Quasi-finite local fibres transfer through quotients and intermediate rings

Let $T$ be an intermediate $R$-subalgebra, so that the given map factors as $R\to T\to S$, and let $\mathfrak{r}=\mathfrak{q}\cap T$. By [L2] the $R$-algebra $S$ is generated by finitely many elements $s_1,\dots ,s_n\in S$; these same elements generate $S$ as a $T$-algebra, so $T\to S$ is of finite type by [L2]. Moreover $\mathfrak{r}$ is a prime of $T$ and $\mathfrak{p}=\mathfrak{r}\cap R$, since $\mathfrak{p}=\mathfrak{q}\cap R=(\mathfrak{q}\cap T)\cap R$.

Now assume in addition that $T$ is of finite type over $R$, that $u\in T\setminus \mathfrak{q}$, and that $T_u=S_u$; set $\mathfrak{r}=\mathfrak{q}\cap T$, so that $u\mathrm{\notin }\mathfrak{r}$ and $\mathfrak{p}=\mathfrak{r}\cap R$. Both $T_{\mathfrak{r}}$ and $S_{\mathfrak{q}}$ are localisations of the common ring $A:=T_u=S_u$: the former is $A$ localised at the multiplicative subset generated by the image of $T\setminus \mathfrak{r}$, the latter at the multiplicative subset generated by the image of $S\setminus \mathfrak{q}$, and every element of $S_u$ is a fraction $a\mathrm{/}{u}^k$ with $a\in T$ by [L3].

Now let $R\to R^{\mathrm{\prime }}$ be a ring map, put $S^{\mathrm{\prime }}=S{\otimes }_R{R}^{\mathrm{\prime }}$ and let ${\mathfrak{q}}^{\mathrm{\prime }}\in \mathrm{Spec}(S^{\mathrm{\prime }})$ lie over $\mathfrak{q}$; set ${\mathfrak{p}}^{\mathrm{\prime }}={\mathfrak{q}}^{\mathrm{\prime }}\cap R^{\mathrm{\prime }}$, so that ${\mathfrak{p}}^{\mathrm{\prime }}\cap R={\mathfrak{q}}^{\mathrm{\prime }}\cap R=\mathfrak{q}\cap R=\mathfrak{p}$. By [L2] write $S=R[s_1,\dots ,s_n]$; then $S^{\mathrm{\prime }}$ is generated as an $R^{\mathrm{\prime }}$-algebra by the images of $s_1,\dots ,s_n$, because $S$ is a quotient of a polynomial ring $R[x_1,\dots ,x_n]$ and tensoring the quotient presentation with $R^{\mathrm{\prime }}$ over $R$ gives a quotient presentation of $S^{\mathrm{\prime }}$ over $R^{\mathrm{\prime }}$ by [L5]. In particular $R^{\mathrm{\prime }}\to S^{\mathrm{\prime }}$ is of finite type by [L2].

Put $E=S_{\mathfrak{q}}\mathrm{/}\mathfrak{p}{S}_{\mathfrak{q}}$. By hypothesis and [L1], $E$ is finite-dimensional over $\kappa (\mathfrak{p})$; choose a basis $x_1,\dots ,x_d$.

Finally let $J\subseteq \mathfrak{q}$ be an ideal of $S$, put $\bar{S}=S\mathrm{/}J$ and let $\bar{\mathfrak{q}}=\mathfrak{q}\mathrm{/}J$. By [L6] the ideal $\bar{\mathfrak{q}}$ is a prime of $\bar{S}$ with $\bar{\mathfrak{q}}\cap R=\mathfrak{p}$; $\bar{S}$ is a quotient of the finite-type $R$-algebra $S$, hence of finite type over $R$ by [L2]. By [L4] there is a canonical isomorphism ${\bar{S}}_{\bar{\mathfrak{q}}}\cong S_{\mathfrak{q}}\mathrm{/}J{S}_{\mathfrak{q}}$ identifying the extensions of $\mathfrak{p}$, so ${\bar{S}}_{\bar{\mathfrak{q}}}\mathrm{/}\mathfrak{p}{\bar{S}}_{\bar{\mathfrak{q}}}\cong S_{\mathfrak{q}}\mathrm{/}(\mathfrak{p}{S}_{\mathfrak{q}}+J{S}_{\mathfrak{q}})$ is a quotient of $S_{\mathfrak{q}}\mathrm{/}\mathfrak{p}{S}_{\mathfrak{q}}$.

The inclusion $\mathfrak{p}\subseteq \mathfrak{r}$ gives $\mathfrak{p}{S}_{\mathfrak{q}}\subseteq \mathfrak{r}{S}_{\mathfrak{q}}$, so the quotient map $S_{\mathfrak{q}}\mathrm{/}\mathfrak{p}{S}_{\mathfrak{q}}\to S_{\mathfrak{q}}\mathrm{/}\mathfrak{r}{S}_{\mathfrak{q}}$ is a surjective $\kappa (\mathfrak{p})$-algebra homomorphism. By hypothesis and [L1] the algebra $S_{\mathfrak{q}}\mathrm{/}\mathfrak{p}{S}_{\mathfrak{q}}$ is finite over $\kappa (\mathfrak{p})$, hence so is its quotient $S_{\mathfrak{q}}\mathrm{/}\mathfrak{r}{S}_{\mathfrak{q}}$.

The homomorphism $T_{\mathfrak{r}}\to S_{\mathfrak{q}}$ induced by the inclusion $T\subseteq S$ is injective. Indeed, let $a\in T$ and $s\in T\setminus \mathfrak{r}$ with $a\mathrm{/}s$ mapping to $0$ in $S_{\mathfrak{q}}$; by [L3] there is $\sigma \in S\setminus \mathfrak{q}$ with $\sigma a=0$ in $S$. By [L3] again write $\sigma =t\mathrm{/}{u}^k$ with $t\in T$, $k\ge 0$. If $t$ lay in $\mathfrak{q}$, then $\sigma =t\mathrm{/}{u}^k$ would lie in the prime $\mathfrak{q}{S}_u$ of $S_u$, contradicting $\sigma \mathrm{\notin }\mathfrak{q}$; hence $t\mathrm{\notin }\mathfrak{q}$, so $t\in T\setminus \mathfrak{r}$. The vanishing $\sigma a=0$ in $S_u$ means $u^{m}ta=0$ in $S$ for some $m\ge 0$ by [L3]; this element lies in $T$, and $u^{m}t\in T\setminus \mathfrak{r}$ is inverted in $T_{\mathfrak{r}}$, so $a\mathrm{/}s=0$.

The composite $S\to S^{\mathrm{\prime }}\to S_{{\mathfrak{q}}^{\mathrm{\prime }}}^{\mathrm{\prime }}$ inverts every element of $S\setminus \mathfrak{q}$, because such an element lies outside ${\mathfrak{q}}^{\mathrm{\prime }}\cap S=\mathfrak{q}$ and hence outside ${\mathfrak{q}}^{\mathrm{\prime }}$. It also kills $\mathfrak{p}$ in the quotient by ${\mathfrak{p}}^{\mathrm{\prime }}{S}_{{\mathfrak{q}}^{\mathrm{\prime }}}^{\mathrm{\prime }}$. Thus the map factors through a $\kappa (\mathfrak{p})$-algebra homomorphism $E\to E^{\mathrm{\prime }}:=S_{{\mathfrak{q}}^{\mathrm{\prime }}}^{\mathrm{\prime }}\mathrm{/}{\mathfrak{p}}^{\mathrm{\prime }}{S}_{{\mathfrak{q}}^{\mathrm{\prime }}}^{\mathrm{\prime }}$, and the residue-field map $\kappa (\mathfrak{p})\to \kappa ({\mathfrak{p}}^{\mathrm{\prime }})$ is induced by $R\to R^{\mathrm{\prime }}$.

The quotient of step 1.5 is therefore finite over $\kappa (\mathfrak{p})=\kappa (\bar{\mathfrak{q}}\cap R)$ by hypothesis and [L1], so $R\to \bar{S}$ is quasi-finite at $\bar{\mathfrak{q}}$ by [L1]. Contrapositively, if $J\subseteq \mathfrak{q}$ and the quotient map $R\to S\mathrm{/}J$ fails to be quasi-finite at $\mathfrak{q}\mathrm{/}J$, then $R\to S$ is not quasi-finite at $\mathfrak{q}$.

The ring $T\mathrm{/}\mathfrak{r}$ is a domain and the composite $T\mathrm{/}\mathfrak{r}\to S_{\mathfrak{q}}\mathrm{/}\mathfrak{r}{S}_{\mathfrak{q}}$ is injective: an element of $T$ has vanishing image in $S_{\mathfrak{q}}\mathrm{/}\mathfrak{r}{S}_{\mathfrak{q}}$ exactly when it lies in $\mathfrak{q}$, and $\mathfrak{q}\cap T=\mathfrak{r}$. Every class $t+\mathfrak{r}$ with $t\in T\setminus \mathfrak{r}$ lies outside $\mathfrak{q}$, hence is a unit of $S_{\mathfrak{q}}$ and therefore a unit of the quotient $S_{\mathfrak{q}}\mathrm{/}\mathfrak{r}{S}_{\mathfrak{q}}$; by [L7] and [L8] the field of fractions $\kappa (\mathfrak{r})=\mathrm{Frac}(T\mathrm{/}\mathfrak{r})$ therefore embeds in $S_{\mathfrak{q}}\mathrm{/}\mathfrak{r}{S}_{\mathfrak{q}}$ as a subring containing the image of $\kappa (\mathfrak{p})$.

The homomorphism $T_{\mathfrak{r}}\to S_{\mathfrak{q}}$ of step 2.2 is also surjective. An element of $S_{\mathfrak{q}}$ is a fraction $\sigma \mathrm{/}\tau$ with $\sigma ,\tau \in S$ and $\tau \mathrm{\notin }\mathfrak{q}$; since $S_u=T_u$ as subrings of $S_u$, and $u\mathrm{\notin }\mathfrak{q}$, we may write $\sigma =t\mathrm{/}{u}^k$ and $\tau =t^{\mathrm{\prime }}\mathrm{/}{u}^l$ with $t,t^{\mathrm{\prime }}\in T$ by [L3]. Then $t^{\mathrm{\prime }}\mathrm{\notin }\mathfrak{q}$, hence $t^{\mathrm{\prime }}\in T\setminus \mathfrak{r}$, and $\sigma \mathrm{/}\tau =t{u}^l\mathrm{/}(t^{\mathrm{\prime }}{u}^k)$ with numerator $t{u}^l\in T$ and denominator $t^{\mathrm{\prime }}{u}^k\in T\setminus \mathfrak{r}$, so $\sigma \mathrm{/}\tau$ is the image of an element of $T_{\mathfrak{r}}$.

Let $F=S{\otimes }_R\kappa (\mathfrak{p})$ and let $\bar{\mathfrak{q}}$ be the prime of this fiber induced by $\mathfrak{q}$, so $E\cong F_{\bar{\mathfrak{q}}}$. The prime ${\mathfrak{q}}^{\mathrm{\prime }}$ induces a prime ${\bar{\mathfrak{q}}}^{\mathrm{\prime }}$ of $F{\otimes }_{\kappa (\mathfrak{p})}\kappa ({\mathfrak{p}}^{\mathrm{\prime }})$ lying over $\bar{\mathfrak{q}}$. By [L9], localization commutes with this scalar extension; localizing further at ${\bar{\mathfrak{q}}}^{\mathrm{\prime }}$ gives the canonical isomorphism $E^{\mathrm{\prime }}\cong (E{\otimes }_{\kappa (\mathfrak{p})}\kappa ({\mathfrak{p}}^{\mathrm{\prime }}){)}_{{\tilde{\mathfrak{q}}}^{\mathrm{\prime }}},$ where ${\tilde{\mathfrak{q}}}^{\mathrm{\prime }}$ is the corresponding prime after the first localization.

