Polynomial algebras over fields have finite integral closures

$R$ is a domain and $F=\mathrm{Frac}(R)$ is a field containing $R$: for $d=0$ the ring $R=K$ is a field, and for $d\ge 1$ it is a polynomial ring over the domain $K$; so [L2] applies, and [L1] gives the fraction field. It is the rational function field $K(x_1,\dots ,x_d)$: a subfield of $F$ containing $K$ and the $x_i$ contains $R$, hence contains every $a\mathrm{/}b$ with $a,b\in R$, $b\mathrm{\ne }0$, that is, all of $F$ by [L1].

The extension $L\mathrm{/}F$ is finite, so by [L3] it has a finite $F$-basis ${\beta }_1,\dots ,{\beta }_m$, which spans $L$, and $L$ is algebraic over $F$; by [L6] and [L4] we may write $L=F({\beta }_1,\dots ,{\beta }_m)$, and for each $j$ the element ${\beta }_j$ has a monic minimal polynomial $f_j\in F[T]$ with $f_j({\beta }_j)=0$, of degree $[F({\beta }_j):F]$.

Let $h:=f_1\cdots f_m\in F[T]$, a nonzero polynomial because each $f_j$ is monic. By [L5] the family $f_1,\dots ,f_m$ has a splitting field over $L$; fix one and call it $M$. Then $h$ splits in $M$ and $M=L(\text{the roots of }h)$, and each ${\beta }_j$ is a root of $h$, so $L=F({\beta }_1,\dots ,{\beta }_m)\subseteq F(\text{the roots of }h)\subseteq M$. Since the middle field contains both $L$ and every root of $h$ in $M$, it contains $L(\text{the roots of }h)=M$; hence $M=F(\text{the roots of }h)$, and $M$ is also a splitting field of $h$ over $F$. The roots of $h$ are algebraic over $L$, so $M\mathrm{/}L$ is finite by [L6], and then $M\mathrm{/}F$ is finite by the tower law [L3].

The finite extension $M\mathrm{/}F$ is algebraic by [L3] and a splitting field of the nonzero polynomial $h$ over $F$, so it is normal by [L5]. Set $G:=\mathrm{Aut}(M\mathrm{/}F)$ and $E:=M^G$, a subfield with $F\subseteq E\subseteq M$ by [L7]. The group $G$ is finite: by [L3] write $M=F({\gamma }_1,\dots ,{\gamma }_r)$ for a basis, each ${\gamma }_i$ has minimal polynomial $m_{{\gamma }_i}$ over $F$ by [L3] and [L4], every $\sigma \in G$ is determined by the images $\sigma ({\gamma }_i)$, which are roots of $m_{{\gamma }_i}$, and by the root bound [L17] there are only finitely many such tuples. Since $G$ is a finite group of automorphisms of $M$, [L7] gives $[M:E]\ge \mathrm{\mid }G\mathrm{\mid }$ and $[M:E]\le \mathrm{\mid }G\mathrm{\mid }$, so $[M:E]=\mathrm{\mid }G\mathrm{\mid }$. Applying [L8] to the finite normal extension $M\mathrm{/}F$: the extension $E\mathrm{/}F$ is finite purely inseparable, $M\mathrm{/}E$ is finite Galois, hence separable, and $E=F$ if $F$ has characteristic $0$.

Let $C$ be the integral closure of $R$ in $E$, i.e. the set of elements of $E$ integral over $R$. By [L14] members of $E$ integral over the nonzero ring $R$ form a subring, so $C$ is a subring of $E$ with $R\subseteq C\subseteq E\subseteq M$, and every element of $C$ is integral over $R$ by definition; that is, $C$ is integral over $R$.

The extension $M\mathrm{/}E$ is finite separable by [L8], so by [L15] it is simple: there is $\theta \in M$ with $M=E(\theta )$.

