Polynomial rings over normal domains are normal

We first prove the field case. Let $K$ be a field and let $z\in K(x)=\mathrm{Frac}(K[x])$ be integral over $K[x]$; here $K[x]$ is a domain by [L4], so [L2] lets us write $z=f\mathrm{/}g$ with $f,g\in K[x]$ and $g\mathrm{\ne }0$. If $f=0$ then $z=0\in K[x]$. Otherwise $f,g$ are not both zero, so [L8] provides $A,B\in K[x]$ and the monic greatest common divisor $d=\gcd (f,g)$ with $d=Af+Bg$, $d\mid f$ and $d\mid g$. Write $f=d{f}_1$ and $g=d{g}_1$ with $f_1,g_1\in K[x]$. The polynomial $d$ is monic, hence nonzero, so cancellation in the domain $K[x]$ of [L4] gives $A{f}_1+B{g}_1=1$ from $d(A{f}_1+B{g}_1)=d\cdot 1$, and the same cancellation in $K(x)$ gives $z=f_1\mathrm{/}{g}_1$.

Let $D$ be a domain with fraction field $L$. Call an element $u\in L$ almost integral over $D$ when there is a nonzero $d\in D$ with $d{u}^n\in D$ for every integer $n\ge 0$. Every element of $D$ is almost integral over $D$ with multiplier $1$, the fraction field $L$ of $D$ being that of [L2] and $D$ being a domain in the sense of [L7]. This is a convention internal to the proof; the following steps establish the three properties of it that the argument uses.

Every ideal $I$ of $\mathbb{Z}$ is an additive subgroup of $(\mathbb{Z},+)$, hence equals $n\mathbb{Z}=(n)$ for some natural number $n$ by [L12] and [L11]. So every ideal of $\mathbb{Z}$ is finitely generated and $\mathbb{Z}$ is Noetherian by [L10].

Now let $C$ be a commutative ring in which every ideal is finitely generated, and let $I\subseteq C[x]$ be an ideal. For $n\ge 0$ let $S_n$ be the set consisting of $0$ and of the leading coefficients of all elements of $I$ of degree $\le n$ [L5], and put $J_n=(S_n)\subseteq C$ [L11]; then $J_0\subseteq J_1\subseteq \cdots$, since $S_n\subseteq S_{n+1}$, and $J={\bigcup }_{n\ge 0}{J}_n$ is an ideal of $C$. By hypothesis $J=(g_1,\dots ,g_m)$ for finitely many $g_i\in C$ [L11], and each $g_i\in J_{n_i}$ for some $n_i$ because $J$ is the union of the $J_n$; with $N=\max \{n_1,\dots ,n_m\}$ this gives $J=J_N$. For each $n\le N$ the ideal $J_n$ is finitely generated and is generated by $S_n$, so finitely many elements of $S_n$, say the leading coefficients of polynomials $p_{n,1},\dots ,p_{n,k_n}\in I$ of degree $\le n$, generate $J_n$ [L11]. Let $W$ be the finite set of all these polynomials for $n\le N$.

Let $C$ be a commutative ring in which every ideal is finitely generated, let $I\subseteq C$ be an ideal and let $K\subseteq C\mathrm{/}I$ be an ideal. By [L14] the preimage $J={\pi }^{-1}(K)$ is an ideal of $C$ containing $I$ and $J\mathrm{/}I=K$, and $J$ is finitely generated, say $J=(c_1,\dots ,c_k)$; by [L13] the projection $\pi$ is surjective, so $K=\pi (J)$ is generated by the images $\pi (c_1),\dots ,\pi (c_k)$ [L11]. Hence every ideal of $C\mathrm{/}I$ is finitely generated.

Let $z^n+c_{n-1}{z}^{n-1}+\cdots +c_0=0$ with $n\ge 1$ and $c_j\in K[x]$ be a monic equation for $z$ over $K[x]$, which exists because $z$ is integral over $K[x]$. Substituting $z=f_1\mathrm{/}{g}_1$ and multiplying by $g_1^n$ gives $f_1^n=-(c_{n-1}{f}_1^{n-1}{g}_1+\cdots +c_0{g}_1^n)$, so $f_1^n\in (g_1)$ by [L11]. Raising $A{f}_1+B{g}_1=1$ to the $n$-th power and expanding by the binomial theorem, every term of the expansion contains a factor $f_1^n$ or a factor $g_1$, so $1=A^{\mathrm{\prime }}{f}_1^n+B^{\mathrm{\prime }}{g}_1$ for suitable $A^{\mathrm{\prime }},B^{\mathrm{\prime }}\in K[x]$; by [L11] both summands lie in the ideal $(g_1)$, so $1\in (g_1)$, say $g_{1}w=1$ with $w\in K[x]$. Then $z=f_1\mathrm{/}{g}_1=f_{1}w\in K[x]$. Hence every element of $K(x)$ integral over $K[x]$ lies in $K[x]$: for every field $K$ the ring $K[x]$ is integrally closed in its fraction field.

