I have been reading "The Geometry of Schemes" by Eisenbud and Harris and have a question about Exercise III-43. There, one should show that there is a bijection between the sets

$\{(n+1)\mbox{-tuples of elements of }A\mbox{ that generate the unit ideal }\}$ and $\{ \mbox{maps} \mbox{ Spec} A \to \mathbb{P}^n_A$ such that the composite $\mbox{Spec} A \to \mathbb{P}^n_A\to \mbox{Spec}A=id\} $, i.e. $A$-valued points of $\mathbb{P}^n_A$.

Now, of course, $(n+1)$-tuples of elements of $A$ give $A$-valued points, but if $A$ is not a ring such that every invertible $A$-module is free of rank one, I don't see why the converse should work:

Let us take, e.g. a number field $K$ such that $A=\mathcal{O}_K$ is not a PID. Then, up to multiplication by a unit, an $A$-valued point corresponds to an invertible $A$-module $P$ and an epimorphism $A^{n+1}\to P$ by the characterization of morphisms from $\mbox{Spec}A$ to $\mathbb{P}^n_{\mathbb{Z}}$ (Corollary III/42 in Eisenbud+Harris).

Starting with an $(n+1)$-tuple generating $A$, I clearly get an epimorphism $A^{n+1}\to A$ and $A$ is a projective $A$-module, so I get an $A$-valued point.

However, if I start with an $A$-valued point corresponding to an epimorphism $A^{n+1}\to P$ and the invertible module $P$ is not free, how can I choose an $(n+1)$-tuple of points of $A$ which generate the unit ideal? Moreover, don't these $A$-valued points give "additional" points, which do not come from $(n+1)$-tuples of elements of $A$?

Most books just consider the case when $A$ is a field, there everything works just fine.

Thanks!