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For a normal Noetherian ring $A$ that is an integral domain, the integral closure in a finite separable field extension of its fraction field is finite as an $A$-module. An integral field basis by denominator clearing gives a finite lattice. Integral traces place the closure in its finite trace-dual lattice; Noetherianity makes that submodule finitely generated.

For an algebraically closed field $k$ and a finitely generated field extension $K/k$ of transcendence degree one, choose $t$ with $K/k(t)$ finite. The integral closures of the two standard coordinate rings of the projective line in $K$ are finite modules and glue to a finite normalization of an algebraic curve of the projective line. A one-dimensional normal local ring is a discrete valuation ring; over the perfect field $k$ the resulting projective curve is smooth. Extension of rational maps gives uniqueness of this smooth projective model.

Let $B$ be the integral closure of a Dedekind domain $R$ in a finite separable extension $L/K$ of degree $n$. For a nonzero prime ideal $\mathfrak{p}$ of $R$, writeand let $f_{\mathfrak{q}/\mathfrak{p}}=[B/\mathfrak{q}:R/\mathfrak{p}]$. Then

For a domain $R$ with fraction field $K$, its integral closure is the intersection of all valuation rings of $K$ containing $R$. Integral elements belong to every such ring because valuation rings are integrally closed. For a nonintegral $x$, the ideal $x^{-1}R[x^{-1}]$ is proper, since its containing $1$ would give a monic equation for $x$. Localize at a maximal ideal containing $x^{-1}$ and apply the valuation domination lemma. The resulting ring contains $R$ but not $x$, proving the reverse inclusion.

For an integral scheme, take on each affine chart the integral closure of its coordinate ring in its function field. Integral closure commutes with localization, so these rings glue to a normal scheme with an integral birational morphism to the original scheme. This is the normalization. It need not be finite without additional hypotheses; for an algebraic curve of finite type over a field it is finite. It is an isomorphism exactly when the original scheme is normal.

At a point $P$ of the smooth algebraic curve $X$, the local ring ${\mathcal{O}}_{X,P}$ is a discrete valuation ring. Embed the projective variety $Y$ in ${\mathbb{P}}^N$. A rational map of algebraic varieties to $Y$ has homogeneous coordinates $[f_0:\cdots :f_N]$ in the function field $k(X)$, not all zero. Let $v_P$ be the local valuation, choose a uniformizer $u$, and set $m={\min }_{f_i\ne 0}{v}_P(f_i)$. The coordinates $u^{-m}{f}_i$ are regular at $P$, and at least one is a unit. On a sufficiently small open subset containing $P$, all these functions are regular and that coordinate stays nonzero. Their ratios therefore define a morphism of algebraic varieties into the corresponding affine chart of ${\mathbb{P}}^N$. Every homogeneous equation defining $Y$ vanishes on the original dense open, hence vanishes identically on this neighborhood. The local extension consequently lands in $Y$.

Two such extensions agree wherever the original rational map of algebraic varieties is defined. Their equalizer is closed, because $Y$ is a separated variety, and contains a dense open. Thus they agree throughout their overlap and glue. The rational map has a unique extension to a morphism on all of $X$. Projectivity is needed for the target; the source need not be projective.

For the function field correspondence, let $K/k$ be a finitely generated field extension of transcendence degree one. Choose a transcendence basis consisting of one element $t$. The extension $K/k(t)$ is finite: finitely many algebraic generators each have finite degree. Take the integral closures of $k[t]$ and $k[t^{-1}]$ in $K$. The finiteness assumption allowed in the paper makes both finite modules over their original coordinate rings. Integral closure commutes with localization, so these two rings agree over $k[t,t^{-1}]$ and glue to the finite normalization model of a one-variable function field $C_K\to {\mathbb{P}}^1$. This is a finite morphism, hence projective, and its function field is $K$. Its one-dimensional normal local rings are discrete valuation rings, so it is regular. Since the algebraically closed field $k$ is perfect, regularity here implies smoothness. Thus $C_K$ is a smooth projective curve.

Conversely the function field of a smooth projective curve is finitely generated over $k$ and has transcendence degree one. Two such curves with isomorphic function fields have mutually inverse birational maps. The extension argument just proved makes both maps morphisms; their composites equal the identity on a dense open, and hence everywhere by separatedness. This proves uniqueness of the smooth projective model. More precisely, pullback gives a contravariant correspondence between nonconstant morphisms of smooth projective curves and $k$-embeddings of their function fields. A field embedding defines the map on affine coordinates near the generic point, and the extension argument completes it. Constant morphisms are excluded from this correspondence because they do not induce embeddings of function fields.

