Theorem
Question
I don't understand the normalization step. We start out with $S = R - \mathfrak p$, and it's clear that $\mathfrak p$ is the only prime ideal for $S^{-1} R$ because it is minimal prime. But then $S$ is changed, and yet we continue to assume that $\mathfrak p$ is the only prime ideal of $R$.
Does normalization mean that we're now saying $R = R_{\mathfrak p}$? If so I think I understand, except I note that the new $S$ is technically $\{ \frac 1 1, \frac x 1, \frac {x^2} {1}, \ldots \}$.