Normalization Finiteness for Affine Domains

Suppose first that $z$ is integral over $A$. By [L1] there is a monic polynomial $f\in A[X]$ with $f(z)=0$. Since $A\subseteq B$, the same polynomial, viewed in $B[X]$, is monic and has the same root $z$; by [L1] again, $z$ is integral over $B$.

Conversely assume that $z$ is integral over $B$. By [L1] there are an integer $n\ge 1$ and coefficients $b_0,\dots ,b_{n-1}\in B$ with $z^n+b_{n-1}{z}^{n-1}+\cdots +b_{1}z+b_0=0.$

Each $b_i$ lies in $B$, and $B$ is integral over $A$, so every $b_i$ is integral over $A$ by [L2]. Let $C:=A[b_0,\dots ,b_{n-1}]$ be the subring of $L$ generated over $A$ by these coefficients. By [L3], applied to the ring extension $A\subseteq L$ (legitimate by [L5]), the elements of $L$ integral over $A$ form a subring of $L$; it contains $A$ and every $b_i$, hence contains the subring $C$ these elements generate. Therefore the inclusion $A\to C$ is an integral ring map.

The equation of step 1.2 has all its coefficients in $C$, so by [L1] the element $z$ is integral over $C$. Applying [L3] to the ring extension $C\subseteq C[z]$ (again $C\mathrm{\ne }0$ by [L5]) shows that the elements of $C[z]$ integral over $C$ form a subring containing $C$ and $z$, hence containing the subring $C[z]$ that they generate; therefore the inclusion $C\to C[z]$ is an integral ring map.

By steps 2.1 and 3.1 the maps $A\to C$ and $C\to C[z]$ are both integral, so the composite inclusion $A\to C[z]$ is integral by [L4]; every element of $C[z]$ is thus integral over $A$ by [L2], and in particular $z$ is integral over $A$. Together with step 1.1 this proves both directions of the equivalence.

Choose a finite $F$-basis of $L$ and enumerate it as $(u_1,\dots ,u_r)$; by [L2] it is an $F$-spanning set, so every element of $L$ is an $F$-linear combination of the $u_j$, hence lies in the subfield $F(u_1,\dots ,u_r)$ generated by them, while conversely $F(u_1,\dots ,u_r)\subseteq L$; thus $L=F(u_1,\dots ,u_r)$ by [L3].

Evaluation at $x_1,\dots ,x_d$ gives a homomorphism $K[X_1,\dots ,X_d]\to F$ that is injective by [L14], with image the subring generated by $K$ and the $x_i$; since $F$ is the field generated by these elements and contains $K[x_1,\dots ,x_d]$, [L4] gives $F=\mathrm{Frac}(K[x_1,\dots ,x_d])$, the fraction field of the domain $K[x_1,\dots ,x_d]$ of [L5]. In particular a nonzero polynomial in the $x_i$ with coefficients in $K$ is not zero in $F$.

By [L1] applied to each generator $u_j$ of step 1.1 there are $e_j\in \mathbb{N}$ with $u_j^{p^{e_j}}\in F$. If $r\ge 1$ put $e:=\max \{e_1,\dots ,e_r\}$ and otherwise put $e:=0$; set $q:=p^e$ and ${\gamma }_j:=u_j^q\in F$ for $j=1,\dots ,r$.

By step 1.2 each ${\gamma }_j$ is a fraction, so there are $h_j,k_j\in K[x_1,\dots ,x_d]$ with $k_j\mathrm{\ne }0$ and ${\gamma }_j=h_j\mathrm{/}{k}_j$. Let $S\subseteq K$ be the finite set of all coefficients occurring in the polynomials $h_1,k_1,\dots ,h_r,k_r$ and enumerate $S=\{c_1,\dots ,c_s\}$.

