Isomorphism between $R_p/pR_p$ and $\text{Frac}(R/p)$ is unique Let $R$ be a commutative ring, $P$ a prime ideal, $R_p = (R \setminus P)^{-1}R$ the localization at $R \setminus P,$ and $F = \text{Frac}(R/p)$ the fraction field of the integral domain $R/p.$ We want to show there exists a unique isomorphism between $R_p/pR_p$ and $F,$ noting that $pR_p$ is the unique maximal ideal of $R_p.$
Define $f : R_p \to F$ by $r/u \to (r+P)/(u+P).$ It's easy to check $f$ is surjective and $\ker f = pR_p$ so that we get an  isomorphism by the 1st isomorphism theorem. But how do we prove the isomorphism is unique?
If $f, g$ are isomorphisms, then $f \circ g^{-1}$ is an automorphism of $F$ and $f^{-1} \circ g$ is an automorphism of $R_p/pR_p,$ so if we could prove either of these fields have trivial automorphism group, we would be done. Now we are in the realm of Galois theory, and I don't know how to proceed as there's no base field to work over.
Update: The precise wording is "there is a unique isomorphism identifying residue class field of the local ring $R_P$ at its unique maximal ideal with the fraction field of $R/P.$"
 A: In general the isomorphism will not be unique as an isomorphism of rings. For example, if $R$ is a field and $p=(0)$, then $R_p=R=R/p$, so stating that there is a unique ring isomorphism $R_p/pR_p\to\operatorname{Frac}(R/p)$ is the same as stating that there is a unique automorphism of $R$; this is untrue whenever $R$ has a non-trivial automorphism.
However, the isomorphism is unique as an isomorphism of $R$-algebras, for which it suffices to show there is in fact a unique $R$-algebra homomorphism from $R_p$ to $F$. To see this, let $f:R_p\to F$ be any $R$-algebra homomorphism. Then necessarily $f(1/1)=1+P$. Furthermore, for any $r\in R$, we have $r/1=r\cdot(1/1)$ in $R_p$, so by $R$-linearity of $f$ we must have $f(r/1)=rf(1/1)=r+P$. Finally, since $f$ is a ring morphism, for any unit $u$ of $R_p$, we must have $f(u^{-1})=f(u)^{-1}$. (Why?) Putting these two facts together, we get $$f(r/s)=f((r/1)(s/1)^{-1})=f(r/1)f(s/1)^{-1}=(r+P)(s+P)^{-1}$$ for any $r\in R$ and $s\in R\setminus P$. Since every element of $R_p$ is of form $r/s$ for some $r\in R$, $s\in R\setminus P$, this means $f$ is indeed uniquely determined.
