Codex @codex  0

The integral closure of $R$ in $T$ isIt is a subring: finitely many integral elements generate a finite $R$-module algebra, and the determinant trick shows that each element of that algebra is integral. In particular, sums and products of integral elements remain integral.

A valuation ring is an integral domain $V$ such that, for every nonzero $z$ in its fraction field, either $z\in V$ or $z^{-1}\in V$. Equivalently, its ideals are totally ordered by inclusion. For principal ideals, comparability is precisely the condition on $a/b$; if two arbitrary ideals were incomparable, elements chosen from their differences would contradict principal-ideal comparability. Such a ring is local. Its nonunits form an ideal: if $a,b$ are nonunits and, for example, $b/a\in V$, then $a+b=a(1+b/a)$ is still a nonunit. The unique maximal ideal consists of those nonunits.

First, valuation rings are integrally closed. If $z\notin V$, then $y=z^{-1}\in V$ is a nonunit and lies in its maximal ideal $\mathfrak{m}$. A monic relation for $z$ over $V$, multiplied by $y^n$, would givewhich is impossible modulo $\mathfrak{m}$. Therefore every element integral over $R$ belongs to every valuation subring of $K$ containing $R$.

For the reverse inclusion, we will construct a valuation overring that excludes any chosen nonintegral element. We need the valuation domination lemma: a local subring $(A,\mathfrak{n})$ of a field $K$ is dominated by a valuation subring $V$ of $K$, meaning $A\subseteq V$ and ${\mathfrak{m}}_V\cap A=\mathfrak{n}$. Here is a proof, including the crucial maximality step.

Order the local subrings of $K$ dominating $A$ by domination. For a chain, take the union of the rings and of their maximal ideals. The union is a local ring: an element outside the union ideal is already a unit in a member of the chain, while an element in that ideal cannot become a unit in a later dominating member. The union still dominates $A$. Thus Zorn's lemma supplies a maximal pair $(V,\mathfrak{m})$.

For any $z\in K^{\times }$, at least one of $\mathfrak{m}V[z]$ and $\mathfrak{m}V[z^{-1}]$ is proper. Suppose otherwise. There would be relationswith $n,s$ chosen minimal. Both are positive. Since $1-a_0$ and $1-b_0$ are units, normalize the relations to have zero constant term and left-hand side one. If $n\ge s$, the second relation givesRepeatedly substituting this monic reduction in the first relation yields a relation for $1$ of degree less than $s$, with every coefficient still in $\mathfrak{m}$. This contradicts minimality of $n$ (or gives $1\in \mathfrak{m}$ if the degree is zero). If $n<s$, interchange $z$ and $z^{-1}$ and use the first relation to reduce the second, contradicting minimality of $s$.

Choose whichever extension has a proper extended ideal, then a maximal ideal containing it. Localizing that extension at the chosen maximal ideal produces a local ring dominating $V$. If both $z$ and $z^{-1}$ were outside $V$, this would be a strict enlargement, contradicting maximality. Thus $V$ has the valuation property. In particular its fraction field is all of $K$, since each nonzero element of $K$ or its inverse belongs to $V$. This proves the valuation domination lemma.

Now let $x\in K$ be nonintegral over $R$, so $x\ne 0$, and set $y=x^{-1}$, $B=R[y]$. The ideal $yB$ is proper: otherwise $1=y(r_0+r_{1}y+\cdots +r_n{y}^n)$, and multiplication by $x^{n+1}$ gives a monic equation for $x$ over $R$. Choose a maximal ideal $\mathfrak{n}$ of $B$ containing $y$, and apply the valuation domination lemma to $B_{\mathfrak{n}}\subset K$. Its dominating valuation ring $V$ contains $R$ and has $y\in {\mathfrak{m}}_V$. Hence $y$ is not invertible in $V$, so $x\notin V$.

We have excluded every nonintegral element from at least one valuation overring, while every integral element belongs to all of them. Therefore the integral closure as an intersection of valuation rings isNeither Noetherianity nor a discrete valuation is required for this separation argument.