We define domination as follows: for any two local rings $(A,\mathfrak{m})$ and $(B,\mathfrak{n})$, we say $(B,\mathfrak{n})$ dominates $(A,\mathfrak{m})$ if $A\subseteq B$ and $\mathfrak{n}\cap A = \mathfrak{m}$ (which then allows us to apply Zorn's Lemma-based arguments). Now if $(A,\mathfrak{m})$ is maximal in $K$ w.r.t. domination, then I wish to show that $A$ is a valuation ring in $K$. I'm assuming there exists some $x\in K^\times$ such that $x,x^{-1}\notin A$, and I have succeeded in showing that $x$ must be integral over $A$. But how do I now derive a contradiction?
Let $K$ be a field, and $(A,\mathfrak{m})$ a local ring maximal in $K$ w.r.t. domination. Then $A$ is integrally closed.
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abstract-algebra
algebraic-geometry
ring-theory
commutative-algebra
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0Set $B=A[x]$. Since $A\subset B$ is an integral extension there is a prime ideal $P\subset B$ lying over $\mathfrak m$, and $B_P$ dominates $A$. – 2017-02-04
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1The title suggests that this is going to be a problem about proving $A$ is integrally closed, but the actual request is "I wish to show that $A$ is a valuation ring in $K$." The body of the Question is poorly formatted to highlight this issue, so perhaps tweaking of the title and breaking up the "block of text" in the body will be helpful to Readers. – 2017-02-04