Put ${\mathrm{\Omega }}_0:=F$. For $i=1,\dots ,d$ choose, using [L7], a monic irreducible factor $g_i\in {\mathrm{\Omega }}_{i-1}[T]$ of $T^q-x_i$, set ${\mathrm{\Omega }}_i:={\mathrm{\Omega }}_{i-1}[T]\mathrm{/}(g_i)$ and $t_i:=T+(g_i)$; then ${\mathrm{\Omega }}_i$ is a field containing ${\mathrm{\Omega }}_{i-1}$ and $t_i^q=x_i$.

For $i=1,\dots ,s$ choose a monic irreducible factor $g_{d+i}\in {\mathrm{\Omega }}_{d+i-1}[T]$ of $T^q-c_i$, set ${\mathrm{\Omega }}_{d+i}:={\mathrm{\Omega }}_{d+i-1}[T]\mathrm{/}(g_{d+i})$ and ${\alpha }_i:=T+(g_{d+i})$. Then $\mathrm{\Omega }:={\mathrm{\Omega }}_{d+s}$ is a field containing $F$ and ${\alpha }_i^q=c_i$ for every $i$.

The subfield $K^{\mathrm{\prime }}:=K({\alpha }_1,\dots ,{\alpha }_s)$ of $\mathrm{\Omega }$ contains $K$ and is finite over $K$, and every element of $K^{\mathrm{\prime }}$ has its $p^{es}$-th power in $K$: each ${\alpha }_i$ is algebraic over the preceding field with a power basis of length at most $q$ by [L11], so raising an element of $K({\alpha }_1,\dots ,{\alpha }_i)$ to the $p^e$-th power uses [L6], ${\alpha }_i^q=c_i\in K$ and additivity of Frobenius to land in $K({\alpha }_1,\dots ,{\alpha }_{i-1})$, and $s$ such steps land in $K$; the same degree bounds give $[K^{\mathrm{\prime }}:K]\le q^s$ by [L12]. Hence $K^{\mathrm{\prime }}\mathrm{/}K$ is finite purely inseparable by [L1], [L12] and [L13].

Initial embedding: the inclusion ${\phi }_0:F\to \mathrm{\Omega }$, ${\phi }_0(z)=z$, is an injective field homomorphism fixing $F$ pointwise.

Moreover $\mathrm{\Omega }=K^{\mathrm{\prime }}(t_1,\dots ,t_d)$ by [L3], and $\mathrm{\Omega }$ has characteristic $p$ by [L13].

Inductive claim. Let $1\le j\le r$ and suppose that ${\phi }_{j-1}:F_{j-1}\to \mathrm{\Omega }$ is an injective field homomorphism fixing $F$ pointwise, where $F_{j-1}:=F(u_1,\dots ,u_{j-1})$. If $u_j\in F_{j-1}$, then $F_j=F_{j-1}$ and ${\phi }_{j-1}$ itself is the required extension.

In the remaining case $u_j\mathrm{\notin }{F}_{j-1}$ put $D_j:=\{n\in \mathbb{N}:u_j^{p^n}\in F_{j-1}\}$. This set is nonempty because $e_j\in D_j$ by step 2.1, so by [L8] it has a least element $d_j$; here $d_j\ge 1$ because $u_j\mathrm{\notin }{F}_{j-1}$, and $d_j\le e_j\le e$. Put ${\beta }_j:=u_j^{p^{d_j}}\in F_{j-1}$.

In the situation of step 7.1 the element ${\beta }_j$ is not a $p$th power in $F_{j-1}$: if ${\beta }_j=b^p$ with $b\in F_{j-1}$, then $(u_j^{p^{d_j-1}}{)}^p={\beta }_j=b^p$, so injectivity of Frobenius over $F_{j-1}$, available by [L6] and [L13], gives $u_j^{p^{d_j-1}}=b\in F_{j-1}$, contradicting the minimality of $d_j$. Hence $m_j(T):=T^{p^{d_j}}-{\beta }_j$ is monic irreducible over $F_{j-1}$ by [L9], and $m_j(u_j)=0$.

