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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?

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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.

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  • $\begingroup$ On your first paragraph: the usual theorem states: if $B$ is integral over $A$, then we can extend $f:A \to \Omega$ to $\overline f:B \to \Omega$. but notice $\alpha \in K\setminus V$ is not in $V$, which is integrally closed, so $\alpha$ is not integral over $V$, and therefore, $V[\alpha]$ is not integral over $V$, so it is not clear how you can extend the homomorphism from theorem 2, so it is not clear to me $V[\alpha] = K$ $\endgroup$
    – shintuku
    Jul 19, 2023 at 13:51
  • $\begingroup$ May i know the problem with extending like $\hat{f}(\alpha) = \alpha$ and $\hat{f}|_A = f|_A$ and $\hat{f} (\sum_i a_i \alpha^i) = \sum_i f(a_i) \alpha^i $? $\endgroup$
    – Balaji sb
    Jul 19, 2023 at 14:10
  • $\begingroup$ If this is possible, there is no guarantee $V$ was indeed maximal. It seems to work for a DVR, but no guarantee for an arbitrary valuation ring. We furthermore know that valuation rings are not necessarily maximal subrings of $K$. $\endgroup$
    – shintuku
    Jul 19, 2023 at 14:19
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    $\begingroup$ OK but think about my suggestion for extending ring homomorphism to $V[\alpha]$ and let me know if you find a counter argument. Cheers ! Lets see about other's opinions also in this post. $\endgroup$
    – Balaji sb
    Jul 19, 2023 at 14:40
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    $\begingroup$ Of interest: if V is maximal wrt. Theorem 2, and t is not in V, then the correct statement is: either V[t] = K, or t is not in the maximal ideal of V[t], and we're guaranteed V[t] is a valuation ring $\endgroup$
    – shintuku
    Jul 19, 2023 at 18:44

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