I am trying to understand the formulation of Proposition 2.2 in Eisenbud's "Commutative algebra with the view...". Here is the place that is not clear to me
Proposition 2.2. Let $\varphi: R\to R[U^{-1}]$ be the natural map $r\to r/1$.
a. For any ideal $I\subset R[U^{-1} ]$ we have $I=\varphi^{-1}(I)R[U^{-1}]$. ...
Do I understand correctly that instead of $I=\varphi^{-1}(I)R[U^{-1}]$ we should have $I=\varphi^{-1}(I)[U^{-1}]$? Namely that we localize $\varphi^{-1}(I)$ at $U$? If my interpretation is wrong, how am I supposed to understand the notation $\varphi^{-1}(I)R[U^{-1}]$?
Disclaimer: I am unfamiliar with Eisenbud's book.
The ring $R[U^{-1}]$ can be naturally viewed as an $R$ module: you can define $s\frac{a}{b}$ as $\frac{s}{1}\frac{a}{b}$, which is the same as $\varphi(s)\frac{a}{b}$. Now remember that for any morphism $f: R\to S$ and any ideals $I\subset R$, $J\subset S$ we have ideals $f^{-1}(J)\subset R$ (contraction) and $f(I)S\subset S$ (extension).
What Eisenbud seems to be saying is that $I$ coincides with the extension of its contraction, which shows that any ideal of the localized ring is an extension of some ideal of the original ring.
But you are also not wrong: this extension coincides with the localization of the ideal. If you have access to Atiyah's Commutative Algebra, you can look up Proposition 3.11 and the text just preceding it.