In the situation of step 7.1 define, inside $\mathrm{\Omega }$, the elements $H_j:={\sum }_a{\alpha }_{c_a}{t}^a,K_j:={\sum }_b{\alpha }_{d_b}{t}^b,$ where the sums run over the finitely many exponent vectors $a$ and $b$ occurring in $h_j$ and in $k_j$, with $t^a:=t_1^{a_1}\cdots t_d^{a_d}$ and likewise for $b$; then $H_j,K_j\in \mathrm{\Omega }$.

Frobenius in $\mathrm{\Omega }$, licit by step 6.1 and [L6], gives $H_j^q={\sum }_a{\alpha }_{c_a}^q{t}^{aq}={\sum }_a{c}_a(t^q{)}^a=h_j(x_1,\dots ,x_d)$ and likewise $K_j^q=k_j(x_1,\dots ,x_d)\mathrm{\ne }0$ by step 1.2, so $K_j\mathrm{\ne }0$ and the quotient $W_j:=H_j\mathrm{/}{K}_j\in \mathrm{\Omega }$ is defined and satisfies $W_j^q={\gamma }_j$.

Since ${\beta }_j=u_j^{p^{d_j}}$ and $q=p^e$ with $d_j\le e$ by step 7.1, one has ${\beta }_j^{p^{e-d_j}}=u_j^{p^e}={\gamma }_j$; since ${\phi }_{j-1}$ fixes $F$ pointwise and ${\gamma }_j\in F$ by step 2.1, step 9.1 gives ${\phi }_{j-1}({\beta }_j{)}^{p^{e-d_j}}={\phi }_{j-1}({\beta }_j^{p^{e-d_j}})={\phi }_{j-1}({\gamma }_j)={\gamma }_j=W_j^{p^e}=(W_j^{p^{d_j}}{)}^{p^{e-d_j}},$ and injectivity of the $(e-d_j)$-fold Frobenius power on $\mathrm{\Omega }$ gives ${\phi }_{j-1}({\beta }_j)=W_j^{p^{d_j}}$: that is, $W_j$ is a root in $\mathrm{\Omega }$ of the transported polynomial ${\sigma }_{\ast }(m_j)=T^{p^{d_j}}-{\phi }_{j-1}({\beta }_j)$, where $\sigma :={\phi }_{j-1}$ is regarded as an isomorphism $F_{j-1}\to {\phi }_{j-1}(F_{j-1})$.

Applying [L10] to $\sigma :={\phi }_{j-1}$, to the monic irreducible $m_j\in F_{j-1}[T]$ of step 8.1, to the root $u_j$ of $m_j$ in the extension $F_j\mathrm{/}{F}_{j-1}$, and to the root $W_j$ of ${\sigma }_{\ast }(m_j)$ in the extension $\mathrm{\Omega }\mathrm{/}{\phi }_{j-1}(F_{j-1})$, we obtain a field isomorphism $F_j\to {\phi }_{j-1}(F_{j-1})(W_j)\subseteq \mathrm{\Omega }$ extending ${\phi }_{j-1}$ and sending $u_j\mapsto W_j$; viewed as a map into $\mathrm{\Omega }$ it is an injective field homomorphism ${\phi }_j$ fixing $F$ pointwise.

Steps 5.2, 6.2 and 11.1 give, by induction on $j=0,1,\dots ,r$, injective field homomorphisms ${\phi }_j:F(u_1,\dots ,u_j)\to \mathrm{\Omega }$ fixing $F$ pointwise; in particular ${\phi }_r:L\to \mathrm{\Omega }$ is an embedding of $L$ over $F$.

