ s \subset X \setminus Z$ (see reference above). The open $X_ s = X_{s'} \cap X_{s_1}$ is affine, see Properties, Lemma 28.26.4. Consider the ring map
\[ \text{Sym}(V)_{(s)} \longrightarrow \mathcal{O}_ X(X_ s) \]
defining the morphism $X_ s \to D_+(s)$. Because $X_{s'} \to D_+(s')$ is a closed immersion, the images of the elements
\[ \frac{s_1 \otimes t'}{s_1 \otimes s'} \]
where $t' \in V'$ generate the image of $\mathcal{O}_ X(X_{s'}) \to \mathcal{O}_ X(X_ s)$. Since $X_ s \to X_{s'}$ is an open immersion, this implies that $X_ s \to D_+(s)$ is an immersion of affine schemes (see below). Thus $\varphi _ n$ is an immersion by Morphisms, Lemma 29.3.5.
Let $a : A' \to A$ and $c : B \to A$ be ring maps such that $\mathop{\mathrm{Spec}}(a)$ is an immersion and $\mathop{\mathrm{Im}}(a) \subset \mathop{\mathrm{Im}}(c)$. Set $B' = A' \times _ A B$ with projections $b : B' \to B$ and $c' : B' \to A'$. By assumption $c'$ is surjective and hence $\mathop{\mathrm{Spec}}(c')$ is a closed immersion. Whence $\mathop{\mathrm{Spec}}(c') \circ \mathop{\mathrm{Spec}}(a)$ is an immersion (Schemes, Lemma 26.24.3). Then $\mathop{\mathrm{Spec}}(c)$ has to be an immersion because it factors the immersion $\mathop{\mathrm{Spec}}(c') \circ \mathop{\mathrm{Spec}}(a) = \mathop{\mathrm{Spec}}(b) \circ \mathop{\mathrm{Spec}}(c)$, see Morphisms, Lemma 29.3.1.
$\square$
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.
Lemma 33.47.3. In Situation 33.47.2 assume
$X$ is smooth over $k$,
the image of $\psi : V \to \Gamma (X, \mathcal{L})$ generates $\mathcal{L}$,
the corresponding morphism $\varphi _{\mathcal{L}, \psi } : X \to \mathbf{P}(V)$ is an immersion.
Then for general $v \in V \otimes _ k k'$ the scheme $H_ v$ is smooth over $k'$.
Proof.
(We observe that $X$ is separated and finite type as a locally closed subscheme of a projective space.) Let us use the notation introduced above the statement of the lemma. We consider the projections
\[ \xymatrix{ \mathbf{A}^ r_ k \times _ k X \ar[d] & H_{univ} \ar[l] \ar[ld]^ p \ar[r] \ar[rd]_ q & \mathbf{A}^ r_ k \times _ k X \ar[d] \\ X & & \mathbf{A}^ r_ k } \]
Let $\Sigma \subset H_{univ}$ be the singular locus of the morphsm $q : H_{univ} \to \mathbf{A}^ r_ k$, i.e., the set of points where $q$ is not smooth. Then $\Sigma $ is closed because the smooth locus of a morphism is open by definition. Since the fibre of a smooth morphism is smooth, it suffices to prove $q(\Sigma )$ is contained in a proper closed subset of $\mathbf{A}^ r_ k$. Since $\Sigma $ (with reduced induced scheme structure) is a finite type scheme over $k$ it suffices to prove $\dim (\Sigma ) < r$ This follows from Lemma 33.20.4. Since dimensions aren't changed by replacing $k$ by a bigger field (Morphisms, Lemma 29.28.3), we may and do assume $k$ is algebraically closed. By dimension theory (Lemma 33.20.4), it suffices to prove that for $x \in X \setminus Z$ closed we have $p^{-1}(\{ x\} ) \cap \Sigma $ has dimension $< r - \dim (X')$ where $X'$ is the unique irreducible component of $X$ containing $x$. As $X$ is smooth over $k$ and $x$ is a closed point we have $\dim (X') = \dim \mathfrak m_ x/\mathfrak m_ x^2$ (Morphisms, Lemma 29.34.12 and Algebra, Lemma 10.140.1). Thus we win if
\[ \dim p^{-1}(x) \cap \Sigma < r - \dim \mathfrak m_ x/\mathfrak m_ x^2 \]
for all $x \in X$ closed.
Since $V$ globally generated $\mathcal{L}$, for every irreducible component $X'$ of $X$ there is a nonempty Zariski open of $\mathbf{A}^ r$ such that the fibres of $q$ over this open do not contain $X'$. (For example, if $x' \in X'$ is a closed point, then we can take the open corresponding to those vectors $v \in V$ such that $\psi (v)$ does not vanish at $x'$. This open will be the complement of a hyperplane in $\mathbf{A}^ r_ k$.) Let $U \subset \mathbf{A}^ r$ be the (nonempty) intersection of these opens. Then the fibres of $q^{-1}(U) \to U$ are effective Cartier divisors on the fibres of $U \times _ k X \to U$ (because a nonvanishing section of an invertible module on an integral scheme is a regular section). Hence the morphism $q^{-1}(U) \to U$ is flat by Divisors, Lemma 31.18.9. Thus for $x \in X$ closed and $v \in V = \mathbf{A}^ r_ k(k)$, if $(x, v) \in H_{univ}$, i.e., if $x \in H_ v$ then $q$ is smooth at $(x, v)$ if and only if the fibre $H_ v$ is smooth at $x$, see Morphisms, Lemma 29.34.14.
Consider the image $\psi (v)_ x$ in the stalk $\mathcal{L}_ x$ of the section corresponding to $v \in V$. We have
\[ x \in H_ v \Leftrightarrow \psi (v)_ x \in \mathfrak m_ x\mathcal{L}_ x \]
If this is true, then we have
\[ H_ v\text{ singular at }x \Leftrightarrow \psi (v)_ x \in \mathfrak m_ x^2\mathcal{L}_ x \]
Namely, $\psi (v)_ x$ is not contained in $\mathfrak m_ x^2\mathcal{L}_ x$ $\Leftrightarrow $ the local equation for $H_ v \subset X$ at $x$ is not contained in $\mathfrak m_ x^2$ $\Leftrightarrow $ $\mathcal{O}_{H_ v, x}$ is regular (Algebra, Lemma 10.106.3) $\Leftrightarrow $ $H_ v$ is smooth at $x$ over $k$ (Algebra, Lemma 10.140.5). We conclude that the closed points of $p^{-1}(x) \cap \Sigma $ correspond to those $v \in V$ such that $\psi (v)_ x \in \mathfrak m_ x^2\mathcal{L}_ x$. However, as $\varphi _{\mathcal{L}, \psi }$ is an immersion the map
\[ V \longrightarrow \mathcal{L}_ x/\mathfrak m_ x^2\mathcal{L}_ x \]
is surjective (small detail omitted). By the above, the closed points of the locus $p^{-1}(x) \cap \Sigma $ viewed as a subspace of $V$ is the kernel of this map and hence has dimension $r - \dim \mathfrak m_ x/\mathfrak m_ x^2 - 1$ as desired.
$\square$
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