Finite-variable polynomial algebras over fields are integrally closed

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 $R[x]$.

Let $R$ be a UFD. Then $R[x]$ is a UFD in which every irreducible is prime: existence of factorizations into irreducibles is step 3.2, and primeness of irreducibles is step 4.1, so the uniqueness clause follows from step 1.1 with (P1) $=$ step 3.2 and (P2) $=$ step 4.1.

We prove by induction on $d$ that $K[x_1,\dots ,x_d]$ is a UFD in which every irreducible element is prime. For $d=0$ the ring is the field $K$ by [L8], a UFD in which there are no irreducible elements by [L13] and [L1]. For $d=1$ the ring is $K[x]$, a UFD by [L5] in which every irreducible is prime by [L5], each of these two cases being a base case. For the induction step, if $K[x_1,\dots ,x_d]$ is a UFD, then $K[x_1,\dots ,x_{d+1}]=K[x_1,\dots ,x_d][x_{d+1}]$ by [L8] is a UFD with prime irreducibles by step 5.1, so the property holds for every $d$.

Every ring $K[x_1,\dots ,x_d]$ is a domain by [L9], in the case $d=0$ by [L13]. It is integrally closed: for $d\ge 1$ it is a UFD with prime irreducibles by step 6.1, so it satisfies (P1) and (P2) of step 1.1 and step 2.1 makes it integrally closed; for $d=0$ the ring is the field $K$ by [L8], and every element of $\mathrm{Frac}(K)$ is a fraction $a\mathrm{/}b$ with $a,b\in K$, $b\mathrm{\ne }0$ by [L7], that is, the unit multiple $a{b}^{-1}$ of an element of $K$ by [L13], and each element of $K$ is a root of the monic polynomial $T-a\in K[T]$, so every element of $\mathrm{Frac}(K)$ integral over $K$ lies in $K$.