Let $S$ be an integral scheme and $X \to S$ be a smooth scheme of finite type over $S$. Let $\mathcal{E}$ be a coherent sheaf on $X$, and $\eta$ be the generic point of $S$. Assume that restriction $\mathcal{E}|_{X_\eta}$ is torsion free. Is it true that there exists an open subset $U$ of $S$ such that $\mathcal{E}|_{X_s}$ is torsion free for every $s \in U$?

  • 4
    $\begingroup$ By generic flatness, there is a dense open subscheme of $S$ over which $\mathcal{E}$ is flat. Replacing $S$ by this open, the torsion-free locus is open in the domain $X$, cf. EGA IV_3, Th'eor`eme 12.1.1(ii), p. 174. Thus, the finitely many embedded primes of $\mathcal{E}$ are closed subsets of $X$ that are disjoint from the central fiber. The images of these subsets in $S$ are constructible subsets that do not contain the generic point of $S$. Thus, they are nowhere dense. The union of the closures is a proper closed subset. The open $U$ is the complement. $\endgroup$ Jun 27, 2017 at 17:20
  • $\begingroup$ Typo correction: "central fiber" --> "generic fiber". $\endgroup$ Jun 27, 2017 at 17:43

2 Answers 2


I am just posting my comment as an answer. Since $f:X\to S$ is smooth of finite type, in particular it is of finite presentation. By limit theorems, after replacing $S$ by the open in an open affine covering, there exists a Cartesian diagram, $$\begin{array}{ccc} X & \xrightarrow{f} & S \\ u~\downarrow & & \downarrow~v \\ X_0 & \xrightarrow{f_0} & S_0\end{array}, $$ such that $f_0$ is smooth and finite type, and such that $S_0$ is integral and Noetherian (even a finite type affine scheme over $\text{Spec}\ \mathbb{Z}$). Up to replacing $S_0$ by a finite type $S_0$-scheme through which $v$ factors, also there exists $\mathcal{E}_0$ on $X_0$ whose pullback by $u$ equals $\mathcal{E}$. Finally, up to replacing $S_0$ by the closure of the image of $v$, assume that $S_0$ is integral and that $v$ is dominant. For every $s$ in $S$ with image point $s_0$ in $S$, the fiber $X_s=\text{Spec}\ \kappa(s)\times_S X$ is the base change of the fiber $X_{s_0} = \text{Spec}\ \kappa(s_0)\times_{S_0} X_0$ via the field extension $\kappa(s_0)\hookrightarrow \kappa(s)$. If the pullback of $\mathcal{E}_0$ to $X_{s_0}$ is torsion-free, then also the flat base change by $\kappa(s_0)\hookrightarrow \kappa(s)$ is also torsion-free. Therefore it suffices to prove the result for $f_0$ and $\mathcal{E}_0$. Without loss of generality, assume that $S$ is Noetherian.

By the Generic Flatness Theorem, there exists a dense open subset of $S$ over which $\mathcal{E}$ is $S$-flat. Up to replacing $S$ by this dense open subset, assume that $\mathcal{E}$ is $S$-flat. By Théorème 12.1.1(iii) on p. 174 of the following,

MR0217086 (36 #178)
Grothendieck, A.
Éléments de géométrie algébrique. IV.
Étude locale des schémas et des morphismes de schémas. III.
Inst. Hautes Études Sci. Publ. Math. No. 28 1966 255 pp.

the following set of points $x$ in $X$ is an open subset $V$ of $X$: those $x$ such that the restriction of $\mathcal{E}$ to the fiber $X_{f(s)}$ has no embedded primes containing the point $x$ of the fiber $X_{f(s)}$.

The complement $C=X\setminus V$ is a closed subset. By hypothesis, $C$ is disjoint from the fiber $X_\eta$ over the generic point $\eta$ of $S$. Thus, $f(C)$ is a constructible subset of $S$ that does not contain $\eta$. Therefore the closure of $f(C)$ is a closed subset of $S$ that does not contain $\eta$. For the open complement $U$ in $S$ of the closure of $f(C)$, the inverse image $f^{-1}(U)$ is contained in $V$. Therefore, for every $s\in U$, the restriction of $\mathcal{E}$ to the fiber $X_s$ has no embedded primes, i.e., the restriction is torsion-free.


The answer is yes, and this is proved also in [Maruyama, M. Openness of a family of torsion free sheaves. J. Math. Kyoto Univ. 16-3 (1976), 627-637], prop. 2.1


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy