If $R_{\mathfrak{p}}$ is an integral domain then $R_f$ is also an integral domain for some $f\notin \mathfrak{p}$ 
Suppose $R$ is a Noetherian ring. I want to show that if there is some prime ideal $\mathfrak{p}\subset R$ such that $R_{\mathfrak{p}}$ is an integral domain then there exists some $f\in R\setminus\mathfrak{p}$ such that $R_f$ is also an integral domain.

All that I have so far is that $R_{\mathfrak{p}}$ is an integral domain implies that for all $r_1,r_2\in R$, we have that $Ann(r_1)\subset\mathfrak{p}$ and $Ann(r_2)\subset\mathfrak{p}$ then $Ann(r_1r_2)\subset\mathfrak{p}$. 
If we could somehow find an $f$ such that $f\notin\mathfrak{p}$ and $f^k\in Ann(r)$ for all $r$ such that $Ann(r)$ is not a subset of $\mathfrak{p}$ we would be good to go.  But is seems unlikely this is possible. How else do we proceed? It seems the fact that we have finite associated primes should be relevant, but I can't see a way to use it directly.
 A: Let $P_1, \dotsc, P_n$ be the associated primes of $R$. Then the associated primes of $R_\mathfrak p$ are those $P_i$ that are contained in $\mathfrak p$, lets call them $P_1, \dotsc, P_m$ with $m \leq n$.
On the other hand, we have $\operatorname{Ass} R_\mathfrak p = \{ 0 \}$, i.e. for all $1 \leq i \leq m$, we have that $P_iR_\mathfrak p=0$, i.e. we find some $f_i \notin \mathfrak p$, such that $f_iP_i=0$ (Here the noetherian property is needed, because you find such an element for each generator of $P_i$ and then you mutiply the finitely many findings).
For $i > m$ we have that $P_i$ is not contained in $\mathfrak p$, i.e. find some $f_i$ with $f_i \in P_i \setminus \mathfrak p$. It seems reasonable to give $f = f_1 \dotsb f_n$ a chance and it turns out to work:
For $i = 1, \dotsc, m$ we have $fP_i=0$ because $fP_i \subset f_iP_i=0$. This shows $P_iR_f=0$. Thus all possible associated primes within $P_1R_f, \dotsc, P_mR_f$ are zero.
For $i > m$ we have $f \in P_i$, i.e. $P_iR_f$ is not an associated prime of $R_f$, because $P_i$ meets $\{f^n | n \in \mathbb N\}$.
Conclusion: $$\operatorname{Ass} R_f= \{P_iR_f ~|~ f \notin P_i \} \subset \{P_1R_f, \dotsc, P_mR_f\} = \{0\}.$$ This is equivalent to the desired claim.