The ring $C$ is a finite $R$-module. If $K$ has characteristic $0$, then $E=F$ by [L8] and $C$ is the integral closure of $R$ in $\mathrm{Frac}(R)$; since $R$ is integrally closed by [L11], $C=R$, generated as an $R$-module by $1$. If $K$ has characteristic $p>0$, then $E\mathrm{/}F$ is finite purely inseparable by [L8], so [L9] provides a finite purely inseparable $K^{\mathrm{\prime }}\mathrm{/}K$, an exponent $e$ with $q=p^e$, and an $F$-embedding $\eta :E\to E^{\mathrm{\prime }}:=K^{\mathrm{\prime }}(x_1^{1\mathrm{/}q},\dots ,x_d^{1\mathrm{/}q})$, the $x_i$ being algebraically independent over $K$. The image field $\eta (E)$ satisfies $K(x_1,\dots ,x_d)=F\subseteq \eta (E)\subseteq E^{\mathrm{\prime }}$, so the second clause of [L10] shows that the integral closure of $R$ in $\eta (E)$ is a finite $R$-module. The map $\eta$ fixes $F$ and hence $R$, so a $z\in E$ satisfies a monic equation over $R$ exactly when $\eta (z)$ does; thus $\eta$ restricts to an $R$-linear bijection from $C$ onto the integral closure of $R$ in $\eta (E)$, and finite generation transfers, so $C$ is a finite $R$-module in this case too.

$\mathrm{Frac}(C)=E$. Let $e\in E$. The extension $E\mathrm{/}F$ is finite by [L8], so $e$ is algebraic over $F$ and has a minimal polynomial $m_e=T^r+a_{r-1}{T}^{r-1}+\cdots +a_0\in F[T]$ of degree $r=[F(e):F]\ge 1$ by [L3] and [L4]. Each coefficient $a_i$ lies in $F=\mathrm{Frac}(R)$ and so is a fraction $u_i\mathrm{/}{v}_i$ with $u_i,v_i\in R$ and $v_i\mathrm{\ne }0$ by [L1]; put $c_1:=v_0{v}_1\cdots v_{r-1}$, a nonzero element of $R$ because $R$ is a domain. Then $c_1{a}_i\in R$ for every $i$, and $e_1:=c_{1}e$ is a root of the monic polynomial $T^r+c_1{a}_{r-1}{T}^{r-1}+c_1^2{a}_{r-2}{T}^{r-2}+\cdots +c_1^r{a}_0\in R[T]$, so $e_1$ is integral over $R$, that is $e_1\in C$; and $e=e_1\mathrm{/}{c}_1$ with $c_1\in C\setminus \{0\}$, so $e\in \mathrm{Frac}(C)$. Hence $E\subseteq \mathrm{Frac}(C)\subseteq E$ and $\mathrm{Frac}(C)=E$.

By [L3] and [L4] the element $\theta$ has a monic minimal polynomial $m_{\theta }=T^n+b_{n-1}{T}^{n-1}+\cdots +b_0\in E[T]$ of degree $n=[E(\theta ):E]=[M:E]$, and the powers ${\theta }^0,\dots ,{\theta }^{n-1}$ are an $E$-basis of $M$, so by [L4] every $z\in M$ has a unique expansion $z={\sum }_{i<n}{z}_i{\theta }^i$ with $z_i\in E$. Since $E=\mathrm{Frac}(C)$ by step 5.2 and $C$ is a domain by [L14], [L1] writes each $b_i$ as a fraction of elements of $C$; choose one nonzero $c\in C$ clearing all denominators, so $c{b}_i\in C$ for every $i$, and put ${\theta }_0:=c\theta$. Then ${\theta }_0$ is a root of the monic polynomial $T^n+c{b}_{n-1}{T}^{n-1}+c^2{b}_{n-2}{T}^{n-2}+\cdots +c^n{b}_0\in C[T]$, hence is integral over $C$; since $0\mathrm{\ne }c\in E$ we have $E({\theta }_0)=E(\theta )=M$, so by [L4] applied to ${\theta }_0$ every $z\in M$ has a unique expansion $z={\sum }_{i<n}{z}_i{\theta }_0^i$ with $z_i\in E$, and $[M:E]=[E({\theta }_0):E]=n$.

