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Let $X$ be an irreducible affine variety, $A$ be its coordinate ring, $M$ be an $A$-module. Suppose that for any maximal ideal $m$ of $A$, the localization $M_m$ is a free module of rank $n$ (finite and fixed) over $A_m$. Can we deduce that $M$ is free over $A$?

I am just a beginner in this field. I know from Atiyah & Macdonald's book that $M$ is always torsion-free. But I wonder if the stronger statement is still true in the specific case above. If not, what if replacing maximal ideals by prime ideals? Thanks!

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    $\begingroup$ this module should be projective, but not necessarily free $\endgroup$
    – Chen Jiang
    Oct 18, 2014 at 14:43
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    $\begingroup$ Take a smooth projective curve of genus $g$ (say, over $\mathbb{C}$) minus one point. This is an affine variety, and the group of projective, rank 1 modules is isomorphic to $(\mathbb{R}/\mathbb{Z})^{2g}$. $\endgroup$
    – abx
    Oct 18, 2014 at 16:25

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Let $k$ be a field, $A=k[t^2,t^3]$, $\alpha$ a non-zero element of $k$, and $M$ the set of polynomials of the form $a_0+\alpha a_0 t+a_2 t^2 + ...+ a_nt^n$.

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  • $\begingroup$ Thank you and Chen Jiang,abx above. I was not familar with projective modules. And I just did some wiki and found that projective modules are always locally free. So my question seems to be a silly one. $\endgroup$
    – Mingchen Xia
    Oct 18, 2014 at 23:06

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