Finally let $f:C\to D$ be nonconstant. Its image is dense, and pullback gives a finite extension $k(D)\subset k(C)$: both fields have transcendence degree one and the latter is finitely generated. Normalize $D$ in $k(C)$, applying the allowed finite-integral closure result on affine charts. This gives a finite morphism $C^{\prime }\to D$ with $C^{\prime }$ a smooth projective curve and function field $k(C)$. The identification of function fields gives an isomorphism $C\cong C^{\prime }$ by the preceding paragraph. Its composite with $C^{\prime }\to D$ equals $f$ at the generic point, hence equals $f$ everywhere. Every nonconstant morphism of these curves is finite.

The Absolute Frobenius morphism is the identity on the underlying topological space and acts on the structure sheaf by $a\mapsto a^p$. On an affine scheme it is induced by the ring homomorphism $A\to A$, $a\mapsto a^p$. This is additive in characteristic $p$, and the inverse image of a prime ideal $\mathfrak{p}$ is again $\mathfrak{p}$, since $a^p\in \mathfrak{p}$ exactly when $a\in \mathfrak{p}$. These affine maps agree under localization and hence glue.

A perfect ring of characteristic p is one whose Frobenius endomorphism is bijective. Since the Absolute Frobenius morphism fixes points, its map on the stalk at $x$ is precisely the Frobenius endomorphism of ${\mathcal{O}}_{X,x}$. A morphism of sheaves is an isomorphism exactly when it is an isomorphism on all stalks. ThereforeIn the reverse direction the inverse sheaf map, together with the identity on points, gives the inverse morphism of locally ringed spaces. This is perfection in the Frobenius sense, not the unrelated homological use of the word perfect.

For the function field assertions below, interpret the normal schemes as integral, as is implicit in writing a single field $k(X)$ or $k(Y)$. General normal schemes need not be irreducible; without this interpretation those fields are not defined globally as single fields. Set $K=k(Y)\subseteq L=k(X)$. For an affine open $V=\mathrm{Spec}A$ in $Y$, finiteness gives $f^{-1}V=\mathrm{Spec}B$, with $A\subseteq B$ finite and both rings normal domains. In fact $B$ is the integral closure of $A$ in $L$: any element of $L$ integral over $A$ is also integral over $B$ and therefore belongs to $B$. This local description yields the two forms of Frobenius factorization through a finite normal cover.

Write $f$ as the ring homomorphism $A\to B$. By the permitted assumption, $B$ is a finitely generated $A$-algebra. We will prove that every element of $B$ is integral over $A$, including any nilpotent contribution. If $B=0$, there is nothing to prove.

We use the integral closure as an intersection of valuation rings in its general form: if an integral domain $R$ lies in a field $K$, then its integral closure inside $K$ is the intersection of all valuation rings of $K$ containing $R$. Here $K$ need not be the fraction field of $R$. One direction follows from valuation rings are integrally closed. For the other, if $0\ne x\in K$ is not integral, the ideal $x^{-1}R[x^{-1}]$ is proper: an expression for $1$ in that ideal, multiplied by a sufficiently large power of $x$, would be a monic equation for $x$. Choose a maximal ideal containing this ideal, and apply the valuation domination lemma to the corresponding local ring inside $K$. The resulting valuation ring $V$ contains $R$ and has $x^{-1}$ in its maximal ideal, so $x\notin V$. This proves the claimed intersection criterion. The valuation domination lemma is obtained by choosing, using Zorn's lemma, a maximal local overring dominating the given local ring; the usual alternative between adjoining $z$ and adjoining $z^{-1}$ forces this maximal overring to be a valuation ring.

The Noetherian ring $B$ has finitely many minimal prime ideals ${\mathfrak{q}}_1,\dots ,{\mathfrak{q}}_r$. SetTake any valuation ring $V$ of $K_i$ containing $R_i$. The maps $B\to D_i\to K_i$ and $A\to R_i\to V$ form a square for the valuative criterion for properness. Its lift gives $B\to V$, whose composite with $V\hookrightarrow K_i$ is the original map. Consequently the image $D_i$ is contained in $V$. The intersection criterion proves that every element of $D_i$ is integral over $R_i$.