Let $D$ be the integral closure of $C$ in $M$, a subring of $M$ containing $C$ by [L14]; by step 6.1 the element ${\theta }_0$ lies in $D$. Every $\sigma \in G$ fixes $E=M^G$ pointwise by [L7] and hence fixes $C\subseteq E$, so applying $\sigma$ to a monic equation for an element of $D$ over $C$ exhibits $\sigma$ of that element as again integral over $C$; in particular the elements ${\alpha }_j:={\sigma }_j({\theta }_0)$ lie in $D$ once $G=\{{\sigma }_0,\dots ,{\sigma }_{n-1}\}$ is an enumeration of $G$ with ${\sigma }_0=\mathrm{id}$, which is possible because $\mathrm{\mid }G\mathrm{\mid }=n$ by step 3.1. The elements ${\alpha }_0,\dots ,{\alpha }_{n-1}$ are pairwise distinct: if ${\alpha }_i={\alpha }_j$, then ${\sigma }_j^{-1}{\sigma }_i$ fixes $E$ pointwise and fixes ${\theta }_0$, hence fixes $M=E({\theta }_0)$ elementwise, so ${\sigma }_i={\sigma }_j$.

Fix $z\in D$ and write $z={\sum }_{i<n}{c}_i{\theta }_0^i$ with $c_i\in E$ by step 6.1. Each ${\sigma }_j$ fixes $E$ pointwise and sends ${\theta }_0$ to ${\alpha }_j$, so ${\sigma }_j(z)={\sum }_{i<n}{c}_i{\alpha }_j^i$; this lies in $D$ because $z\in D$ and ${\sigma }_j$ preserves integrality over $C$ as in step 7.1. Let $V\in M_n(M)$ be the matrix with entries $V_{ji}:={\alpha }_j^i$ (row $j$, column $i$), let $c:=(c_0,\dots ,c_{n-1})\in M^n$ and let $w:=({\sigma }_0(z),\dots ,{\sigma }_{n-1}(z))\in M^n$. The displayed equations say exactly $Vc=w$, and every entry of $V$ and of $w$ lies in $D$ by step 7.1.

By [L16] (Cramer's rule over the commutative ring $M$) the solution $c$ of $Vc=w$ satisfies $\det (V)c_i=\det (V_i)$ for every $i<n$, where $V_i$ is $V$ with column $i$ replaced by $w$. Every entry of $V_i$ lies in the subring $D$ of $M$, and the determinant is a finite signed sum of products of entries by [L16], so $\det (V_i)\in D$; with $\delta :=\det (V)$ this gives $\delta c_i\in D$ for all $i<n$.

$\delta \mathrm{\ne }0$. Suppose $x=(x_0,\dots ,x_{n-1})\in M^n$ satisfies $Vx=0$. Then the polynomial $P(T):={\sum }_{i<n}{x}_i{T}^i\in M[T]$ has $P({\alpha }_j)={\sum }_{i<n}{x}_i{\alpha }_j^i=0$ for every $j<n$, that is, $P$ vanishes at the $n$ pairwise distinct elements ${\alpha }_0,\dots ,{\alpha }_{n-1}$ of the field $M$ (step 7.1). If $P$ were nonzero, then $\mathrm{deg}P<n$ would contradict the root bound [L17]; hence $P=0$ and therefore $x=0$. So the map $x\mapsto Vx$ has trivial kernel and [L16] makes $V$ invertible over the field $M$, so its determinant $\delta$ is a unit of $M$, in particular $\delta \mathrm{\ne }0$.

