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?

  • 2
    $\begingroup$ 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"?) $\endgroup$ Jan 27, 2021 at 19:56
  • $\begingroup$ @user2831784, yes, I did mean locally isomorphic as sheaves of $\pi_\ast \mathcal{O}_Y$-modules, thank you! $\endgroup$ Jan 27, 2021 at 20:09

1 Answer 1


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 $A_{r_i}$.

  • $\begingroup$ Thank you so much! I'm amazed that this question is so exactly answered there. $\endgroup$ Jan 27, 2021 at 20:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.