Finite algebras over a strongly transcendental variable are nowhere quasi-finite

By the hypothesis of module-finiteness there are $t_1,\dots ,t_m\in S$ with $S=R[x]t_1+\cdots +R[x]t_m$. Hence $S=R[x,t_1,\dots ,t_m]$ is generated as an $R$-algebra by the finitely many elements $x,t_1,\dots ,t_m$ and is of finite type over $R$ by [L2]; consequently the notion of quasi-finiteness of [L1] is defined for $R\to S$, and it remains to show that no prime of $S$ satisfies it.

We first treat the case in which $R$ and $S$ are domains; here, by [L5], the strong transcendence of $x$ over $R$ says exactly that $x$ is transcendental over $K=\mathrm{Frac}(R)$. Let $\mathfrak{q}\in \mathrm{Spec}(S)$, put $\mathfrak{p}=\mathfrak{q}\cap R$ and $\mathfrak{r}=\mathfrak{q}\cap R[x]$, and suppose for contradiction that $R\to S$ is quasi-finite at $\mathfrak{q}$.

Now let $R\subseteq S$ be reduced and let $\mathfrak{q}\in \mathrm{Spec}(S)$. By [L25] the localisation $S_{\mathfrak{q}}=(S\setminus \mathfrak{q}{)}^{-1}S$ is nonzero, because $0\in \mathfrak{q}$ excludes $0\in S\setminus \mathfrak{q}$ and $S_{\mathfrak{q}}=0$ would force $0\in S\setminus \mathfrak{q}$ by [L24]; so the zero ideal is a proper ideal of $S_{\mathfrak{q}}$, and by [L23] there is a prime $P$ of $S_{\mathfrak{q}}$ minimal over $(0)$. By [L22] every prime of $S_{\mathfrak{q}}$ is the extension of a prime of $S$ contained in $\mathfrak{q}$, so $P={\mathfrak{q}}_0{S}_{\mathfrak{q}}$ for the prime ${\mathfrak{q}}_0=P\cap S\subseteq \mathfrak{q}$, and the inclusion-preserving bijection shows that ${\mathfrak{q}}_0$ is a minimal prime of $S$.

First consider the normal-domain case, so assume here that $R$ is a domain and integrally closed in $K=\mathrm{Frac}(R)$. Then $R[x]$ is an integral domain by [L7] and is integrally closed by [L8] and [L9]. Moreover $R[x]\subseteq S$ is an integral extension: every element of the module-finite $R[x]$-algebra $S$ is integral over $R[x]$ by [L10].

By [L1] the algebra $S_{\mathfrak{q}}\mathrm{/}\mathfrak{p}{S}_{\mathfrak{q}}$ is finite over $\kappa (\mathfrak{p})$, and by [L3] the residue field $\kappa (\mathfrak{q})$ is $S_{\mathfrak{q}}\mathrm{/}\mathfrak{q}{S}_{\mathfrak{q}}$, which is a quotient of $S_{\mathfrak{q}}\mathrm{/}\mathfrak{p}{S}_{\mathfrak{q}}$ because $\mathfrak{p}\subseteq \mathfrak{q}$. Hence $\kappa (\mathfrak{q})$ is finite over $\kappa (\mathfrak{p})$.

For the general domain case drop the normality of $R$: let $\bar{R}$ be the integral closure of $R$ in $K=\mathrm{Frac}(R)$ as in [L9], which is a subring of $K$ by [L13] and is integrally closed by [L9]; put $L=\mathrm{Frac}(S)$ and let $\bar{S}=\bar{R}[S]$ be the subring of $L$ generated by $\bar{R}$ and $S$, a domain because it is a subring of the field $L$.

By [L21] the image $\bar{x}$ of $x$ in $S\mathrm{/}{\mathfrak{q}}_0$ is strongly transcendental over $R\mathrm{/}{\mathfrak{p}}_0$, where ${\mathfrak{p}}_0={\mathfrak{q}}_0\cap R$. Both $R\mathrm{/}{\mathfrak{p}}_0$ and $S\mathrm{/}{\mathfrak{q}}_0$ are domains by [L4], so by [L5] the element $\bar{x}$ is transcendental over $\kappa ({\mathfrak{p}}_0)=\mathrm{Frac}(R\mathrm{/}{\mathfrak{p}}_0)$.

The quotient $S\mathrm{/}{\mathfrak{q}}_0$ is module-finite over $(R\mathrm{/}{\mathfrak{p}}_0)[\bar{x}]$: it is the image of the module-finite $R[x]$-algebra $S$ under the quotient map, and the image of $R[x]$ is the subalgebra generated by $\bar{x}$ over $R\mathrm{/}{\mathfrak{p}}_0$.

