The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

Lemma 15.81.20. Let $(R \to R', f)$ be a glueing pair. Let $M$ be an $R$-module which is not necessarily glueable for $(R \to R', f)$. Then $M$ is a finite projective $R$-module if and only if $M \otimes _ R R'$ is finite projective over $R'$ and $M_ f$ is finite projective over $R_ f$.

Proof. Assume that $M \otimes _ R R'$ is a finite projective module over $R'$ and that $M_ f$ is a finite projective module over $R_ f$. Our task is to prove that $M$ is finite projective over $R$. We will use Algebra, Lemma 10.77.2 without further mention. By Lemma 15.81.19 we see that $M$ is flat. By Lemma 15.81.5 we see that $M$ is finite. Choose a short exact sequence $0 \to K \to R^{\oplus n} \to M \to 0$. Since a finite projective module is of finite presentation and since the sequence remains exact after tensoring with $R'$ (by Algebra, Lemma 10.38.12) and $R_ f$, we conclude that $K \otimes _ R R'$ and $K_ f$ are finite modules. Using the lemma above we conclude that $K$ is finitely generated. Hence $M$ is finitely presented and hence finite projective. $\square$

Second proof. Assume that $M \otimes _ R R^\wedge $ is a finite projective module over $R^\wedge $ and that $M_ f$ is a finite projective module over $R_ f$. Our task is to prove that $M$ is finite projective over $R$.

Case I: Assume that $M \otimes _ R R^\wedge \cong (R^\wedge )^{\oplus n}$ and $M_ f \cong R_ f^{\oplus n}$ for some $n$. Choose a presentation

\[ \bigoplus \nolimits _{i \in I} R \to R^{\oplus m} \to M \to 0 \]

This is possible because $M$ is a finite $R$-module by the first case of the lemma. Let $A = (a_{ij})$ be the matrix of the first map so that $\text{Fit}_ k(M)$ is generated by the $(m - k) \times (m - k)$-minors of $A$. By our assumption we see that $\text{Fit}_{n - 1}(M)$, resp. $\text{Fit}_ n(M)$ generates the zero, resp. unit ideal in $R^\wedge $ and $R_ f$, see Lemma 15.8.7. Since $\mathop{\mathrm{Spec}}(R^\wedge ) \amalg \mathop{\mathrm{Spec}}(R_ f) \to \mathop{\mathrm{Spec}}(R)$ is surjective (Lemma 15.81.4) we conclude that $\text{Fit}_ n(M)$ generates the unit ideal in $R$. Since $R \to R^\wedge \oplus R_ f$ is injective, we see that $\text{Fit}_{n - 1}(M)$ is zero. Hence $M$ is finite locally free of rank $n$ by the lemma cited above.

Case II: general case. Choose an $n$ and isomorphisms $(R^\wedge )^{\oplus n} = M \otimes _ R R^\wedge \oplus C'$ and $R_ f^{\oplus n} = M_ f \oplus C_1$. Then we have

\[ (C')_ f \oplus (R^\wedge )_ f^{\oplus n} = (C')_ f \oplus M_ f \otimes _ R R^\wedge \oplus C_1 \otimes _ R R^\wedge = (R^\wedge )_ f^{\oplus n} \oplus C_1 \otimes _ R R^\wedge \]

In other words, the $(R^\wedge )_ f$-modules $(C')_ f$ and $C_1 \otimes _ R R^\wedge $ become isomorphic after adding $n$ copies of the free module. This gives a glueing datum

\[ ((C')_ f \oplus (R^\wedge )^{\oplus n}, C_1 \oplus R_ f^{\oplus n}, \alpha _1) \]

such that in $\text{Glue}(R \to R^\wedge , f)$ we have

\[ ((R^\wedge )^{\oplus 2n}, R_ f^{\oplus 2n}, \alpha _1 \oplus \text{can}) = ((C')_ f \oplus (R^\wedge )^{\oplus n}, C_1 \oplus R_ f^{\oplus n}, \alpha _1) \oplus (M \otimes _ R R^\wedge , M_ f, \text{can}) \]

By Case I we see that $H^0((R^\wedge )^{\oplus 2n}, R_ f^{\oplus 2n}, \alpha _1 \oplus \text{can})$ is a finite projective $R$-module and hence the summand $\tilde M = H^0(M \otimes _ R R^\wedge , M_ f, \text{can})$ is a finite projective $R$-module as well. By Remark 15.81.18 we see that $M \to \tilde M$ is surjective. Let $K$ be the kernel of this map; since $\tilde M$ is a flat $R$-module, we must have $K \otimes _ R (R^\wedge \oplus R_ f) = 0$. By Lemma 15.81.3, this forces $K = 0$, so $M \cong \tilde M$. $\square$

Comments (0)

There are also:

  • 4 comment(s) on Section 15.81: The Beauville-Laszlo theorem

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0BP6. Beware of the difference between the letter 'O' and the digit '0'.