What is the relationship between domination of local rings and extension of ring homomorphisms to an algebraically closed field?
Let $K$ be a field, $A,B$ local rings contained in $K$. We say $B$ dominates $A$ if $A \subset B$ and $\mathfrak m_A = A \cap \mathfrak m_B$. Consider the following two theorems:
Theorem 1. (Stacks Project) Let $K$ be a field, $A$ be a local subring. Then there exists a valuation ring with fraction field $K$ dominating $A$.
Theorem 2. (Atiyah-Macdonald) Let $K$ be a field, $\Omega$ an algebraically closed field. Let $\Sigma$ be the set of all pairs $(A,f)$ where $A$ is a subring of $K$ and $f$ is a homomorphism of $A$ to $\Omega$. We partially order $\Sigma$ as follows: $(A,f) < (A',f') \iff A \subseteq A' \land f'|A = f$. Then if $(V,\phi)$ is a maximal element of $\Sigma$, we have $V$ is a valuation ring of $K$.
Theorem $1$ and $2$ can both be used to prove that, if $x$ is an element of $K$ that is not integral over $A$, then there exists a valuation ring that contains $A$ and not $x$. That both of these theorems have the same use and are both used to find specific valuation rings suggests there might be some relationship between extensions of ring homomorphisms mapping to algebraically closed fields and the domination of local rings. Is there any deeper relationship here beyond the superficial appearance?
Theorem 2:
If $V$ is a maximal element then for an element $\alpha \in K \setminus V$, $V[\alpha] = K$ or else we will be able to extend the chain to $V[\alpha]$.
Let $v \in V$ be a non-unit, then by choosing $\alpha = v^{-1}$, we have $V[v^{-1}] = K$. Hence field of fractions of $V$ is equal to $K$. Its clear that $V[v^{-1}]$ is a local ring with unique maximal ideal $\langle v \rangle$. Hence $V$ is a valuation ring.
Theorem 1:
In this also $A$ is such that its field of fractions is equal to $K$. Further $A$ is a valuation ring.
Theorem 2:
Since $V[v^{-1}] = K \implies $ $V$ has inverses of all elements except $v^i$.
Theorem 1:
Take $A$ and choose $S = A \setminus \langle a \rangle$ where $a \in A$, is a prime element.
Now localize, $A$ w.r.t $S$ i.e., form $S^{-1}A$. My guess is $S^{-1}A$ is a maximal element w.r.t Theorem 2 and a local ring with unique maximal ideal $\langle a \rangle$.
On the other hand if $A$ is a local ring with maximal ideal of the form $\langle a \rangle$ and field of fractions of $A$ is $K$ then $A$ is a maximal element by the above explanation.