The kernel of $R[x]\to S\mathrm{/}\mathfrak{q}$ is exactly $\mathfrak{r}$, so $R[x]\mathrm{/}\mathfrak{r}$ is a subring of the domain $S\mathrm{/}\mathfrak{q}$ by [L4]; by [L3] the residue fields are $\kappa (\mathfrak{r})=\mathrm{Frac}(R[x]\mathrm{/}\mathfrak{r})$ and $\kappa (\mathfrak{q})=\mathrm{Frac}(S\mathrm{/}\mathfrak{q})$, and by [L12] the inclusion of domains extends to an injection $\kappa (\mathfrak{r})\to \kappa (\mathfrak{q})$ of fields carrying the image of $\kappa (\mathfrak{p})$ onto a subfield contained in $\kappa (\mathfrak{r})$. Since a $\kappa (\mathfrak{p})$-linearly independent subset of $\kappa (\mathfrak{r})$ is also $\kappa (\mathfrak{p})$-linearly independent in $\kappa (\mathfrak{q})$, step 2.2 shows that $\kappa (\mathfrak{r})$ is finite over $\kappa (\mathfrak{p})$.

Since $R\subseteq \bar{R}\subseteq K$ and $K=\mathrm{Frac}(R)$, their fraction fields satisfy $K\subseteq \mathrm{Frac}(\bar{R})\subseteq K$, hence $\mathrm{Frac}(\bar{R})=K$. The element $x$ is transcendental over $K$ by [L5]: if a nonzero polynomial over $K$ vanished at $x$, clearing its finitely many denominators would give a nonzero $h\in R[X]$ with $h(x)=0$, and strong transcendence with multiplier $1$ would force every coefficient of $h$ to be zero, a contradiction. Hence $x$ is transcendental over $\mathrm{Frac}(\bar{R})$ as well.

The algebra $\bar{S}$ is integral over $S$: by [L13] the set ${\mathrm{Int}}_S(\bar{S})$ is a subring of $\bar{S}$ containing $S$, and it contains every element of $\bar{R}$, which is integral over $R$ by [L9] and hence over $S$ by the same monic equation; being a subring containing $S$ and $\bar{R}$, it contains the subring they generate, namely $\bar{S}$, so $\bar{S}={\mathrm{Int}}_S(\bar{S})$ is integral over $S$.

The algebra $\bar{S}$ is module-finite over $\bar{R}[x]$: from $S=R[x]t_1+\cdots +R[x]t_m$ of step 1.1 we get $\bar{S}=\bar{R}[x][t_1,\dots ,t_m]$, and each $t_j$ is integral over $R[x]$ by module-finiteness and [L10], hence over $\bar{R}[x]\supseteq R[x]$ by the same monic equation, so [L14] makes $\bar{S}$ module-finite over $\bar{R}[x]$.

Consequently $\mathfrak{r}\mathrm{\ne }\mathfrak{p}R[x]$: if equality held, then $R[x]\mathrm{/}\mathfrak{r}=(R\mathrm{/}\mathfrak{p})[x]$ and $\kappa (\mathfrak{r})=\kappa (\mathfrak{p})(\bar{x})$ where $\bar{x}$ is the class of $x$. Then $\bar{x}$ is transcendental over $\kappa (\mathfrak{p})$: if $0\mathrm{\ne }h\in \kappa (\mathfrak{p})[X]$ satisfied $h(\bar{x})=0$, then writing $h=d^{-1}g$ with $0\mathrm{\ne }d\in R\mathrm{/}\mathfrak{p}$ and $g\in (R\mathrm{/}\mathfrak{p})[X]$ would give $g(\bar{x})=0$, that is $g=0$ because evaluation at the indeterminate is the identity by [L6], contradicting $g\mathrm{\ne }0$ in the domain $(R\mathrm{/}\mathfrak{p})[x]$ by [L7]. But then the powers $1,\bar{x},\dots ,{\bar{x}}^d$ for $d={\dim }_{\kappa (\mathfrak{p})}\kappa (\mathfrak{r})<\mathrm{\infty }$ would be $\kappa (\mathfrak{p})$-linearly dependent, yielding a nonzero polynomial relation over $\kappa (\mathfrak{p})$ of degree $\le d$ satisfied by $\bar{x}$, a contradiction.

Since $\mathfrak{p}R[x]⊊\mathfrak{r}$ and $\mathfrak{q}$ lies over $\mathfrak{r}$ along the integral extension $R[x]\subseteq S$ of step 2.1, going down [L11] provides a prime ${\mathfrak{q}}^{\mathrm{\prime }}\subseteq \mathfrak{q}$ of $S$ with ${\mathfrak{q}}^{\mathrm{\prime }}\cap R[x]=\mathfrak{p}R[x]$; then ${\mathfrak{q}}^{\mathrm{\prime }}\mathrm{\ne }\mathfrak{q}$, and ${\mathfrak{q}}^{\mathrm{\prime }}\cap R=({\mathfrak{q}}^{\mathrm{\prime }}\cap R[x])\cap R=\mathfrak{p}R[x]\cap R=\mathfrak{p}$.