Fix $b\in B$. For each $i$, choose a monic polynomial $P_i(T)\in A[T]$ whose value at the image of $b$ in $D_i$ is zero; coefficients over $R_i$ can be lifted to $A$. The product $P(T)={\prod }_i{P}_i(T)$ is monic and satisfiesThe nilradical of a Noetherian ring is nilpotent. Explicitly, choose finitely many generators $n_j$ with $n_j^{e_j}=0$; every product of $1+{\sum }_j(e_j-1)$ generators then contains a vanishing power. Thus $(\sqrt{(0)}{)}^N=0$ for some $N$, and $P(T{)}^N$ is a monic polynomial annihilating $b$. Hence $B$ is integral over $A$.

Finally choose algebra generators $b_1,\dots ,b_s$, with monic equations of degrees $d_1,\dots ,d_s$. Reducing powers in these equations shows that the finitely many monomialsspan $B$ as an $A$-module. Therefore $B$ is a finite $A$-module, proving

Put $K=\mathrm{Frac}(A)$ and let $B$ be the integral closure of $A$ in $L$. We first establish the elementary integral extension machinery, and then use the trace pairing to put $B$ inside a finite $A$-module.

An integral element $x$ satisfies a monic polynomial over the base ring. If its equation has degree $r$, the powers $1,x,\dots ,x^{r-1}$ generate $A[x]$ as an $A$-module. More generally, adjoining finitely many integral elements gives a finitely generated module: reduce the exponent of each generator using its monic equation. Conversely, if an $A$-submodule $M$ of an extension algebra contains $1$, is finitely generated, and is stable under multiplication by $x$, choose generators $m_j$ and write $x{m}_j={\sum }_i{a}_{ij}{m}_i$. Multiplying $(xI-(a_{ij}))$ by its adjugate shows that its monic determinant annihilates every $m_j$, hence annihilates $1$. This finite-module criterion for integrality is the determinant trick.

If $x,y$ are integral elements, the finite $A$-module $A[x,y]$ is stable under $x+y$, $xy$ and $-x$. The criterion proves that these are integral elements. Thus the integral closure really is a subring. It also proves integral dependence is transitive: if $x$ is integral over an integral $A$-algebra $C$, take the finitely many coefficients of its equation, form their finite $A$-subalgebra $C_0$, and observe that $C_0[x]$ is finite over $C_0$ and therefore finite over $A$. The criterion applies to $x$. In particular, an algebra generated by finitely many integral elements is a finite integral extension, even though an arbitrary integral extension need not be finite.

Choose a $K$-basis $v_1,\dots ,v_d$ of $L$. Each basis element is algebraic over $K$. Ifchoose a nonzero $a_i\in A$ clearing all coefficient denominators. Then $e_i=a_i{v}_i$ satisfies a monic equation with coefficients $a_i^j{c}_{ij}\in A$. Hence all $e_i$ belong to $B$ and still form a $K$-basis. This is integral field basis by denominator clearing.

We next prove the trace of an integral element over a normal domain property. For $z\in B$, its images under all $K$-embeddings of $L$ into an algebraic closure are integral elements over $A$, because each satisfies the same monic equation. Their sum is integral by the subring property just proved. Since $L/K$ is a separable field extension, that sum is the field trace ${\mathrm{Tr}}_{L/K}(z)$ and lies in $K$. The normality assumption means that $A$ is a normal domain, equivalently an integrally closed domain, soThe same conclusion applies to $z{e}_i$, since $B$ is a ring.

The trace pairing of a finite separable field extension is nondegenerate. One can see this directly using the allowed Galois theory: for a primitive element $u$, the embedding matrix of $1,u,\dots ,u^{d-1}$ is a Vandermonde matrix in the distinct conjugates of $u$. Its determinant is nonzero, and the trace Gram matrix is its transpose times itself. Nondegeneracy is unchanged by a change of basis.

Let $M={\sum }_{i}A{e}_i$. Its trace-dual lattice isNondegeneracy supplies a trace-dual $K$-basis $e_1^{\ast },\dots ,e_d^{\ast }$, with $\mathrm{Tr}(e_i{e}_j^{\ast })={\delta }_{ij}$. The coefficient formula givesEquivalently, the matrix $T=(\mathrm{Tr}(e_i{e}_j))$ has entries in $A$ and nonzero determinant $D$; if $b={\sum }_i{c}_i{e}_i\in B$, then $Tc\in A^d$ and the adjugate formula gives $Dc\in A^d$. Thus $B\subseteq D^{-1}M$ as well. Both descriptions exhibit a finite free ambient $A$-module.

