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The Stacks project

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 generates \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

  1. X is smooth over k,

  2. the image of \psi : V \to \Gamma (X, \mathcal{L}) generates \mathcal{L},

  3. 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

[1] For example we could consider the condition that H_ v is smooth over k', or geometrically irreducible over k'.

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