What does it mean when elements act as units on a set? I'm reading Eisenbud's Commutative Algebra with a View Toward Algebraic Geometry, and I am confused by one of his proofs.
The setup is that $R$ is a commutative ring, $U$ is a multiplicatively closed subset, and $\varphi$ is the natural map $\varphi: R \rightarrow R[U^{-1}]$ sending $r$ to $r/1$.
He says that if an ideal $J \subset R$ is of the form $\varphi^{-1}(I)$, where $I \subset R[U^{-1}]$ is an ideal, then "since the elements of $U$ act as units on $R[U^{-1}]/I$, they act as nonzerodivisors on the $R$-submodule $R/J$." 
What does he mean by "acting as units on" and "acting as nonzerodivisors on"?
 A: Let $\phi\colon R \rightarrow R[U^{-1}]/I$ be the canonical map.
I think he meant that $\phi(u)$ is a unit, i.e. an invertible element of $R[U^{-1}]/I$ for every $u \in U$ when he said $U$ acts as units on $R[U^{-1}]/I$.
$U$ acts as nonzerodivisors on $R/J$ means that $ux = 0$ implies $x = 0$ for $u \in U$ and $x \in R/J$.
More generally, let $M$ be an $R$-module.
Let $End_R(M)$ be the endomorphism ring.
Let $\psi\colon R \rightarrow End_R(M)$ be the canonical map.
Let $r \in R$.
We say $r$ acts on $M$ as a unit when $\psi(r)$ is invertible in $End_R(M)$, i.e. $\psi(r)$ is bijective.
We say $r$ acts on $M$ as a nonzerodivisor when $\psi(r)$ is injective, i.e. $rx = 0$ implies $x = 0$.
A: $R[U^{-1}]$ is an $R$-module through the map $\varphi.$ An element $u\in U$ acts on an element $r/s\in R[U^{-1}]$ by $u\cdot (r/s)=(u/1)(r/s)=ur/s,$ and $\varphi:R\to R[U^{-1}]$ is universal with respect to the property $\varphi(U)\subseteq R[U^{-1}]^\times$.
Let $I\subseteq R[U^{-1}]$ be an ideal such that $J=\varphi^{-1}(I).$ Suppose that $u\cdot\overline r=\overline{ur}=0$ for some $\overline r\in R/J.$ Then $ur\in J,$ which implies that $\varphi(ur)=ur/1\in I.$ But $u/1$ is a unit, so $r/1\in I.$ Thus, we see that $r\in J,$ so $\overline r=0.$ Hence, multiplication by any $u\in U$ cannot kill nonzero elements of $R/J.$
