
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 $M \cong \tilde M$. $\square$

There are also:

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

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).