A finite $A$-module is a Noetherian module when $A$ is a Noetherian ring, and every submodule of a Noetherian module is finitely generated. Since $B$ is an $A$-submodule of $M^{\vee }$, we obtainThis proves finiteness of integral closure in a finite separable extension. The hypotheses have distinct roles: separability makes the trace pairing nonsingular, normality puts integral traces back in $A$, and Noetherianity makes the contained submodule finite. Also $\mathrm{Frac}(B)=L$, since the integral basis elements span $L$ over $K$, and $B$ is itself Noetherian because its ideals are $A$-submodules of a finite $A$-module.

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.

An integral domain $A$ is an integrally closed domain if every element of its fraction field which is an integral element over $A$ belongs to $A$. Explicitly, if $z\in \mathrm{Frac}A$ satisfiesthen $z\in A$. Thus $\boxed{A\text{ is integrally closed exactly when its integral closure in its fraction field is }A}$.

Suppose $A$ is a unique factorization domain and $z=a/b$ is integral, with $a,b\in A$, $b\ne 0$, and with no common irreducible element as a factor. Multiplying its monic equation by $b^d$ givesHence $b$ divides $a^d$. Every irreducible element dividing $b$ is a prime element, so it would divide $a$, contrary to the choice of the fraction. Therefore $b$ is a unit, and $z\in A$. We have proved

The going-down theorem is the following. Let $A\subseteq B$ be an integral extension of integral domains with $A$ an integrally closed domain. IfthenWe first prove this for a module-finite ring extension and then remove that extra hypothesis. We use the Lying-over theorem, the going-up theorem, and the incomparability theorem for integral extensions: primes exist over every base prime, prime chains can be extended upwards, and distinct primes over the same base prime cannot contain one another. The going-up theorem follows by applying the Lying-over theorem to the quotient by the lower prime. These results concern arbitrary integral extensions; none assumes the going-down theorem.

For the finite case, let $K=\mathrm{Frac}A$ and $L=\mathrm{Frac}B$. Since $B$ is module-finite, $L/K$ is a finite field extension. Take a finite normal field extension $E/K$ containing $L$, obtained as a splitting field of the minimal polynomials of finitely many field generators. The extension $E/K$ need not be a separable field extension. Let $C$ be the integral closure of $A$ in $E$. Then $B\subseteq C$, and $C$ is integral over both $A$ and $B$.

We need one auxiliary fact: the finite group $G={\mathrm{Aut}}_K(E)$ acts transitively on the prime ideals of $C$ over any fixed prime ideal $\mathfrak{p}$ of $A$. First observe that an element $w\in E$ fixed by all of $G$ is a purely inseparable algebraic element over $K$. Indeed, all $K$-embeddings of $K(w)$ into an algebraic closure extend to $K$-automorphisms of the normal field extension $E$. Hence the minimal polynomial of an algebraic element $w$ has only one distinct root. In characteristic zero this says $w\in K$; in characteristic $p>0$ it says $w^{p^e}\in K$ for some $e\ge 0$.

Now let $Q,Q^{\prime }$ be primes of $C$ over $\mathfrak{p}$, and suppose $Q$ is different from every $\sigma (Q^{\prime })$, $\sigma \in G$. By the incomparability theorem for integral extensions, $Q$ is contained in none of these conjugate primes. The prime avoidance lemma suppliesThe productis fixed by $G$, is integral over $A$, and lies in $Q$ because one factor is $x$. Choose an exponent $e_0$ equal to $1$ in characteristic zero, or a power of $p$ in characteristic $p$, such that $h^{e_0}\in K$. Since $A$ is an integrally closed domain, $h^{e_0}\in A$. It also lies in $Q\cap A=\mathfrak{p}$, hence in $Q^{\prime }$. Because $Q^{\prime }$ is a prime ideal, some factor $\sigma (x)$ lies in $Q^{\prime }$. This means $x\in {\sigma }^{-1}(Q^{\prime })$, contradicting our choice. Thus $Q=\sigma (Q^{\prime })$ for some $\sigma$, proving transitivity. The power $e_0$ accounts for the possible purely inseparable field extension and makes the argument valid in every characteristic.

