Let $U$ be a dense open subscheme of an integral noetherian scheme $X$ and let $E$ be a vector bundle on $U$. Suppose that the complement $Y$ of $U$ has codimension $\textrm{codim}(Y,X) \geq 2$. Let $F$ be a vector bundle on $X$ extending $E$, i.e., $F|_{U} = E$.

Is any extension of $E$ to $X$ isomorphic to $F$?


4 Answers 4


This is true if $X$ satisfies Serre's condition $S_2$, i.e. $\mathcal O_X$ is $S_2$. Then a vector bundle is $S_2$ since locally it is isomorphic to $\mathcal O_X^n$.

More generally, a coherent sheaf $F$ on a Japanese scheme (for example: $X$ is of finite type over a field) which is $S_2$ has a unique extension from an open subset $U$ with $\operatorname{codim} (X\setminus U)\ge 2$. This follows at once from the cohomological characterization of $S_2$.

Thus, another name for the $S_2$-sheaves: they are sheaves which are saturated in codimension 2, and another name for the $S_2$-fication: saturation in codimension 2.

P.S. Of course, by Serre's criterion, normal = $S_2+R_1$. So the above statement is true for any normal (e.g. smooth) variety.

P.P.S. And of course, Gorenstein implies Cohen-Macaulay implies $S_2$. So the statement is also true for hypersurfaces and complete intersections, which could be very singular and non-reduced.

Edit to define some terms:

  1. A Japanese (or Nagata) ring is a ring obtained from a ring finitely generated over a field or $\mathbb Z$ by optionally applying localizations and completions. The property used here is that for a Japanese ring $R$, its integral closure (normalization) $\tilde R$ is a finitely generated $R$-module. This is important because the $S_2$-fication $S_2(R)$ lies between $R$ and $\tilde R$.

  2. A coherent sheaf $F$ satisfies $S_n$ if for any point $x\in Supp(F)$, one has $$ depth_x (F) \ge \min(\dim_x Supp(F),n) $$ If $F$ locally corresponds to an $R$-module $M$, and $x$ to a prime ideal $p$, then the depth is the length of a maximal regular sequence $(f_1,\dots, f_k)$ of elements of $R_p$ for $M_p$ (so, $f_1$ is a nonzerodivisor in $M_p$, etc.).

  • $\begingroup$ "Japanese scheme"?? :D :D $\endgroup$
    – Qfwfq
    Apr 22, 2010 at 12:43
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    $\begingroup$ Very satisfactory answer! $\endgroup$
    – Qfwfq
    Apr 22, 2010 at 12:52
  • $\begingroup$ Dear VA, sorry for my ignorance: what is the definition of (S_2) for sheaves? $\endgroup$ Apr 22, 2010 at 15:15
  • $\begingroup$ Hi! I wonder if I understand correctly: Could you replace the Japaneseness hypothesis by requiring O_X and F both to satisfy S2? $\endgroup$
    – Lars
    Apr 23, 2010 at 15:03
  • $\begingroup$ Lars: No. What is needed is this property: the normalization of $R$ is a finite $R$-module. Japanese implies that. $\endgroup$
    – VA.
    Apr 23, 2010 at 18:05

Let $i:U \to X$ be the embedding. Assume that $i^*F = E$. Then by the adjunction we have a map $F \to i_*E$ which is an embedding, since the kernel is zero on $U$, so it is a torsion sheaf, and a vector bundle doesn't have torsion subsheaves (if $X$ doesn't have embedded components). So, we have an exact triple $0 \to F \to i_*E \to G \to 0$ for some $G$, which is supported on $X \setminus U$. If $X$ satisfies and $codim_X (X\setminus U) \ge 2$, then (provided $S_2$ condition) we have $G^* = {\mathcal Ext}^1(G,O_X) = 0$, so dualizing the sequence we see that $F^* = (i_* E)^\ast$, hence $F = (i_* E)^{\ast\ast}$, so $F$ has to be the reflexive envelope of $i_* E$. This proves the uniquencess. This also allows to construct an example of a vector bundle on $U$ which does not extend to a vector bundle on $X$.

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    $\begingroup$ Hi, Sasha, can you explain how to construct a an example of a vector bundle on $U$ which does not extend to a vector bundle on $X$. Thanks! $\endgroup$
    – Fei YE
    Apr 26, 2010 at 19:59

This is false as stated; for example, if $X$ is obtained from a projective geometrically connected smooth surface over a field $k$ by gluing two points together and $U$ is the complement of the singular point, then the kernel of the restriction map from the Picard group of $X$ to the Picard group of $U$ is easily seen to be $k^*$. You need to assume that the depth of all the components of the complement of $U$ in $X$ is at least 2; then the statement is correct. If $j\colon U \to X$ is the embedding, then $j_*j^*F = F$.


Edit: The following argument cannot work always, given the counterexample by Angelo, but maybe it works if $X$ is smooth? Edit2: Yes, this argument should work for $W$ smooth ("smooth" being, as noted in another answer, a particular case of "Serre S2").

The transition functions of an extension $F$ of $E$ are morphisms on open (affine) subsets $V$ of $X$ with values in $GL_n$ where $n=rank(F)$. Their restrictions to $W:=V\cap U$ are the transition functions of $E$. So, the problem reduces to proving that a morphism from a the complement of a codimension $\geq 2$ subvariety $Y$ of a quasi-affine variety $W$, with values in $GL_n$, extends at most in one way to the whole variety $W$. This is true, since you can assume $W$ is a subset of affine space, and each component $W-Y \rightarrow GL_n \hookrightarrow \mathbb{A}^{n\times n} \rightarrow \mathbb{A}^1$ is a regular function, and polynomials extend uniquely through codimension $\geq 2$ subvarieties. So, if you have an extension, it has to be unique.

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    $\begingroup$ The fact that regular functions extend over to codim $\geq 2$ subvarieties doesn't mean that the morphism to $GL_n$ will extend; you can only claim that a morphism to $\mathbb{A}^{n^2+1}$ will extend, but the extension might well wander off the embedded $GL_n$. (I'll just leave this comment here in case someone gets confused while reading the answers, as I did) $\endgroup$ Mar 7, 2012 at 20:45
  • $\begingroup$ @DimaSustretov: that is not the problem. If a dense subset lands in a closed subvariety, then everything does (note that $\operatorname{GL}_n$ is affine, by the closed embedding into $\mathbb A^{n^2+1}$). The problem is that Angelo's example is not $S_2$, hence not normal. "Hartog's theorem" holds for normal schemes. $\endgroup$ May 30, 2017 at 12:40

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