Let $A$ be a commutative ring. An $A$-module $M$ is called a finite projective module if there exists a module $N$ and a natural number $n\in\mathbb N$ such that $M\oplus N\cong A^n$, and an $A$-module $M$ is called (Zariski-)locally free if there exists a finitely many elements $f_1,\dots,f_n\in A$ such that $f_1,\ldots,f_n$ generates the unital ideal, that is to say, $(f_1,\dots,f_n)=A$, and that the localizations $M_{f_i}:=M\otimes A_{f_i}$ of the module $M$ at $f_i\in A$ are free for $i=1,\ldots,n$.
Remark: The terminology (Zariski-)locally free is reflected by the fact that the maps $\operatorname{Spec}(A_{f_i})\to\operatorname{Spec}A$ constitute a Zariski cover.
Question: Is it true that all finite projective modules over a commutative ring are locally free without axiom of choice?
It is known that this is true if certain version of axiom of choice is allowed, essentially the existence of prime ideals, which is equivalent to a weaker version of axiom of choice, for the equivalence see https://mathoverflow.net/questions/98549/ or https://mathoverflow.net/questions/27163/.
It is worthwhile to mention that finite projective modules are also free over noncommutative local rings. See Weibel's Kbook, Chapter 1 Lemma 2.2.
Now I am asking whether we can avoid axiom of choice to prove this statement? By avoiding AC, it is easier to find analogues for other topoi.