Vector bundles on complete rings

Question

Given a ring $A$ and an ideal $I$, consider the completion $\hat{A}$. What does usually mean by a vector bundle on $\hat{A}$? One way is to consider projective $\hat{A}$-modules. Another one is a system of compatible vector bundles on $A/I^n$. Compatible mean there are isomorphisms of vector bundles when you pullback the vector bundle from $A/I^{n+1}$ to $A/I^n$ with the one that is already defined on $A/I^n$. When do these definitions coincide?

Answer 1

I will assume that $A$ is Noetherian (if you wish to work in the non-Noetherian setting, I'll see what I can do to modify my answer).

Without loss of generality, we may assume that $A$ is in fact $I$-adically complete. We know that $$ \operatorname{Coh}( A) \stackrel{\sim}{\to} \varprojlim \operatorname{Coh}(A/I^{n+1}). \hspace{1in} (\ast)$$ Note of caution: By $\operatorname{Coh}(A)$, I really mean the category of finitely generated $A$-modules and not the category of coherent sheaves on Spec $A$. The reason for this is because for this question, it is easier to work with modules and avoid the theory of (affine) formal schemes. Similarly, by $\operatorname{Vect}(A)$ we mean the category of finite projectives.

Ok, so back to the problem. Since finite projectives are preserved under tensoring, the functor $\operatorname{Coh}(A) \to \varprojlim \operatorname{Coh}(A/I^{n+1})$ restricts to a functor $$ \operatorname{Vect}(A) \to \varprojlim \operatorname{Vect}(A/I^{n+1})$$ which is necessarily fully faithful since $\operatorname{Vect}(A)$ is a full subcategory of $\operatorname{Coh}(A)$. To see it is an equivalence of categories, we know by the equivalence $(\ast)$ that an adic system of finite projectives $\{P_n\}$ algebraizes to an a priori finite module $P$, which we must now check is projective. But we know that $P \otimes_A A/I^{n+1}$ is projective for all $n \geq 0$, so we must show the following:

Proposition: Let $A$ be an $I$-adically complete Noetherian ring, and $P$ a finite $A$-module such that $P \otimes_A A/I^{n+1}$ is projective for all $n \geq 0$. Then $P$ is projective.

The proof this is essentially the local criterion for flatness: Indeed since $I$ is contained in every prime ideal of $A$, we may reduce to the situation where $A$ is in fact complete Noetherian local with maximal ideal $I$. Then being projective is equivalent to being flat (in fact free!) in which case the result follows from the local criterion for flatness.