The quasi-finite locus of a finite-type algebra is open

We assume the Axiom of Choice as recorded in [L10]. By [L2] the finite type $R$-algebra $S$ has the form $S=R[s_1,\dots ,s_n]$ for some $n\ge 0$ and elements $s_i\in S$, and $S^{\mathrm{\prime }}={\mathrm{Int}}_R(S)$ is an $R$-subalgebra of $S$ containing the image of $R$ by [L4]. We must show that the set $U$ of primes $\mathfrak{q}$ at which $R\to S$ is quasi-finite, in the sense of [L1], is open in the topology of [L14].

We record a general fact about module-finite algebras. Let $B$ be module-finite over a ring $C$ and let $\mathfrak{p}\in \mathrm{Spec}(C)$; choose $b_1,\dots ,b_m\in B$ generating $B$ as a $C$-module. Then the images of $b_1,\dots ,b_m$ generate $B\mathrm{/}\mathfrak{p}B$ as a $C\mathrm{/}\mathfrak{p}$-module, and therefore generate the $C\mathrm{/}\mathfrak{p}$-algebra $(B\mathrm{/}\mathfrak{p}B){\otimes }_{C\mathrm{/}\mathfrak{p}}\kappa (\mathfrak{p})$ as a module over $\kappa (\mathfrak{p})=\mathrm{Frac}(C\mathrm{/}\mathfrak{p})$ by [L9]: localising a module at a multiplicative set multiplies its generating set by scalars and never enlarges it. Hence $(B\mathrm{/}\mathfrak{p}B){\otimes }_{C\mathrm{/}\mathfrak{p}}\kappa (\mathfrak{p})$ is a finite-dimensional $\kappa (\mathfrak{p})$-algebra.

We also record a general fact about localising finite-dimensional algebras. Let $C$ be a finite-dimensional algebra over a field $K$ and let $W\subseteq C$ be a multiplicative subset. For every $u\in W$ the ideals $(u)\supseteq (u^2)\supseteq (u^3)\supseteq \cdots$ of $C$ are $K$-subspaces of the finite-dimensional $K$-vector space $C$, so the chain stabilises: there is $N\ge 1$ with $(u^N)=(u^{N+1})$, hence $u^N=u^{N+1}x$ for some $x\in C$ and therefore $u^N(1-ux)=0$ in $C$. Since $u$ is a unit of $W^{-1}C$ with inverse $1\mathrm{/}u$ by [L8], the image of $u^N$ is invertible in $W^{-1}C$, so the image of $1-ux$ is zero there, that is $x=1\mathrm{/}u$ in $W^{-1}C$. Consequently every element of $W^{-1}C$ is of the form $c\cdot x$ with $c\in C$ and hence lies in the image of the localisation map $C\to W^{-1}C$, which is therefore surjective as a $K$-linear map; so ${\dim }_K{W}^{-1}C\le {\dim }_{K}C$ and $W^{-1}C$ is finite-dimensional over $K$.

Let $\mathfrak{q}\in U$ and put $\mathfrak{p}=\mathfrak{q}\cap R$. By [L3] there is $g\in S^{\mathrm{\prime }}\setminus \mathfrak{q}$ such that the inclusion $S^{\mathrm{\prime }}\to S$ induces an isomorphism $S_g^{\mathrm{\prime }}\to S_g$ of localisations; we fix such a $g$ and view $S_g^{\mathrm{\prime }}=S_g$ as subrings of the common ring $S_g$.

The localisation $S_g$ is a finitely generated $R$-algebra: its elements are the fractions with numerator in $S$ and denominator a power of $g$ by [L12], so the images of $s_1,\dots ,s_n$ together with $1\mathrm{/}g$ generate $S_g$ over $R$ by [L2]. Hence there are finitely many elements $z_1,\dots ,z_N\in S_g$ generating $S_g$ as an $R$-algebra; taking them to be the explicit generators just listed.

