Suppose we have a (commutative, unital) ring $R$ and a (commutative, unital) $R$-algebra $A$ such that $A$ is projective of constant rank $n$ as an $R$-module. This condition is equivalent to there existing $r\_1,\dots, r\_k\in R$ that together generate the unit ideal and for which each localization $A\_{r\_i}$ is isomorphic to $R\_{r\_i}^{\oplus n}$ as an $R\_{r\_i}$-module, so in practice if we want to work with such objects, we switch to a localization in which $A$ has an $R$-module basis.


Now suppose we also have an $A$-module $M$ that is projective of constant rank $m$ as an $A$-module, so that we have a "tower" of locally-free modules $M$ over $A$ over $R$. I want to use the localization trick to be able to assume simultaneously that $A$ has an $R$-basis $x\_1,\dots, x\_n$ and that $M$ has an $A$-basis $y\_1,\dots,y\_m$, as in [this related question](https://mathoverflow.net/questions/379331/identity-relating-iterated-determinant-line-bundles). However, I'm suddenly not sure this is possible: it seems to me that by localizing $R$, we can ensure that $A$ is free as an $R$-module, but to ensure that $M$ is free as an $A$-module we need to be able to localize $A$ by elements of $A$ generating the unit ideal, and it's not clear to me how that can be done merely by localizing $R$ by elements generating its unit ideal. So the question is:



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> If $A$ is an $R$-algebra isomorphic to $R^n$ as $R$-modules, and $M$ is a locally free $A$-module of rank $m$, then do there exist $r\_1,\dots, r\_k\in R$ generating the unit ideal such that $M\_{r\_i} \cong A\_{r\_i}^m$ as $A\_{r\_i}$-modules?
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As an example of the type of situation I'm hoping is the general case, consider $R = \mathbb{Z}$, $A = R[\sqrt{-5}]$, and $M = (3, 1+\sqrt{-5}) \subseteq A$. Then $M$ is a locally free $A$-module of rank $1$, meaning that there exist $a\_1,\dots, a\_k\in A$ generating the unit ideal such that $M$ becomes principal over each localization. But we can also choose those $a\_i$ to belong to $R$: choosing $a\_1 = 2$ and $a\_2 = 3$ works.


If you're more of an algebraic geometer, consider that we have a finite, flat, finite presentation morphism $\pi: Y\to X$ and a sheaf $\mathcal F$ of $\mathcal O\_Y$-modules that is locally isomorphic to $\mathcal O\_Y^m$. Then are $\pi\_\ast\mathcal F$ and $\pi\_\ast \mathcal O\_Y^m$ also locally isomorphic as sheaves of $\pi\_\ast\mathcal O\_Y$-modules? Do we get extra mileage from the fact that $\pi$ is a covering map in the fppf topology?



Is a tower of locally-free modules locally a tower of free modules?
The answer to your question is "yes." See EGA II, Prop. 6.1.12.


That proposition tells you something a bit more general:


Let $A$ be a finite $R$-algebra (finite as $R$-module), and let $M$ be an $A$-module. Then $M$ is locally free of rank $m$ over $A$ if and only if there is a list $r\_1, \dots, r\_k$ of elements of $R$ generating the unit ideal in $R$ such that each $M\_{r\_i}$ is free of rank $m$ over $A\_{r\_i}$.


