A more concrete description of the functor of points of projective space. The functor $\mathbb{P}^n_{\mathbb{Z}}: \text{CRing}\to \text{Sets}$ corresponding to projective $n$-space over $\mathbb{Z}$ has the following description: For each commutative ring $A$, $\mathbb{P}^n_{\mathbb{Z}}(A)$ is the set of all equivalence classes of pairs $(L, p)$ where $L$ is an invertible $A$-module and $p: A^{n + 1}\to L$ is a surjective $A$-module homomorphism. $(L, p)$ and $(L', p')$ are considered equivalent if there exists an $A$-module isomorphism $\varphi: L\to L'$ such that $\varphi\circ p = p'$.
If $A$ is a ring such that every invertible $A$-module is free (such as a local ring or a PID), then $\mathbb{P}^n_{\mathbb{Z}}(A)$ is just the set of all $(n+1)$-tuples $(a_0, \dots, a_n)$ of elements of $A$ which generate the unit ideal, up to multiplication by a unit. This is an example of what I would call a concrete description of $\mathbb{P}^n_{\mathbb{Z}}(A)$ because it involves $(n + 1)$-tuples of elements of $A$.
Now consider $A = \mathbb{Z}[\sqrt{-5}]$. Then $\mathfrak{a} = (2, 1 + \sqrt{-5})$ is an example of an invertible $A$-module which is not free. So $\mathfrak{a}$ together with the generators $2$ and $1 + \sqrt{-5}$ corresponds to an element of $\mathbb{P}^1_{\mathbb{Z}}(A)$ but $\mathfrak{a}$ is not isomorphic to $A$ as an $A$-module. This example is sometimes used to show that the homogeneous coordinate definition of projective space cannot be used for general rings. However, what is wrong with saying that the homogeneous coordinates are $(2 : 1 + \sqrt{-5})$? Is there another counter-example where this kind of logic breaks down?
In general, is there any way to say that $\mathbb{P}^n_{\mathbb{Z}}(A)$ is the set of all $(n+1)$-tuples $(a_0, \dots, a_n)$ of elements of $A$ such that [insert condition on $(a_0, \dots, a_n)$ here], up to [insert equivalence relation here]? The condition on $(a_0, \dots, a_n)$ should be something related to not having any nontrivial common divisors, but I'm not sure how to formulate that for a general ring.
 A: Here is an "elementary" description of $\mathbb{P}^n (A)$.
It is essentially the "obvious" definition in terms of homogeneous coordinates, except that one has to apply Zariski sheafification and this makes things rather messy.
First, let us define a partial element of $\mathbb{P}^n (A)$ to be a tuple $(s, a_0, \ldots, a_n)$ of elements of $A$ such that $s$ is in the ideal generated by $a_0, \ldots, a_n$.
We say two partial elements $(s, a_0, \ldots, a_n)$ and $(t, b_0, \ldots, b_n)$ are compatible if there is a natural number $r$ such that, for all $(i, j)$, $s^r t^r a_i b_j = s^r t^r a_j b_i$.
Suppose $(s, a_0, \ldots, a_n)$ and $(t, b_0, \ldots, b_n)$ are compatible partial elements.
By definition, there are $c_0, \ldots, c_n$ in $A$ such that $a_0 c_0 + \cdots + a_n c_n = s$.
The equations $s^r t^r a_i b_j = s^r t^r a_j b_i$ then imply
$$s^{r+1} t^{r+1} b_j = s^r t^{r+1} (a_0 c_0 + \cdots + a_n c_n) b_j = s^r t^{r+1} (b_0 c_0 + \cdots + b_n c_n) a_j$$
so taking $u = t (b_0 c_0 + \cdots + b_n c_n)$ we have $u \in A$ such that $s^r t^r a_j u = s^{r+1} t^{r+1} b_j$ for all $j$.
By symmetry, we have $v \in A$ such that $s^{r+1} t^{r+1} a_i = s^r t^r b_i v$ for all $i$, and it is easy to check that $s^{r+1} t^{r+1} u v = s^{r+2} t^{r+2}$.
Conversely, given partial elements $(s, a_0, \ldots, a_n)$ and $(t, a_0, \ldots, b_n)$ and $u, p, r$ such that $s^r t^r a_j u = s^{r+p} t^{r+p} b_j$ for all $j$, we have
$$s^{r+p} t^{r+p} a_i b_j = s^r t^r a_i a_j u = s^{r+p} t^{r+p} a_j b_i$$
hence $(s, a_0, \ldots, a_n)$ and $(t, b_0, \ldots, b_n)$ are compatible.
In short: $(s, a_0, \ldots, a_n)$ and $(t, b_0, \ldots, b_n)$ are compatible if and only if there is an element $w$ in $A [s^{-1} t^{-1}]$ such that $a_j w = b_j$ for all $j$; furthermore, any such $w$ is automatically a unit in $A [s^{-1} t^{-1}]$.