Since $\kappa (\mathfrak{r})$ is a $\kappa (\mathfrak{p})$-subspace of the finite-dimensional $\kappa (\mathfrak{p})$-vector space $S_{\mathfrak{q}}\mathrm{/}\mathfrak{r}{S}_{\mathfrak{q}}$ of step 2.1, the field extension $\kappa (\mathfrak{r})\mathrm{/}\kappa (\mathfrak{p})$ is finite. The algebra $S_{\mathfrak{q}}\mathrm{/}\mathfrak{r}{S}_{\mathfrak{q}}$ of step 2.1 is a module over the field $\kappa (\mathfrak{r})$ by step 3.1, and a $\kappa (\mathfrak{r})$-linearly independent subset of it is $\kappa (\mathfrak{p})$-linearly independent, so $S_{\mathfrak{q}}\mathrm{/}\mathfrak{r}{S}_{\mathfrak{q}}$ is finite-dimensional over $\kappa (\mathfrak{r})$; by [L1] the map $T\to S$ is quasi-finite at $\mathfrak{q}$, which is assertion (1).

By steps 2.2 and 3.2 the inclusion induces an isomorphism $T_{\mathfrak{r}}\cong S_{\mathfrak{q}}$; it carries $\mathfrak{p}{T}_{\mathfrak{r}}$ onto $\mathfrak{p}{S}_{\mathfrak{q}}$ because it is an isomorphism of $R$-algebras. Hence $T_{\mathfrak{r}}\mathrm{/}\mathfrak{p}{T}_{\mathfrak{r}}\cong S_{\mathfrak{q}}\mathrm{/}\mathfrak{p}{S}_{\mathfrak{q}}$ is finite over $\kappa (\mathfrak{p})$ by hypothesis and [L1], and since $\mathfrak{p}=\mathfrak{r}\cap R$ this says by [L1] that $R\to T$ is quasi-finite at $\mathfrak{r}$, which is assertion (2).

The algebra $E{\otimes }_{\kappa (\mathfrak{p})}\kappa ({\mathfrak{p}}^{\mathrm{\prime }})$ is finite-dimensional over $\kappa ({\mathfrak{p}}^{\mathrm{\prime }})$, since the images of the basis in step 1.4 span it. Any localization $A_{\mathfrak{r}}$ of a finite-dimensional algebra $A$ over a field at a prime $\mathfrak{r}$ is finite-dimensional: for $s\mathrm{\notin }\mathfrak{r}$, the descending chain of vector subspaces $(s^n)\subseteq A$ stabilizes, so for some $N\ge 0$ and $a\in A$ one has $s^N=s^{N+1}a$; in $A_{\mathfrak{r}}$ this gives $1=sa$, so the inverse of every denominator is already in the image of $A$. Hence $A\to A_{\mathfrak{r}}$ is surjective and its target is finite-dimensional. Applying this to the localization in step 3.3 shows $E^{\mathrm{\prime }}$ is finite-dimensional over $\kappa ({\mathfrak{p}}^{\mathrm{\prime }})$.

Thus the $\kappa ({\mathfrak{p}}^{\mathrm{\prime }})$-algebra $S_{{\mathfrak{q}}^{\mathrm{\prime }}}^{\mathrm{\prime }}\mathrm{/}{\mathfrak{p}}^{\mathrm{\prime }}{S}_{{\mathfrak{q}}^{\mathrm{\prime }}}^{\mathrm{\prime }}$ is generated as a $\kappa ({\mathfrak{p}}^{\mathrm{\prime }})$-module by $y_1,\dots ,y_d$, hence is finite over $\kappa ({\mathfrak{p}}^{\mathrm{\prime }})$ by [L1]; that is, $R^{\mathrm{\prime }}\to S^{\mathrm{\prime }}$ is quasi-finite at ${\mathfrak{q}}^{\mathrm{\prime }}$ by [L1], which is assertion (3).

Assertion (1) is step 4.1, assertion (2) is step 4.2, assertion (3) is step 5.1 and assertion (4) with its contrapositive form is step 2.4; all four reduce to the single finite-dimensionality condition of [L1] at the relevant prime, and no Noetherian hypothesis and no form of the Axiom of Choice was used. ∎