According to the authors of uk.arxiv.org/abs/1012.3235 "no explicit triangulation of $CP^3$ was known so far".

An elliptic curve has (affine) equation $y^2=f(x)$ where $f$ is a cubic or a quartic. In either cases there is a map from $E$ to $P^1$ given by $(x,y)\mapsto x$. This is a double cover, ramified at the zeros of $f$, and also at $\infty$ when $f$ is cubic.

And also in Macdonald's Symmetric Functions and Hall Polynomials. I don't know off-hand if they have results like this though.

And others think it's the usual proof in disguise :-)

For finitely generated free modules over commutative rings, the result is elementary. For finitely generated free modules over non-commutative rings, the assertion can fail.

So why does $p/q$ being in lowest terms entail that $p^2/q^2$ is? If "in lowest terms" means having no common factors save units, then this implication doesn't hold in all integral domains.

Clockwise and anticlockwise don't mean anything in three dimensions.