By step 6.1 the field $\mathrm{\Omega }$ equals $K^{\mathrm{\prime }}(t_1,\dots ,t_d)$ with $t_i^q=x_i$; writing $x_i^{1\mathrm{/}q}:=t_i$, this subfield is $K^{\mathrm{\prime }}(x_1^{1\mathrm{/}q},\dots ,x_d^{1\mathrm{/}q})$, and $K^{\mathrm{\prime }}\mathrm{/}K$ is finite purely inseparable by step 5.1. Together with step 12.1 this exhibits the required embedding of $L$ over $F$ into $K^{\mathrm{\prime }}(x_1^{1\mathrm{/}q},\dots ,x_d^{1\mathrm{/}q})$, so the lemma is proved.

Let $R$ be a domain with the two properties

Let $R$ be a UFD, which by [L1] is a domain, and let $0\mathrm{\ne }f\in R[x]$ have nonzero coefficients $a_1,\dots ,a_s$. Write $\mathcal{P}$ for the set of associate classes of irreducibles of $R$. For $C\in \mathcal{P}$ and $0\mathrm{\ne }a\in R$, let $v_C(a)$ be the exponent of any representative of $C$ in a factorization of $a$; this is independent of the chosen representative and factorization by the uniqueness clause of [L1]. Set $v_C(f):={\min }_i{v}_C(a_i)$. Only finitely many classes have $v_C(f)>0$, because each $a_i$ has only finitely many irreducible factors. For each such class choose one representative $p_C$, and put $c(f):={\prod }_{C\in \mathcal{P},v_C(f)>0}{p}_C^{v_C(f)}\in R\setminus \{0\},f^{\ast }:=f\mathrm{/}c(f)\in R[x]\mathrm{.}$ The product is finite; its associate class does not depend on the representatives chosen. Each $v_C(f)$ is the exponent of $C$ in $c(f)$, so $c(f)$ divides every coefficient of $f$ and $f=c(f)f^{\ast }$. For every class $C$, some coefficient $a_i$ has $v_C(a_i)=v_C(f)$, so that coefficient of $f^{\ast }$ is not divisible by a representative of $C$. Thus no irreducible divides every coefficient of $f^{\ast }$, and $f^{\ast }$ is primitive. If also $0\mathrm{\ne }c\in R$ and $0\mathrm{\ne }h\in R[x]$ satisfies $f=ch$ with $h$ primitive, then for each $C$ the coefficientwise identity $v_C(ca)=v_C(c)+v_C(a)$ gives $v_C(f)=v_C(c)+v_C(h)=v_C(c)$; hence $c$ and $c(f)$ have the same exponent in every associate class and are associates by [L1, L3]. In particular a polynomial is primitive exactly when its content is a unit.

Let $R$ be an integral domain. Then $f\in R[x]$ is a unit of $R[x]$ if and only if $f$ is a constant and a unit of $R$; and if $p\in R$, then $p$ is irreducible in $R[x]$ if and only if $p$ is irreducible in $R$. Indeed, if $fg=1$ in $R[x]$ then $f,g\mathrm{\ne }0$ and [L10] gives $\mathrm{deg}f+\mathrm{deg}g=\mathrm{deg}1=0$, so $\mathrm{deg}f=\mathrm{deg}g=0$ and $fg=1$ holds in $R$; conversely units of $R$ are units of $R[x]$. The cases $p=0$ or $p$ a unit are excluded from irreducibility in both rings. For a nonzero nonunit $p$, if $p=fg$ with $\mathrm{deg}p=0$ then $\mathrm{deg}f=\mathrm{deg}g=0$ by [L10], so the factorization takes place in $R$, while a factorization in $R$ is one in $R[x]$.

