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Suppose that $S$ is smooth and that $U\subset S$ is a dense open subscheme. Let $X$ be a scheme (not necessarily smooth) and let $f:X\to U$ be a finite flat morphism. I would like to know whether this finite flat family can be extended over $S$, and whether such an extension is unique. More precisely:

  1. Does there always exist a scheme $Y$ and a finite flat morphism $g:Y\to S$ such that $g^{-1}(U)$ is isomorphic to $X$ as a scheme over $U$?

  2. Given two such extensions, do they have to be isomorphic?

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  • $\begingroup$ not very useful remark: since the map is finite flat then $f_* O_X$ is a vector bundle (of rk = #fibre) and moreover has the structure of an $O_U$-algebra. So first you can ask whether just the vector bundle has an extension and the codimension of $S - U$ is a first obstruction. $\endgroup$ – bananastack Apr 30 '15 at 16:19
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    $\begingroup$ Regarding (2), both the projection $Spec(k[x] \times k[x]) \to Spec(k[x])$ and $Spec(k[x,y]/(x^2-y^2)) \to Spec(k[x])$ are finite flat extensions of the map $Spec(k[x^{\pm 1}] \times k[x^{\pm 1}]) \to Spec(k[x^{\pm 1}])$ when $2$ is invertible in $k$. $\endgroup$ – Tyler Lawson Apr 30 '15 at 20:34
  • $\begingroup$ @bananastack And, vice-versa, if $V$ is a vector bundle on $U$ such that $V \oplus \mathbb{C}$ doesn't extend to $S$, then we can realize $V \oplus \mathbb{C}$ as the push forward of a structure sheaf. Let $I$ be the ideal sheaf of the zero section in the bundle $V^{\ast}$, and consider $\mathcal{O}_{V^{\ast}}/I^2$. $\endgroup$ – David E Speyer Apr 30 '15 at 21:01
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$\def\cO{\mathcal{O}}\def\cE{\mathcal{E}}$I will show that the answer to (1) is yes if and only if the vector bundle $f_{\ast} \cO_X$ on $U$ extends to a vector bundle on $S$. Moreover, if $S \setminus U$ is codimension $\geq 2$ in $S$, I will show that the extension is unique as well.

This condition is obviously necessary.


Here is an example of a case where $f_{\ast} \cO_X$ does not extend. Start with Steven Landsburg's example (well, Swan's example, but Landsburg is easier to cite) of a vector bundle $V$ on $U:=\mathbb{A}^3 \setminus \{ 0 \}$ which doesn't extend $S:=\mathbb{A}^3$. If we take global sections of $V$ over $U$, we get $M:= \{ (f,g,h) \in k[x,y,z]^3 : xf+yg+zh=0 \}$. Note that, if $V$ did extend to a bundle on $S$, then every section of $V$ over $U$ would extend to a section of $S$ by Hartog, so $M$ would be a projective $k[x,y,z]$ module, but it isn't; this gives a proof that $V$ doesn't extend. $V$ doesn't quite do the job for us; we need $V \oplus T$, where $T$ is a trivial bundle. But that doesn't extend either for the same reason: If it did, then $M \oplus k[x,y,z]$ would be a projective $k[x,y,z]$ module, but then its summand $M$ would be, a contradiction.

Now, let $V^{\ast}$ be the dual bundle to $V$. Let $\mathcal{I}$ be the ideal sheaf of the zero section of $V^{\ast} \to U$ and set $X = \mathcal{O}_{V^{\ast}}/\mathcal{I}^2$. Then $f_{\ast} \cO_X \cong \mathcal{V} \oplus \cO_U$, where $\mathcal{V}$ is the sheaf of sections of $V$. In other words, $f_{\ast} \cO_X$ is the sheaf of sections of $V \oplus T$.


I now prove that, if the vector bundle extends, then so does $X$.

Set $Z = S \setminus U$ and assume that $f_{\ast} \cO_X$ extends to a vector bundle on $S$. I'll write $E$ for the total space of the bundle and $\cE$ for the sheaf of sections.