Let $u,v\in L$ be almost integral over $D$ with multipliers $d,e\in D\setminus \{0\}$. Then $de\mathrm{\ne }0$ by [L7], and for every $n\ge 0$ one has $(de)(uv{)}^n=(d{u}^n)(e{v}^n)\in D$ and $(de)(u+v{)}^n={\sum }_{j=0}^n\left(\genfrac{}{}{0px}{}{n}{j}\right)(d{u}^j)(e{v}^{n-j})\in D$, each summand being a product of two elements of $D$. So the almost integral elements of $L$ form a subring of $L$ containing $D$.

Let $u\in L$ be integral over $D$. By [L9] the ring $D[u]$ is a finitely generated $D$-module; choose generators $h_1,\dots ,h_N\in D[u]$ and write $h_i=a_i\mathrm{/}{d}_i$ with $a_i,d_i\in D$ and $d_i\mathrm{\ne }0$ by [L2]. Then $d=d_1\cdots d_N\mathrm{\ne }0$ by [L7] and $d{h}_i=a_i{\prod }_{j\mathrm{\ne }i}{d}_j\in D$ for every $i$, so $d\cdot D[u]\subseteq D$; in particular $d{u}^n\in D$ for all $n\ge 0$. Hence every element of $L$ integral over $D$ is almost integral over $D$.

Suppose every ideal of $D$ is finitely generated and $u\in L$ is almost integral over $D$ with multiplier $d\mathrm{\ne }0$. Then $D[u]\subseteq d^{-1}D=\{c\mathrm{/}d:c\in D\}$, since $d{u}^n\in D$ for every $n\ge 0$ and $d^{-1}D$ is a $D$-submodule of $L$. Multiplication by $d$ is an injective $D$-module map $D[u]\to D$, because $L$ is a field and $d\mathrm{\ne }0$, and its image $dD[u]$ is an ideal of $D$; by hypothesis $dD[u]=(y_1,\dots ,y_k)$ for some $y_j\in D$ [L11]. Then $D[u]=(d^{-1}{y}_1,\dots ,d^{-1}{y}_k)$ inside $L$: each $d^{-1}{y}_j$ lies in $D[u]$ because $y_j=d{x}_j$ for some $x_j\in D[u]$ and cancellation in $L$ gives $d^{-1}{y}_j=x_j$, while for $x\in D[u]$ the identity $dx={\sum }_j{c}_j{y}_j$ with $c_j\in D$ gives $x={\sum }_j{c}_j(d^{-1}{y}_j)$. So $D[u]$ is a finitely generated $D$-module and [L9] makes $u$ integral over $D$. Consequently, over a domain in which every ideal is finitely generated, almost integral and integral elements of the fraction field coincide.

We show that the finite set $W$ generates $I$. Let $f\in I$ be nonzero of degree $m$ and leading coefficient $b\mathrm{\ne }0$ [L5], and put $m^{\ast }=\min \{m,N\}$. The leading coefficient $b$ lies in $J_m$, and $J_m=J_{m^{\ast }}$: for $m\le N$ this is $m^{\ast }=m$, and for $m>N$ it holds because $J_N\subseteq J_m\subseteq J=J_N$. So $b={\sum }_j{r}_j\mathrm{lc}(p_{m^{\ast },j})$ with $r_j\in C$ [L11]. Every $p_{m^{\ast },j}$ has degree $\le m^{\ast }\le m$, and the polynomial $f-{\sum }_j{r}_j{x}^{m-\mathrm{deg}{p}_{m^{\ast },j}}{p}_{m^{\ast },j}$ lies in $I$, is congruent to $f$ modulo $(W)$, and has degree $<m$ by [L6]. Induction on $m$ therefore exhibits every element of $I$ as an element of $(W)$; hence $I$ is finitely generated, and every ideal of $C[x]$ is finitely generated.

Let $C$ be a commutative ring in which every ideal is finitely generated, let $A$ be a commutative $C$-algebra and let $t_1,\dots ,t_d\in A$. The $C$-subalgebra of $A$ generated by $t_1,\dots ,t_d$ is the image of the evaluation homomorphism $C[x_1,\dots ,x_d]\to A$ with $x_i\mapsto t_i$, hence is a quotient of $C[x_1,\dots ,x_d]$; iterating step 1.4 and applying step 1.5, every ideal of it is finitely generated. In particular, taking $C=\mathbb{Z}$ and using step 1.3, every $\mathbb{Z}$-subalgebra of a commutative ring that is generated by finitely many elements is Noetherian by [L10].

