Module over an associative ring with unity and axioms of projective geometry

Question

According to Wikipedia(http://en.wikipedia.org/wiki/Projective_geometry#Axioms_of_projective_geometry), the axioms of projective geometry due to A. N. Whitehead are:

G1: Every line contains at least 3 points

G2: Every two points, A and B, lie on a unique line, AB.

G3: If lines AB and CD intersect, then so do lines AC and BD (where it is assumed that A and D are distinct from B and C).

Now let $V$ be a vector space over a field $K$. We denote by $P(V)$ the set of one-dimensional subspaces of $V$. If $V$ is finite dimensional, this is the usual definition of a projective space over $K$. We say a two-dimensional subspace of $V$ a line of $P(V)$. Then points and lines of $P(V)$ satisfy the above axioms(see Whitehead's axioms of projective geometry and a vector space over a field)

Now let $M \neq 0$ be a left module over an associative ring $R$ with unity. Suppose $M$ has a a composition series. We also suppose that $M$ is not a finite set to avoid trivial cases. Then every submodule $N$ has a composition series. We call its length the dimension of $N$ and denote it by dim $N$. Let $P(M)$ be the set of one-dimensional submodules of $M$. An element of $P(M)$ is called a point. We say a two-dimensional submodule a line of $P(M)$. Then it is clear that $P(M)$ satisfies Axiom G2. Does $P(M)$ satisfy Axiom G1 or G3 or both? If not, can you find counter-examples?

Answer 1

It's clear that $G3$ always holds. If $A,B,C,D$ are four distinct simple submodules of $M$ such that $(A+B)\cap (C+D)\neq\{0\}$, then $A+B+C+D$ has length $3$. (You can see this by noting that $(A+B+C+D)/(A+B)\cong (C+D)/((A+B)\cap(C+D))$ and that the length of the right hand side is $1$ and then it must be that $len(A+B+C+D)-2=1$.

Then it is impossible for $(A+C)\cap (B+D)=\{0\}$, for then the sum $A+B+C+D$ would be direct, and we'd have a length $4$ module. So, the intersection is nontrivial.

$G1$ need not be true. Let $R=M=F_2\times F_3$, the product of fields of two and three elements, respectively. It has two isotypes of simple modules: $S_2=F_2\times\{0\}$ and $S_3=\{0\}\times F_3$ are representatives, and you can see that $M$ is just the line through those two points.

But $R$, as a cyclic group of order $6$, has exactly one subgroup of order $2$ and exactly one subgroup of order $3$. Therefore there can't be a third point on that line other than the two points we already spotted.