Each $z_j$ lies in $S_g^{\mathrm{\prime }}=S_g$, so by [L12] and [L8] it can be written $z_j=y_j\mathrm{/}{g}^{m_j}$ with $y_j\in S^{\mathrm{\prime }}$ and some exponent $m_j\ge 0$. Put $T:=R[g,y_1,\dots ,y_N]\subseteq S^{\mathrm{\prime }}$. Then $T_g=S_g$: the inclusion $T\subseteq S^{\mathrm{\prime }}$ gives $T_g\subseteq S_g^{\mathrm{\prime }}=S_g$, while each generator $z_j=y_j\mathrm{/}{g}^{m_j}$ of $S_g$ over $R$ lies in $T_g$, so $S_g\subseteq T_g$. Also $T$ is an $R$-subalgebra of $S$ containing the image of $R$.

Every generator $g,y_1,\dots ,y_N$ of $T$ lies in $S^{\mathrm{\prime }}$ and is therefore integral over $R$ by [L4]. Writing $A$ for the image of $R$ in $S$, the elements $g,y_1,\dots ,y_N$ are integral over the subring $A\subseteq S$, so [L5] shows that $T=A[g,y_1,\dots ,y_N]$ is module-finite over $A$; the same finite list generates $T$ as an $R$-module, so $T$ is module-finite over $R$, and $T$ is of finite type over $R$ by [L2].

Now let ${\mathfrak{q}}^{\mathrm{\prime }}\in \mathrm{Spec}(S)$ with $g\mathrm{\notin }{\mathfrak{q}}^{\mathrm{\prime }}$, and put ${\mathfrak{q}}^{\mathrm{\prime }\mathrm{\prime }}:={\mathfrak{q}}^{\mathrm{\prime }}\cap T$, ${\mathfrak{p}}^{\mathrm{\prime }}:={\mathfrak{q}}^{\mathrm{\prime }}\cap R={\mathfrak{q}}^{\mathrm{\prime }\mathrm{\prime }}\cap R$ and $T_g$-prime ${\mathfrak{q}}^{\mathrm{\prime }\mathrm{\prime }}{T}_g\subseteq T_g$, ${\mathfrak{q}}^{\mathrm{\prime }}{S}_g\subseteq S_g$. The primes ${\mathfrak{q}}^{\mathrm{\prime }\mathrm{\prime }}{T}_g$ and ${\mathfrak{q}}^{\mathrm{\prime }}{S}_g$ correspond under the $R$-algebra isomorphism $T_g\cong S_g$ of step 3.1, because contraction along $T\to T_g$ and $S\to S_g$ recovers ${\mathfrak{q}}^{\mathrm{\prime }\mathrm{\prime }}$ and ${\mathfrak{q}}^{\mathrm{\prime }}$ by [L11] and ${\mathfrak{q}}^{\mathrm{\prime }}\cap T={\mathfrak{q}}^{\mathrm{\prime }\mathrm{\prime }}$.

