Let $R$ be an integral domain with unique maximal ideal $\mathfrak{m}$. Let $F:=\operatorname{Frac}(R)$ be the field of fraction of $R$.
I know that $F/k(B)$ where $k(B):=R/\mathfrak{m}$ is a residue field is not always true. For instance, take the integers $\mathbb{Z}$ and localize at $p\mathbb{Z}$. The residue field is $\mathbb{Z}_p/p\mathbb{Z}_p$ while the quotient field of $\mathbb{Z}_p$ has no zero-divisors.
However, I was wondering if there were contexts in which this could be true. For instance, if $B$ were a finitely generated algebra over some field (or maybe even if $B$ is the localization of some such algebra). Something along this line is mentioned here and this is the context I would love for it to be true in.
A disproof/proof of this last statement would be greatly helpful (I'll need to keep thinking about it while this post is up). I will even accept a reference.
I can at least explain the situation you linked to.
Let $R$ be a finitely generated algebra over an algebraically closed field $k$. Let $\mathfrak{m}$ be a maximal ideal of $R$, and let $B = R_{\mathfrak{m}}$.
Now the quotient $B/\mathfrak{m}B$ is a field which is also a finitely generated $k$-algebra, so by Zariski's Lemma $B/\mathfrak{m}B$ is a finite extension of $k$. But now since $k$ is algebraically closed, this forces $B/\mathfrak{m}B$ to just be $k$. So, $B/\mathfrak{m}B$ embeds in $B$, and thus also in $\operatorname{Frac}(B)$.