Put $c:={\prod }_{j<n}{\sigma }_j(\delta )\in M$. Since $\delta \in D$ by step 9.1 and each ${\sigma }_j$ preserves integrality over $C$ as in step 7.1, every factor ${\sigma }_j(\delta )$ lies in $D$, so $c\in D$ because $D$ is a subring; and for $\tau \in G$ the assignment $\sigma \mapsto \tau \sigma$ is a bijection of $G$, so $\tau (c)={\prod }_{j<n}(\tau {\sigma }_j)(\delta )={\prod }_{j<n}{\sigma }_j(\delta )=c$, showing that $c$ is fixed by every element of $G$, that is $c\in E=M^G$ by [L7]. Since $\delta \mathrm{\ne }0$ by step 9.2 and $M$ is a field, each factor ${\sigma }_j(\delta )$ is nonzero, so $c\mathrm{\ne }0$.

$D\subseteq N:={\sum }_{i<n}C\cdot ({\theta }_0^i\mathrm{/}c)$. Let $z\in D$ with expansion $z={\sum }_{i<n}{c}_i{\theta }_0^i$ as in step 8.1. Since ${\sigma }_0=\mathrm{id}$, the element $c$ of step 10.1 factors as $c=\delta \cdot {\prod }_{j\ge 1}{\sigma }_j(\delta )$, so $c{c}_i=({\prod }_{j\ge 1}{\sigma }_j(\delta ))(\delta c_i)\in D$ because both factors lie in $D$ by steps 10.1 and 9.1, and $c{c}_i\in E$ because $c\in E$ and $c_i\in E$; hence $c{c}_i\in D\cap E$. Now $D\cap E=C$: by [L13] applied to the domains $R\subseteq C\subseteq M$, whose middle term is integral over $R$ by step 4.1, an element of $M$ is integral over $R$ exactly when it is integral over $C$; so an $x\in E$ lies in $D$ (integral over $C$) exactly when $x\in C$ (integral over $R$). Therefore $c{c}_i\in C$ for every $i$, and since $0\mathrm{\ne }c\in E=\mathrm{Frac}(C)$ by steps 10.1 and 5.2 each element ${\theta }_0^i\mathrm{/}c$ lies in $M$ and $z={\sum }_{i<n}(c{c}_i)\cdot ({\theta }_0^i\mathrm{/}c)$ exhibits $z$ as an element of $N$. Hence $D\subseteq N$.

$N$ is a finite $R$-module: it is generated as a $C$-module by the $n$ elements ${\theta }_0^0\mathrm{/}c,\dots ,{\theta }_0^{n-1}\mathrm{/}c$, and $C$ is a finite $R$-module by step 5.1, so [L12] (transitivity of module finiteness) makes $N$ a finite $R$-module.

$D$ is a finite $R$-module and a finite $C$-module. The set $D$ is closed under addition, and under multiplication by $R\subseteq C\subseteq D$ because it is a subring containing $C\supseteq R$; so $D$ is an $R$-submodule of the finite $R$-module $N$ by step 11.1. Since $R$ is Noetherian by [L12], the submodule lemma of [L12] makes $D$ a finitely generated $R$-module; its finitely many $R$-generators generate it over $C$ as well, since $R\subseteq C\subseteq D$ and $D$ is closed under multiplication by $C$.

Let $x\in L$. By [L13] applied to $R\subseteq C\subseteq M$, whose middle term is integral over $R$ by step 4.1, the element $x$ is integral over $R$ exactly when it is integral over $C$, that is, exactly when $x\in D$; hence the integral closure of $R$ in $L$ equals $D\cap L$. That set is an $R$-submodule of the finitely generated $R$-module $D$ (step 13.1), so the submodule lemma of [L12] makes it a finite $R$-module: the integral closure of $R=K[x_1,\dots ,x_d]$ in the finite extension $L$ of $K(x_1,\dots ,x_d)$ is finite over $R$. Every selection in the proof was made from a finite list — the finite basis ${\beta }_1,\dots ,{\beta }_m$ of $L$, its minimal polynomials, the finitely many roots of their product, the finite group $G$ and its enumeration, one common denominator clearing the coefficients of $m_e$, one primitive element $\theta$, its finitely many coefficients and one further common denominator, and the finite sums inside the determinant computation — and all inductions run over finite data, so no choice principle is used.