Motivation for the definition of the projective line over a ring

Question

Given a ring (with identity) R, the definition for the projective line over a ring I have is as follows:

On $R \times R$ define an equivalence relation as $(u, v) \sim (a, b)$ if and only if there is some unit $r \in R$ such that $(r a, r b) = (u, v)$.

Two elements $a, b \in R$ are relatively prime if the ideal in $R$ that they generate is the whole of $R$; that is, $R a + R b = R$. The projective line over R is the set of equivalence classes for $\sim$ on pairs of relatively prime elements:

$$P(R) = \{(a,b) \in R \times R : R a + R b = R \}/{\sim}.$$

Comparing the definition with the definition for projective lines over fields, I see that we would want to restrict the multipliers $r$ to be units so that $\sim$ is symmetric. My question is, why do we want to restrict the entries in pairs $(a,b)$ to be relatively prime? Is there perhaps some particular ring $R$ that sheds light on this, that is, where it's clear for some reason that we should only consider relatively prime pairs?

Answer 1

The following are equivalent:

• $a$ and $b$ are relatively prime elements of $R$
• $(f(a), f(b)) \neq (0,0)$ for every homomorphism $f : R \to F$ to a field $F$

Incidentally, if $R$ is a field, then the only pair of elements that is not relatively prime is $(0,0)$.

Answer 2

For example, if $R=\mathbb{Z}$, then you don't want to consider (2,4) as a point because (1,2) is already the "same" point. In the case of a domain, you could get around this by defining an equivalence on all non-zero pairs as $(a,b) \sim (c,d)$ iff $ad=bc$, but this won't work in a non-domain (it won't be transitive). I guess excluding non-relatively prime pairs is the best you can do for a general ring $R$.