Consequently, for the module-finite $R$-algebra $T$ of step 4.1 and any prime $\mathfrak{r}\in \mathrm{Spec}(T)$ with contraction $\mathfrak{p}=\mathfrak{r}\cap R$, the algebra $T_{\mathfrak{r}}\mathrm{/}\mathfrak{p}{T}_{\mathfrak{r}}$ is finite-dimensional over $\kappa (\mathfrak{p})$. Indeed, $\mathfrak{p}T\subseteq \mathfrak{r}$, so [L7] gives $T_{\mathfrak{r}}\mathrm{/}\mathfrak{p}{T}_{\mathfrak{r}}\cong (T\mathrm{/}\mathfrak{p}T{)}_{\bar{\mathfrak{r}}}$ for the prime $\bar{\mathfrak{r}}=\mathfrak{r}\mathrm{/}\mathfrak{p}T$; put $C:=(T\mathrm{/}\mathfrak{p}T){\otimes }_{R\mathrm{/}\mathfrak{p}}\kappa (\mathfrak{p})$, which is finite-dimensional over $\kappa (\mathfrak{p})$ by step 1.2, and let $S_0$ be the image of $(R\mathrm{/}\mathfrak{p})\setminus \{0\}$ in $T\mathrm{/}\mathfrak{p}T$ and $\bar{U}$ the image of $(T\mathrm{/}\mathfrak{p}T)\setminus \bar{\mathfrak{r}}$. Since $\bar{\mathfrak{r}}\cap (R\mathrm{/}\mathfrak{p})=0$, we have $S_0\subseteq (T\mathrm{/}\mathfrak{p}T)\setminus \bar{\mathfrak{r}}$; applying [L6] to the multiplicative subsets $S_0$ and $(T\mathrm{/}\mathfrak{p}T)\setminus \bar{\mathfrak{r}}$ of the ring $T\mathrm{/}\mathfrak{p}T$ shows that $(T\mathrm{/}\mathfrak{p}T{)}_{\bar{\mathfrak{r}}}\cong {\bar{U}}^{-1}C$, a localisation of the finite-dimensional $\kappa (\mathfrak{p})$-algebra $C$ at a multiplicative subset, which is finite-dimensional over $\kappa (\mathfrak{p})$ by step 1.3.

Hence the canonical $R$-algebra homomorphism $T_{{\mathfrak{q}}^{\mathrm{\prime }\mathrm{\prime }}}\to S_{{\mathfrak{q}}^{\mathrm{\prime }}}$ is an isomorphism: both rings are localisations of the common ring $T_g=S_g$ at the primes ${\mathfrak{q}}^{\mathrm{\prime }\mathrm{\prime }}{T}_g$ and ${\mathfrak{q}}^{\mathrm{\prime }}{S}_g$ of step 4.2, and [L6] exhibits each of them as a localisation at the multiplicative subset of the common ring generated by $\{g^n\}$ together with the elements outside that prime.

Therefore $T_{{\mathfrak{q}}^{\mathrm{\prime }\mathrm{\prime }}}\mathrm{/}{\mathfrak{p}}^{\mathrm{\prime }}{T}_{{\mathfrak{q}}^{\mathrm{\prime }\mathrm{\prime }}}\cong S_{{\mathfrak{q}}^{\mathrm{\prime }}}\mathrm{/}{\mathfrak{p}}^{\mathrm{\prime }}{S}_{{\mathfrak{q}}^{\mathrm{\prime }}}$ is finite-dimensional over $\kappa ({\mathfrak{p}}^{\mathrm{\prime }})$ by step 5.1, and since $S$ is of finite type over $R$ by step 1.1 and ${\mathfrak{p}}^{\mathrm{\prime }}={\mathfrak{q}}^{\mathrm{\prime }}\cap R$, the map $R\to S$ is quasi-finite at ${\mathfrak{q}}^{\mathrm{\prime }}$ by [L1]. Thus $D_S(g)\subseteq U$.

It remains to draw the topological conclusion. By [L13] and [L14] each set $D_S(g)=\mathrm{Spec}(S)\setminus V((g))$ is the complement of a closed subset, hence open in the topology of [L14]. Step 2.1 attaches to every $\mathfrak{q}\in U$ an element $g\in S$ with $\mathfrak{q}\in D_S(g)$, and step 6.1 shows $D_S(g)\subseteq U$; thus every point of $U$ has an open neighbourhood contained in $U$, which is the defining property of an open subset, so $U$ is open (no selection from infinitely many points is needed, the argument being applied to one prime at a time).

The proof used the Axiom of Choice only in step 2.1, through [L3]; every other step selected finitely many elements (the generators $s_i$, the generators $z_j$, the numerators $y_j$, the module generators $b_i$, the exponent $N$ and the elements $u,x$ of step 1.3). ∎