I'm reading the Paper "What makes a complex exact?" by Eisenbud and Buchsbaum. On page 266 it says: Thus we may assume $0 \neq \operatorname{rank}(\phi_n,L) < \operatorname{rank}(F_n)$ and $I(\phi_n,L)$ contains a nonzerodivisor $r$ on $L$. After that they claim:
Inverting $r$, we may assume $I(\phi_n,L)=R$.
How can they invert $r$? I first thought that they actually meant localizing the ring $R$ at $\{r^n|n\in \mathbb N_0 \}$. But even that is not possible in general, is it? $r$ being a nonzerodivisor on $L$ does not imply that $r$ is a nonzerodivisor on $R$ which would be necessary to localize at the above set.
Can someone enlighten me?
To the ones not knowing the paper:
$R$ is a commutative, unital and noetherian Ring, $\phi: F_n \to F_{n-1}$ a monomorphism of free $R$ modules and $L$ a submodule of a finitely generated $R$ module $M$.
$\operatorname{rank}(\phi_m,L)$ is the rank of $\phi_n$ on $L$ meaning the largest integer $n$ such that $\Lambda^n(\phi \otimes 1_L) \neq 0$.
Let $k$ be the rank of $\phi \otimes 1_L$. Then $I(\phi_n,L)$ is a generalization of the ideal of $k\times k$ subdeterminantes of $\phi$. Explicitly: $$I(\phi,L) := im \left( \left( \Lambda^k(F_{n-1} \otimes L) \right)^* \otimes \Lambda^k (F_n \otimes L) \to R \right)$$ which is induced by $\phi_n \otimes 1_L$.
You can invert $r$ as long as $r$ is not nilpotent (otherwise the multiplicative set generated by $r$ contains $0$ and the localizations are also $0$). Since $r$ is a non-zero divisor on $L$ it follows that $r$ is not nilpotent (unless $L=0$ which I think it's not the case due to the rank condition).