Is trace equal to 1 equivalent to free on one generator?

Question

Let $R$ be a commutative ring and $M$ a f.g. projective module. Is it true that if the trace of the identity map of $M$ is equal to 1, then $M \cong R$?

Answer 1

No, $M$ can be any f.g. projective module of rank $1$ (an invertible module / line bundle over $\text{Spec } R$). In this case $M$ is locally free of rank $1$, so that $M_P \cong R_P$ for any prime ideal $P$, meaning the trace of the identity on $M_P$ is locally equal to $1$. I have to admit I don't remember enough commutative algebra to know off the top of my head what the kernel of localization at every prime ideal is but under mild hypotheses, e.g. $R$ an integral domain but this is probably too strong, it follows that the trace of the identity on $M$ is equal to $1$. But I am pretty sure the trace of the identity is $1$ with no further hypotheses.

For a sillier counterexample, $R$ can have positive characteristic, in which case the trace of the identity does not determine the rank for free modules; for example if $R = \mathbb{F}_p$ then we can take $M = \mathbb{F}_p^{p+1}$.