Let $R$ be a domain with (P1) and (P2) of step 1.1. Then $R$ is integrally closed. Indeed let $z\in \mathrm{Frac}(R)$ be integral over $R$; if $z=0$ then $z\in R$, so assume $z\mathrm{\ne }0$. By [L7] there are $a,b\in R$ with $b\mathrm{\ne }0$ and $z=a\mathrm{/}b$. For $0\mathrm{\ne }c\in R$ let $m(c)$ be the number of irreducible factors in a factorization of $c$ when $c$ is a nonunit, and $0$ when $c$ is a unit; by (P1) and step 1.1 this number does not depend on the chosen factorization. Representations $z=a\mathrm{/}b$ exist, so by [L11] we may fix one for which $m(b)$ is least. Suppose $b$ were not a unit; then $b=q_1\cdots q_m$ with $m=m(b)\ge 1$ and all $q_i$ irreducible by (P1). By [L6] there is a monic equation $z^n+r_{n-1}{z}^{n-1}+\cdots +r_0=0$ with $n\ge 1$ and $r_i\in R$; multiplying by $b^n$ gives $a^n+r_{n-1}{a}^{n-1}b+\cdots +r_0{b}^n=0$, so $b\mid a^n$, hence $q_1\mid a^n$. Since $q_1$ is prime by (P2), iterating [L2] yields $q_1\mid a$. Writing $a=q_1{a}^{\mathrm{\prime }}$ and $b=q_1{b}^{\mathrm{\prime }}$ gives a new representation $z=a^{\mathrm{\prime }}\mathrm{/}{b}^{\mathrm{\prime }}$ whose denominator $b^{\mathrm{\prime }}=q_2\cdots q_m$ satisfies $m(b^{\mathrm{\prime }})=m-1$ by step 1.1, contradicting the minimality of $m(b)$. So $b$ is a unit, $z=a{b}^{-1}\in R$, and every element of $\mathrm{Frac}(R)$ integral over $R$ lies in $R$: by [L6], $R$ is integrally closed.

Let $R$ be a UFD and let $0\mathrm{\ne }f,g\in R[x]$. Then the contents satisfy $c(fg)\sim c(f)c(g)$ (associates). Indeed by step 1.2 both $f^{\ast }=f\mathrm{/}c(f)$ and $g^{\ast }=g\mathrm{/}c(g)$ are primitive, so $f^{\ast }{g}^{\ast }$ is primitive by [L4], and applying the uniqueness of contents from step 1.2 to the identity $fg=c(f)c(g)(f^{\ast }{g}^{\ast })$ shows that $c(fg)$ is associate to $c(f)c(g)$.

Let $R$ be a UFD, $K=\mathrm{Frac}(R)$, and let $p\in R[x]$ be irreducible of positive degree. Then $p$ is primitive and irreducible in $K[x]$. If some nonunit $d\in R$ divided every coefficient of $p$, then $p=d(p\mathrm{/}d)$ with $d$ a nonunit and $p\mathrm{/}d$ of positive degree, hence a nonunit of $R[x]$ by step 1.3: this contradicts irreducibility of $p$. So $p$ is primitive, and then [L4] applied to $p$ over the UFD $R$ makes $p$ irreducible in $K[x]$.

Let $R$ be a UFD with $K=\mathrm{Frac}(R)$, let $p\in R[x]$ be primitive, let $a\in R[x]$, and let $q\in K[x]$ satisfy $a=pq$. Then $q\in R[x]$. If $q=0$ this is immediate; otherwise $p$ and $a=pq$ are nonzero, so their contents are defined. Choose $0\mathrm{\ne }u\in R$ with $uq\in R[x]$, which is possible by [L7] applied to the finitely many nonzero coefficients of $q$. Applying step 2.2 in the ring $R[x]$ to $ua=p(uq)$ gives $c(ua)\sim c(p)c(uq)\sim c(uq)$, where we used that $p$ is primitive, so $c(p)\sim 1$ by step 1.2. On the other hand $c(ua)\sim uc(a)$ by the coefficientwise exponent identity of step 1.2. Hence $c(uq)\sim uc(a)$ is divisible by $u$, so $u$ divides every coefficient of $uq$; writing each coefficient of $uq$ as $ur$ with $r\in R$ and cancelling $u$ in $K$ shows that the corresponding coefficient of $q$ equals $r$. Therefore all coefficients of $q$ lie in $R$.

