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Eisenbud's Commutative Algebra with a View Toward Algebraic Geometry, Section 4.6, Exercise 4.12 (a) tells us if $M$ is a finitely presented $R$-module, then $M$ is projective if and only if $M$ is locally free in the stronger sense that there is a finite set of elements $f_1,\ldots,f_n\in R$ such that $(f_1,\ldots,f_n)=R$ and $M[f_i^{-1}]$ is a free $R[f_i^{-1}]$-module for every $i$.

My question is: could we add some finiteness condition on $R$, say $R$ is noetherian, such that in the second half of the above statement the choice of $f_1,\ldots,f_n $ is independent of the module $M$? In other words,

is there a finite set of elements $f_1,\ldots,f_n\in R$ such that $(f_1,\ldots,f_n)=R$ and any $R$-module $M$ is projective if and only if $M[f_i^{-1}]$ is a free $R[f_i^{-1}]$-module for every $i$?

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NO !
Indeed it is even impossible in general to find a single non-invertible $f\in R$ such that for all projective modules $M$ of finite type over $R$ the $R[f^{-1}]$- module $M[f^{-1}]$ is free.

Translated into algebraic geometry this means that in general given an affine scheme $X=\operatorname {Spec }R$ no non-empty open subset $U\subset X$ can be found on which all vector bundles on $X$ become trivial.
Here is an explicit example:
Start with a projective smooth algebraic curve $C$ of genus $\geq 1$ over a non denumerable algebraically closed field, delete a point $Q_0\in C$, take for $X$ the complement $X=C\setminus \{Q_0\}$ of a point $Q_0\in C$ and for $U$ take the complement $U=X\setminus \{Q_1,\dots,Q_r\}=C\setminus \{Q_0,Q_1,\dots,Q_r\}$ of finitely many points $Q_1,\dots,Q_r\in X$.
The ring $R$ such that $X=\operatorname {Spec }R$ is finitely generated over $k$ and thus noetherian but by considering the Jacobian $\operatorname {Jac}(C)$ it is easy to see that only denumerably many of the non denumerable group of line bundles on $X$ have a trivial restriction to $U$, so that most line bundles on $X$ have nontrivial restriction to $U$.

Edit: some details
Given finitely many points $Q_0,Q_1,\dots, Q_r\in C$ and the open subset $U=C\setminus \{Q_0,Q_1,\dots, Q_r\} \subset C$ there is an exact sequence (cf. Hartshorne, slight generalization of II, Prop 6 (c), page 133) $$\mathbb Z^{r+1}\stackrel u \to \operatorname {Pic}(C)\stackrel {\operatorname {res}} \to \operatorname {Pic}(U)\to 0$$ where $u$ sends $(z_0,z_1,\dots ,z_r)$ to the class in $\operatorname {Pic}(C)$ of the divisor $z_0 Q_0+z_1Q_1+\dots+z_r Q_r$, and $\operatorname {res}$ is the restriction to $U$ of a line bundle on $C$.
This proves that $ \operatorname {Pic}(U)$ has same cardinality $\gt \aleph_0$ as $ \operatorname {Pic}(X)$ and is thus nonzero.
But all open subsets of $X=C\setminus \{Q_0\}$ are equal to some such $U$ and thus admit of non-trivial line bundles, i.e. $ \operatorname {Pic}(U)\neq 0$, as claimed in the answer above.

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  • $\begingroup$ Anybody not familiar with algebraic geometry could just Trusssst in me... $\endgroup$ – Georges Elencwajg Mar 10 '15 at 12:29
  • $\begingroup$ Will you be kind enough to explain the last paragraph of your answer a little more? (Or maybe any reference) $\endgroup$ – Krish Mar 16 '15 at 4:45
  • $\begingroup$ Dear @Krish: I have added some explanations in an Edit. $\endgroup$ – Georges Elencwajg Mar 16 '15 at 7:54
  • $\begingroup$ Thank you for such a beautiful answer and the added explanation. $\endgroup$ – Krish Mar 16 '15 at 8:27
  • $\begingroup$ You are welcome, dear @Krish. $\endgroup$ – Georges Elencwajg Mar 16 '15 at 8:43

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