Let $D$ be a domain with fraction field $K_0$ and let $f={\alpha }_0+{\alpha }_{1}x+\cdots +{\alpha }_r{x}^r\in K_0[x]$ with ${\alpha }_r\mathrm{\ne }0$ be almost integral over $D[x]$, with multiplier $h=b_0+b_{1}x+\cdots +b_s{x}^s\in D[x]\setminus \{0\}$ and $b_s\mathrm{\ne }0$. For every $n\ge 0$ the coefficient of $x^{rn+s}$ in the product $h{f}^n$ is $b_s{\alpha }_r^n$ by [L6], and $h{f}^n\in D[x]$, so $b_s{\alpha }_r^n\in D$; thus ${\alpha }_r$ is almost integral over $D$ with multiplier $b_s$. Then ${\alpha }_r{x}^r$ is almost integral over $D[x]$ with the same multiplier, because $b_s({\alpha }_r{x}^r{)}^n=(b_s{\alpha }_r^n)x^{rn}\in D[x]$, and by step 2.2 the difference $f-{\alpha }_r{x}^r\in K_0[x]$, which has degree $<r$, is almost integral over $D[x]$ as well. Induction on $r$ and on the degree of $f$ therefore shows that every coefficient ${\alpha }_i$ of $f$ is almost integral over $D$.

Now return to the given data: $R$ is an integrally closed domain with fraction field $K$ [L1], and $z\in \mathrm{Frac}(R[x])$ is integral over $R[x]$. The rings $R[x]\subseteq K[x]$ are domains by [L4], $K(x)=\mathrm{Frac}(K[x])$ is a field, and the inclusion $R[x]\to K(x)$ carries every nonzero element of $R[x]$ to a unit, so by [L3] it extends uniquely to a unital ring homomorphism $\mathrm{Frac}(R[x])\to K(x)$, $f\mathrm{/}g\mapsto f\mathrm{/}g$; this map is injective because its restriction to the domain $R[x]$ is injective. We therefore regard $z$ as an element of $K(x)$. The monic equation for $z$ over $R[x]$ is in particular a monic equation over $K[x]$, so $z$ is integral over $K[x]$, and step 2.1 applied to the field $K$ gives $z\in K[x]$. Write $z={\alpha }_0+{\alpha }_{1}x+\cdots +{\alpha }_r{x}^r$ with ${\alpha }_i\in K$.

Choose a monic equation $z^n+c_{n-1}{z}^{n-1}+\cdots +c_0=0$ with all $c_j\in R[x]$, and write each ${\alpha }_i=a_i\mathrm{/}{b}_i$ with $a_i,b_i\in R$ and $b_i\mathrm{\ne }0$ by [L2]. Let $R_0\subseteq R$ be the $\mathbb{Z}$-subalgebra generated by the finitely many coefficients of the polynomials $c_j$ together with all the elements $a_i,b_i$. Then $R_0$ is a subring of $R$ containing $1$, hence a domain by [L15] and [L7], and every ideal of $R_0$ is finitely generated by step 2.6 applied with $C=\mathbb{Z}$ and step 1.3. Moreover ${\alpha }_i=a_i\mathrm{/}{b}_i$ with $a_i,b_i\in R_0$ and $b_i\mathrm{\ne }0$, so ${\alpha }_i\in K_0:=\mathrm{Frac}(R_0)\subseteq K$ [L2], and $z\in K_0[x]$; since all coefficients of the $c_j$ lie in $R_0$, the displayed monic equation has coefficients in $R_0[x]$. Thus $z\in K_0[x]$ is integral over $R_0[x]$.

Let $D$ be a domain in which every ideal is finitely generated, put $K_0=\mathrm{Frac}(D)$, and let $f\in K_0[x]$ be integral over $D[x]$. Then every coefficient of $f$ is integral over $D$: the ring $D[x]$ is a domain by [L4] and $f$ lies in its fraction field $K_0(x)=\mathrm{Frac}(K_0[x])$, so step 2.3 makes $f$ almost integral over $D[x]$; step 3.1 makes each coefficient of $f$ almost integral over $D$; and step 2.4 turns almost integrality over $D$ into integrality over $D$.

By step 4.1 applied to the domain $R_0$, in which every ideal is finitely generated, each coefficient ${\alpha }_i$ of $z$ is integral over $R_0$. The monic polynomial over $R_0\subseteq R$ witnessing this has coefficients in $R$, so each ${\alpha }_i$ is integral over $R$; since $R$ is integrally closed in $K=\mathrm{Frac}(R)$ [L1], each ${\alpha }_i$ lies in $R$.

Consequently $z={\alpha }_0+{\alpha }_{1}x+\cdots +{\alpha }_r{x}^r\in R[x]$. Since the homomorphism of step 3.2 is injective and restricts to the identity on $R[x]$, every element of $\mathrm{Frac}(R[x])$ integral over $R[x]$ is already an element of $R[x]$: the polynomial ring $R[x]$ is an integrally closed domain by [L1] and [L4], for every integrally closed domain $R$ and with no Noetherian hypothesis on $R$. ∎