Why are the prime ideals of $R_f$ exactly the prime ideals of $R$ not containing $f$?

Question

Let $R$ be a ring and $D_f=\{\text{primes ideals in } R \text{ not containing } f\}$ be a basic open set. Let $R_f$ be the localization of $R$ at $f$, I am trying to show that $D_f=\operatorname{Spec}R_f$. I can see every prime in $D_f$ certainly gives a prime ideal of $R_f$, but how can I show the other direction?

Answer 1

For any prime ideal $P$ in $R_f$, its inverse image under the canonical map $R \to R_f$ is again a prime ideal. Show that this can not contain $f$.