= Solution
Consider the <kernel of a ring homomorphism>
$$
K=\ker(R\longrightarrow R_{\mathfrak p})
=\{a\in R:sa=0\text{ for some }s\notin\mathfrak p\}.
$$
Since $R$ is a <Noetherian ring>, this <ideal> has a finite <generating set of an ideal>, say $K=(a_1,\ldots,a_t)$. For each $i$, choose $s_i\notin\mathfrak p$ with $s_i a_i=0$, and put
$$
f=\prod_{i=1}^t s_i.
$$
Then $f\notin\mathfrak p$ by the <prime ideal> condition, and $fK=0$. If $K=0$, take the empty product $f=1$.
If $a/f^n$ maps to zero under the <ring homomorphism> $\theta_f$ from <2/a>, then $a/1=0$ in $R_{\mathfrak p}$ because $f/1$ is a <unit> there. Hence $a\in K$, so $fa=0$. The <vanishing criterion in a module localization> now gives $a/f^n=0$ already in $R_f$. Therefore
$$
\boxed{\ker\theta_f=0,\qquad\theta_f:R_f\hookrightarrow R_{\mathfrak p}.}
$$
This is the <finite generation clears localization kernels> argument: the finite <generating set of an ideal> permits a single element outside $\mathfrak p$ to annihilate the entire <kernel of a ring homomorphism>.
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