The prime ideals ${\mathfrak{q}}^{\mathrm{\prime }}{S}_{\mathfrak{q}}\subseteq \mathfrak{q}{S}_{\mathfrak{q}}$ of $S_{\mathfrak{q}}$ are distinct and both contain $\mathfrak{p}{S}_{\mathfrak{q}}$ by step 5.1, so $A:=S_{\mathfrak{q}}\mathrm{/}\mathfrak{p}{S}_{\mathfrak{q}}$ carries a strict chain of primes; $A$ is finite-dimensional over $\kappa (\mathfrak{p})$ by [L1]. But every prime $P$ of a finite-dimensional algebra $A$ over a field $\kappa (\mathfrak{p})$ is maximal: $A\mathrm{/}P$ is a domain by [L4], finite-dimensional over $\kappa (\mathfrak{p})$, so for $0\mathrm{\ne }u\in A\mathrm{/}P$ the $\kappa (\mathfrak{p})$-linear map $a\mapsto ua$ is injective by [L4] and hence surjective by finite dimension, which makes $u$ a unit and $A\mathrm{/}P$ a field. This contradiction shows that, when $R,S$ are domains and $R$ is integrally closed, $R\to S$ is quasi-finite at no prime of $S$, and then also when $R,S$ are any domains, as the next steps show.

Applying the result of step 6.1 to the domains $\bar{R}\subseteq \bar{S}$, where $\bar{R}$ is integrally closed by step 2.3 and $x$ is transcendental over $\mathrm{Frac}(\bar{R})$ by step 3.2 while $\bar{S}$ is module-finite over $\bar{R}[x]$ by step 3.4, we obtain: $\bar{R}\to \bar{S}$ is quasi-finite at no prime of $\bar{S}$.

Now suppose that $R\to S$ is quasi-finite at a prime $\mathfrak{q}\in \mathrm{Spec}(S)$. By [L17] applied to the integral extension $S\subseteq \bar{S}$ of step 3.3 there is $\bar{\mathfrak{q}}\in \mathrm{Spec}(\bar{S})$ with $\bar{\mathfrak{q}}\cap S=\mathfrak{q}$. The $R$-algebra homomorphism $S{\otimes }_R\bar{R}\to \bar{S}$ of [L19] sending $s\otimes r$ to $sr$ is surjective because its image is a subring of $\bar{S}$ containing the images of $S$ and of $\bar{R}$; with $J$ its kernel we have $\bar{S}\cong (S{\otimes }_R\bar{R})\mathrm{/}J$, and by [L18] the prime $\bar{\mathfrak{q}}$ corresponds to a prime $\tilde{\mathfrak{q}}$ of $S{\otimes }_R\bar{R}$ mapping to $\mathfrak{q}$ in $S$. By [L20] the base change $\bar{R}\to S{\otimes }_R\bar{R}$ is quasi-finite at $\tilde{\mathfrak{q}}$, and by [L20] again, applied to the quotient $\bar{S}$, the map $\bar{R}\to \bar{S}$ is quasi-finite at $\bar{\mathfrak{q}}$, contradicting step 7.1. Hence in the domain case $R\to S$ is quasi-finite at no prime of $S$.

By step 8.1 applied to the domains $R\mathrm{/}{\mathfrak{p}}_0\subseteq S\mathrm{/}{\mathfrak{q}}_0$ with the element $\bar{x}$ of step 2.4, the map $R\mathrm{/}{\mathfrak{p}}_0\to S\mathrm{/}{\mathfrak{q}}_0$ is quasi-finite at no prime of $S\mathrm{/}{\mathfrak{q}}_0$; in particular it is not quasi-finite at the prime $\mathfrak{q}\mathrm{/}{\mathfrak{q}}_0$ of $S\mathrm{/}{\mathfrak{q}}_0$, which exists by [L18] because ${\mathfrak{q}}_0\subseteq \mathfrak{q}$ and ${\mathfrak{q}}_0$ is prime.

If $R\to S$ were quasi-finite at $\mathfrak{q}$, then by [L20] applied to the ideal ${\mathfrak{q}}_0\subseteq \mathfrak{q}$ the quotient map $R\to S\mathrm{/}{\mathfrak{q}}_0$ would be quasi-finite at $\mathfrak{q}\mathrm{/}{\mathfrak{q}}_0$, contradicting step 9.1: the actions factor through $R\mathrm{/}{\mathfrak{p}}_0$, and $\kappa (\mathfrak{p}\mathrm{/}{\mathfrak{p}}_0)=\kappa (\mathfrak{p})$, so these two base-ring descriptions give the same local fibre. Hence $R\to S$ is not quasi-finite at $\mathfrak{q}$; since $\mathfrak{q}\in \mathrm{Spec}(S)$ was arbitrary, $R\to S$ is quasi-finite at no prime of $S$.

Steps 1.1 and 10.1 prove the Statement: $R\to S$ is of finite type whenever $S$ is module-finite over $R[x]$, and it is quasi-finite at no prime of $S$ for reduced $R\subseteq S$ and strongly transcendental $x$. The Axiom of Choice was assumed in the Statement and used exactly through [L11] in step 5.1, [L17] in step 8.1, [L21] in step 2.4 and [L23] in step 1.3; all other steps are choice-free. ∎