Extending in codimension $1$ We note that $X$ embeds in $E^{\ast}$ (the dual bundle) such that $f$ is the coordinate projection. The way this works is the following: Let $\mathcal{R}$ be the graded $\cO_U$-algebra $\cO_U \oplus (f_{\ast} \cO_X) \oplus (f_{\ast} \cO_X) \oplus \cdots$. Then $\mathrm{Proj} \mathcal{R} \cong X$. There is a map of $\cO_S$ algebras $\mathrm{Sym}^{\bullet} \cE \to \mathcal{R}$, which gives an embedding $X \to \mathbb{P}(E^{\ast})$. (Vakil, Exercise 7.3.J uses this trick to show finite maps are projective.)

So we get a map from $U$ to the Hilbert scheme of $\mathbb{P}(E^{\ast})/S$. Hilbert schemes are proper, and $S$ is smooth, so this map extends in codimension $1$. Thus, we may assume that $\mathrm{codim} Z \geq 2$.

Beyond codimension $1$ Now assume that $\mathrm{codim} Z \geq 2$. In this case, the extension of $f_{\ast} \cO_X$ to a vector bundle on $S$ can occur in at most one way.

The multiplicative identity of $\cO_X$ gives an section of $E$ over $U$. By Hartog's theorem, this section uniquely extends to a section $e: S \to E$. Similarly, the multiplication map $\mu : \cO_X \otimes_{\cO_S} \cO_X \to \cO_X$ gives a section of $\mathrm{Hom}(E \otimes E, E)$ over $U$, which extends uniquely to a section $\mu$ over $S$. The condition that $\mu$ and $e$ define the structure of a unital commutative algebra on $\cE$ is a closed condition, so they do. This uniquely equips $\cE$ with the structure of a (commutative, unital) $\cO_S$ algebra, and (relative) Spec of that algebra gives your $Y$.

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  • $\begingroup$ Hi David, thanks for the detailed answer! One question: You give a somewhat complicated counterexample to existence of extensions in which $S\smallsetminus U$ has codimension 3. Is it possible that existence holds when the codimension is equal to 1? I guess that Tyler Lawson's example shows that uniqueness can fail when the codimension is 1. $\endgroup$ – Nicholas Proudfoot May 1 '15 at 5:06
  • $\begingroup$ Oh, sorry, I guess that your comment above shows that existence can fail if and only if there is a bundle on $U$ that doesn't extend to $S$. Thanks! $\endgroup$ – Nicholas Proudfoot May 1 '15 at 5:10
  • $\begingroup$ Right. And there always is an extension in codimension $\leq 2$. At least, that is what I am picking up from the conversation in mathoverflow.net/questions/120776 and mathoverflow.net/questions/84597 and section 4 of ams.org/mathscinet-getitem?mr=169877 , although I haven't found an official place that says so. $\endgroup$ – David E Speyer May 1 '15 at 11:03
  • $\begingroup$ Just to be clear, is it correct that the Hilbert scheme that you refer to is non-separated, and this is to blame for the non-uniqueness in Tyler Lawson's example? $\endgroup$ – Nicholas Proudfoot May 4 '15 at 18:29
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    $\begingroup$ No, I guess that what I wrote above is wrong. If I understand correctly, the difference between the two examples in Tyler's comment is the choice of extension of $f_*\mathcal{O}_X$. Such an extension is a subtle object--it is a locally free sheaf $\mathcal{E}$ on $S$ along with an isomorphism $\mathcal{E}|_U\cong f_*\mathcal{O}_X$. In Tyler's two examples, $X$ and $X'$ are isomorphic, as are the implied extensions $\mathcal{E}$ and $\mathcal{E}'$, but the isomorphism between $\mathcal{E}$ and $\mathcal{E}'$ cannot be chosen to be compatible with the isomorphism between $X$ and $X'$. $\endgroup$ – Nicholas Proudfoot May 5 '15 at 17:16

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