We say a set $E$ of partial elements is consistent if every pair of elements of $E$ are compatible, and we say $E$ is total if the first elements of the elements of $E$ generate the unit ideal of $A$.
A complete element of $\mathbb{P}^n (A)$ is then a set of partial elements that is consistent, total, and maximal with respect to set inclusion among consistent total sets of partial elements.
Given three sets $E, F, G$ of partial elements that are consistent and total, if $E \cup F$ and $F \cup G$ are consistent and total, then $E \cup G$ and $E \cup F \cup G$ are also consistent and total.
It follows that every set of partial elements that is consistent and total is contained in a unique complete element of $\mathbb{P}^n (A)$.
Perhaps this looks horrifically complicated.
Let me point out that in a local ring, a finite sum is a unit if and only if one of the summands is a unit.
Thus, if $A$ is a local ring, any complete element of $\mathbb{P}^n (A)$ must contain a partial element of the form $(s, a_0, \ldots, a_n)$ where $s$ is a unit.
But $(s, a_0, \ldots, a_n)$ is compatible with $(1, s^{-1} a_0, \ldots, s^{-1} a_n)$, so we may even assume $s = 1$.
Moreover $(1, a_0, \ldots, a_n)$ and $(1, b_0, \ldots, b_n)$ are compatible if and only if $a_i b_j = a_j b_i$ for all $(i, j)$, which is precisely the equivalence relation one imposes on homogeneous coordinates.
This is where the classical definition survives after sheafification.
Now let us see the connection with $A$-modules.
Let $E$ be a consistent total set of partial elements of $\mathbb{P}^n (A)$.
Let $M (E)$ be the subset of $A^{n+1}$ consisting of those $(m_0, \ldots, m_n) \in A^{n+1}$ such that, for all $(s, a_0, \ldots, a_n) \in E$, there is a natural number $r$ such that, for all $(i, j)$, $s^r a_i m_j = s^r a_j m_i$.
For example, we have $(0, \ldots, 0) \in M (E)$.
It is easy to check that $M (E)$ is an $A$-submodule of $A^{n+1}$.
An argument similar to one used above shows that we have $u \in A$ such that $s^r a_j u = s^{r+1} m_j$.
Furthermore, if we have $x \in A$ such that $s^p a_j x = s^{p+q} m_j$, we get $s^{p+r+1} a_j x = s^{p+q+r+1} m_j = s^{p+q+r} a_j u$, hence $s^{p+r+2} x = s^{p+q+r+1} u$, i.e. $s^{-1} u = s^{-q} x$ in $A [s^{-1}]$.
Thus, $M (E) [s^{-1}]$ is a free $A [s^{-1}]$-module generated by $(a_0, \ldots, a_n)$, so $M (E)$ is locally free of rank 1.
In fact, $M (E) [s^{-1}]$ is a direct summand of $A^{n+1} [s^{-1}]$: we have $a_0 c_0 + \cdots + a_n c_n = s$, so the map $(x_0, \ldots, x_n) \mapsto s^{-1} (x_0 c_0 + \cdots + x_n c_n) (a_0, \ldots, a_n)$ is a retraction of $A^{n+1} [s^{-1}]$ onto $M (E) [s^{-1}]$.
Conversely, given any $A$-submodule $M \subseteq A^{n+1}$ that is locally a direct summand free of rank 1, let $E$ be the set of all $(s, a_0, \ldots, a_n)$ such that $(a_0, \ldots, a_n)$ freely generates $M [s^{-1}]$ and $s$ is in the ideal generated by $a_0, \ldots, a_n$.
Clearly, $E$ is a consistent set of partial elements of $\mathbb{P}^n (A)$, and with a bit of work we see $E$ is also total and maximal.
We may then verify that this defines a natural bijection between $A$-submodules of $A^{n+1}$ that are locally a direct summand free of rank 1 and the set of complete elements of $\mathbb{P}^n (A)$.
Finally, we can translate between the submodule and quotient pictures to get the version you want.
If $M$ is an $A$-submodule of $A^{n+1}$ that is locally a direct summand then the quotient $A^{n+1} / M$ is also locally a direct summand – hence locally free a fortiori.
If $M$ is locally free of finite rank then $M$ is finitely generated.
So if $M$ is locally a direct summand free of finite rank then $A^{n+1} / M$ is finitely presented and locally free – hence projective, and therefore $M$ is a direct summand of $A^{n+1}$.
Dualising yields $M^\vee$ a direct summand of $(A^{n+1})^\vee$, but crucially whereas we had a monomorphism $M \to A^{n+1}$ before, now we have an epimorphism $(A^{n+1})^\vee \to M^\vee$.
Thus we obtain a natural bijection between complete elements of $\mathbb{P}^n (A)$ and isomorphism classes of quotients of $A^{n+1}$ that are locally free of rank 1.