Let $R$ be a UFD and $0\mathrm{\ne }f\in R[x]$ a nonunit. Then $f$ is a product of irreducibles of $R[x]$. If $\mathrm{deg}f=0$ then $f\in R$ is a nonzero nonunit and [L1] factors it into irreducibles of $R$, each irreducible in $R[x]$ by step 1.3. Assume $\mathrm{deg}f\ge 1$ and write $f=c(f)f^{\ast }$ with $c(f)\in R$ and $f^{\ast }$ primitive by step 1.2; then $\mathrm{deg}{f}^{\ast }=\mathrm{deg}f\ge 1$, so $f^{\ast }$ is a nonunit of $R[x]$ by step 1.3. Retain $c(f)$ as a scalar (it may be a unit), and factor the nonzero nonunit $f^{\ast }$ in the UFD $K[x]$ of [L5] as $f^{\ast }=q_1\cdots q_n$ with each $q_j$ irreducible in $K[x]$ and $n\ge 1$. Each $q_j$ has positive degree, since a nonzero constant element of $K[x]$ is a unit there, and each $q_j$ is not a unit because $f^{\ast }$ is not. Choose $0\mathrm{\ne }{a}_j\in R$ with $a_j{q}_j\in R[x]$ and write $a_j{q}_j=d_j{h}_j$ with $d_j\in R$ and $h_j\in R[x]$ primitive, using step 1.2. Then $h_j=(a_j\mathrm{/}{d}_j)q_j$ is a nonzero $K$-multiple of the irreducible $q_j$, hence irreducible in $K[x]$, and it is primitive, so $h_j$ is irreducible in $R[x]$ by [L4]. By [L4] the product $P:=h_1\cdots h_n$ is primitive, and $f^{\ast }=\lambda P$ with $\lambda :={\prod }_j{d}_j\mathrm{/}{a}_j\in K^{\times }$. Choose $b\in R$ and $0\mathrm{\ne }c\in R$ with $\lambda =b\mathrm{/}c$, by [L7]. Then $c{f}^{\ast }=bP$ in $R[x]$, so step 2.2 and the content identity of step 1.2 give $cc(f^{\ast })\sim c(c{f}^{\ast })\sim c(bP)\sim bc(P)\sim b$, because $c(P)\sim 1$ by step 1.2; hence $c$ divides $b$ and $\lambda =b\mathrm{/}c$ lies in $R$. Therefore $f=c(f)\lambda h_1\cdots h_n$ exhibits $f$ as a product of irreducibles of $R[x]$, the nonzero scalar $c(f)\lambda \in R$ itself being a product of irreducibles if it is a nonunit, or being absorbed into $h_1$ if it is a unit; a unit multiple of an irreducible is irreducible.

Let $R$ be a UFD. Then every irreducible element $p$ of $R[x]$ is prime. If $\mathrm{deg}p=0$, then $p\in R$ is irreducible in $R$ by step 1.3. When $p\mid ab$ in $R[x]$, if $a=0$ or $b=0$ then $p$ divides that factor; otherwise both are nonzero, and every coefficient of $ab$ is divisible by $p$. For the associate class $C=[p]$, this gives $v_C(ab)\ge 1$, while step 2.2 gives $v_C(ab)=v_C(a)+v_C(b)$. Hence $v_C(a)\ge 1$ or $v_C(b)\ge 1$, which says exactly that $p$ divides every coefficient of $a$ or of $b$, so $p\mid a$ or $p\mid b$ in $R[x]$. If $\mathrm{deg}p\ge 1$, then $p$ is primitive and irreducible in $K[x]$ by step 2.3, hence prime in $K[x]$ by [L5]. If $p\mid ab$ in $R[x]$, then also $p\mid ab$ in $K[x]$, so $p\mid a$ or $p\mid b$ in $K[x]$; say $a=pq$ with $q\in K[x]$. Since $p$ is primitive, step 3.1 gives $q\in R[x]$, so $p\mid a$ in $R[x]$. Thus [L2] holds for $p$ in $R[x]$.