Nice, I think this will works for any matrix $M$ whose $0$-eigenspace is the same as the generalized $0$-eigenspace, that is the set of all vectors killed by a power of $M$.

Of course, it's debatable whether measurable cardinals exist, but this example at least shows that some hypothesis is needed to prove this version of regularity.

The answer to your first question is rather obviously no, since if $L = G \times H$ with H abelian then every element $l\in H$ has $[l,g]=0\in G$.

If anyone is still interested, a paper on triangulations of $CP^2$ by Bagchi and Datta appeared on the ArXiV today: uk.arxiv.org/abs/1004.3157 .

If $k$ is a field then the ring $R=k[x,y]$ has global dimension $2$ and all modules over $R$ have a projective resolution $$0\to P_2\to P_1\to P_0\to M\to 0.$$ By the Quillen-Suslin theorem we may take the $P_i$ to be free.

Qiaochu, arguably a simpler way to see that the (1-sided) long ray is not homeomorphic to the (2-sided) long line is that removing a point from the former always creates one paracompact component, but removing a point from the latter never does.

I should add that this is a special case of the fact that if the matrix $A$ has minimum polynomial $g$ then there is a vector $v$ such that $f(A)v=0$ iff $g\mid f$. This fact drops straight out of the theory of the rational canonical form of a matrix.