Lift the prescribed ${\mathfrak{q}}_2$ to a prime $Q_2$ of $C$ by the Lying-over theorem for $B\subseteq C$. Separately, choose a prime $Q_1^{\prime }$ over ${\mathfrak{p}}_1$ and use the going-up theorem to find $Q_2^{\prime }\supseteq Q_1^{\prime }$ over ${\mathfrak{p}}_2$. Transitivity gives $\sigma \in G$ with $\sigma (Q_2^{\prime })=Q_2$. ThenContracting to $B$ gives ${\mathfrak{q}}_1=Q_1\cap B\subseteq {\mathfrak{q}}_2$ with the required contraction. This proves the going-down theorem for the module-finite case.

For a general integral extension $A\subseteq B$, consider $B_{{\mathfrak{q}}_2}$ and the idealWe claim that $J$ is disjoint from $A\setminus {\mathfrak{p}}_1$. Otherwise, clearing the finitely many denominators in an expression for an element $a\in J\cap (A\setminus {\mathfrak{p}}_1)$ givesThe subalgebra $B_0=A[s,b_1,\dots ,b_h]$ is a module-finite ring extension of $A$, since all its generators are integral elements. Apply the finite case to ${\mathfrak{q}}_2\cap B_0$: there is a prime $Q_1\subseteq {\mathfrak{q}}_2\cap B_0$ contracting to ${\mathfrak{p}}_1$. The displayed identity puts $sa$ in $Q_1$, but $a\notin Q_1$. Thus $s\in Q_1\subseteq {\mathfrak{q}}_2$, a contradiction.

Localize $B_{{\mathfrak{q}}_2}$ further at the multiplicative subset $A\setminus {\mathfrak{p}}_1$. The extension of $J$ is a proper ideal, by the disjointness just proved. Choose a maximal ideal containing it and contract back to a prime ideal of $B_{{\mathfrak{q}}_2}$. This prime contains $J$ and avoids $A\setminus {\mathfrak{p}}_1$. Its contraction ${\mathfrak{q}}_1$ to $B$ lies inside ${\mathfrak{q}}_2$ by the prime ideal correspondence for localization, and its contraction to $A$ is exactly ${\mathfrak{p}}_1$. This completes the proof of the going-down theorem without any finite generation assumption on $B$.

If $x,y$ are integral over $A$, then $A[x,y]$ is finite over $A$: it is generated by finitely many monomials $x^i{y}^j$. Multiplication by $x+y$, $x-y$, or $xy$ is an endomorphism of this finite module, so the determinant trick gives a monic annihilating polynomial. Hence the integral elements form a subring $C$ containing $A$.

The integral closure of $A$ in $B$ is this subring $C$. The ring $A$ is integrally closed in $B$ when $C=A$, and $B$ is integral over $A$ when $C=B$.

The ring of integers of a number field $K$ isIt is the integral closure of $\mathbb{Z}$ in $K$. Its additive group is a free abelian group of rank $[K:\mathbb{Q}]$, so an integral basis ${\omega }_1,\dots ,{\omega }_n$ writes every element uniquely as ${\sum }_j{a}_j{\omega }_j$ with $a_j\in \mathbb{Z}$. Its field of fractions is $K$.^[ref]

For $K=\mathbb{Q}$, this gives ${\mathcal{O}}_K=\mathbb{Z}$; for $K=\mathbb{Q}(\sqrt{2})$, it gives $\mathbb{Z}[\sqrt{2}]$. The Gaussian integers give the corresponding example inside $\mathbb{Q}(i)$. The ring of integers is determined by the field, and need not equal $\mathbb{Z}[\theta ]$ for a chosen algebraic integer generator: in $\mathbb{Q}(\sqrt{5})$, the larger ring $\mathbb{Z}[(1+\sqrt{5})/2]$ contains $\mathbb{Z}[\sqrt{5}]$ with index two.

This is the setting for divisibility, global units, and prime ideal factorization. A ring of integers is a Dedekind domain, so every nonzero proper ideal factors uniquely into nonzero prime ideals, even when unique factorization of elements fails.^[ref]

For a normal domain $A$ with fraction field $K$, the field trace of an integral element in a finite separable field extension $L/K$ belongs to $A$. Its conjugates are integral, so their sum is integral; it lies in $K$ and hence in $A$ by normality. This places the integral closure inside a trace-dual lattice.