The Stacks project

Lemma 33.46.3. In Situation 33.46.2 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. $\square$


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