# Is a tower of locally-free modules locally a tower of free modules?

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. 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:

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?

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?

• We may work locally on $R$; suppose that $R$ is local, so that $A$ is semilocal; then use that vector bundles over semilocal rings are trivial. (Do you mean "locally isomorphic as sheaves of $\pi_{\ast}\mathcal O_Y$-modules"?) Jan 27 at 19:56
• @user2831784, yes, I did mean locally isomorphic as sheaves of $\pi_\ast \mathcal{O}_Y$-modules, thank you! Jan 27 at 20:09

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 $$A_{r_i}$$.