A quasi-finite algebra factors openly through a finite algebra

We assume the Axiom of Choice as recorded in [L16]. By [L2] the finite type $R$-algebra $S$ has the form $S=R[s_1,\dots ,s_m]$, and $S^{\mathrm{\prime }}={\mathrm{Int}}_R(S)$ is an $R$-subalgebra of $S$ containing the image of $R$ by [L4]. The hypothesis is that $R\to S$ is quasi-finite at every prime of $S$ in the sense of [L1].

For each prime $\mathfrak{q}\in \mathrm{Spec}(S)$ the local theorem [L3] applies and produces an element $g_{\mathfrak{q}}\in S^{\mathrm{\prime }}\setminus \mathfrak{q}$ such that the inclusion $S^{\mathrm{\prime }}\to S$ induces an isomorphism $S_{g_{\mathfrak{q}}}^{\mathrm{\prime }}\to S_{g_{\mathfrak{q}}}$. Since $\mathfrak{q}\in D_S(g_{\mathfrak{q}})=\{\mathfrak{r}:g_{\mathfrak{q}}\mathrm{\notin }\mathfrak{r}\}$, and each $D_S(g_{\mathfrak{q}})$ is open in the topology of [L11] because it is the complement of $V((g_{\mathfrak{q}}))$ by [L10], the family $\{D_S(g_{\mathfrak{q}}){\}}_{\mathfrak{q}}$ is an open cover of $\mathrm{Spec}(S)$.

By compactness [L7] there are finitely many elements $g_1,\dots ,g_n\in S^{\mathrm{\prime }}$ with $\mathrm{Spec}(S)=D_S(g_1)\cup \cdots \cup D_S(g_n),$ and then the ideal generated by $g_1,\dots ,g_n$ is the unit ideal of $S$ by [L8].

For each $i$ the localisation $S_{g_i}$ is a finitely generated $R$-algebra: by [L2] the images of $s_1,\dots ,s_m$ together with the inverse of (the image of) $g_i$ generate it over $R$, and $1\mathrm{/}{g}_i$ is an element of $S_{g_i}$ by [L14]. Hence for each $i$ there are finitely many elements $z_{i1},\dots ,z_{i{N}_i}\in S_{g_i}$ generating $S_{g_i}$ as an $R$-algebra; using the isomorphism $S_{g_i}^{\mathrm{\prime }}\cong S_{g_i}$ of step 2.1 and [L14] we may write $z_{ij}=y_{ij}\mathrm{/}{g}_i^{m_{ij}}\text{with }{y}_{ij}\in S^{\mathrm{\prime }},m_{ij}\ge 0.$ Put $T:=R[g_1,\dots ,g_n,y_{ij}(1\le i\le n,1\le j\le N_i)]\subseteq S^{\mathrm{\prime }}\mathrm{.}$ Then $T_{g_i}=S_{g_i}$ for every $i$: the inclusion $T\subseteq S^{\mathrm{\prime }}$ gives $T_{g_i}\subseteq S_{g_i}^{\mathrm{\prime }}=S_{g_i}$, while every generator $z_{ij}=y_{ij}\mathrm{/}{g}_i^{m_{ij}}$ of $S_{g_i}$ over $R$ lies in $T_{g_i}$, so $S_{g_i}\subseteq T_{g_i}$.

Every generator $g_i$, $y_{ij}$ of $T$ lies in $S^{\mathrm{\prime }}$ and hence is integral over $R$ by [L4]. Writing $A$ for the image of $R$ in $S$, [L5] shows that $T=A[g_i,y_{ij}]$ is module-finite over $A$, and the same finite list generates $T$ as an $R$-module, so $T$ is module-finite over $R$; in particular $T$ is of finite type over $R$ by [L2], and it is a subalgebra of $S^{\mathrm{\prime }}$ by step 3.2.

Let $\phi :\mathrm{Spec}(S)\to \mathrm{Spec}(T)$ be the contraction map along the inclusion $T\subseteq S$, which is continuous by [L12]. For $h\in T\subseteq S$ one has ${\phi }^{-1}(D_T(h))=D_S(h)$, because a prime $\mathfrak{r}\in \mathrm{Spec}(S)$ contracts to a prime containing $h$ exactly when $\mathfrak{r}$ itself contains $h$. Consequently ${\phi }^{-1}(U)={\bigcup }_i{\phi }^{-1}(D_T(g_i))={\bigcup }_i{D}_S(g_i)=\mathrm{Spec}(S)$ for $U:=D_T(g_1)\cup \cdots \cup D_T(g_n)$, so $\phi$ maps $\mathrm{Spec}(S)$ into $U$; and $U$ is open in $\mathrm{Spec}(T)$ by [L10] and [L11].

For each $i$ the restriction of $\phi$ to $D_S(g_i)$ is a homeomorphism onto $D_T(g_i)$. Indeed [L9] identifies $D_S(g_i)$ with $\mathrm{Spec}(S_{g_i})$ and $D_T(g_i)$ with $\mathrm{Spec}(T_{g_i})$ through the contraction maps of the localisations, and the ring isomorphism $T_{g_i}\cong S_{g_i}$ of step 3.2 induces a homeomorphism $\mathrm{Spec}(S_{g_i})\to \mathrm{Spec}(T_{g_i})$; by [L13] the inverse of each of these identifications is the extension of a prime, and contracting a localised prime back to $T$ recovers the contraction of the original prime, so the composite is exactly the restriction of $\phi$.

For every $i$ the inclusion induces an isomorphism $T_{g{g}_i}\to S_{g{g}_i}$: localising the isomorphism $T_{g_i}\to S_{g_i}$ of step 3.2 at the element $g$, and rewriting $(T_{g_i}{)}_g\cong T_{g_{i}g}$ and $(S_{g_i}{)}_g\cong S_{g_{i}g}$ by [L6], gives the claim.

Now let $g\in T$ with $D_T(g)\subseteq U$, where the subalgebra $T$, the elements $g_1,\dots ,g_n$ and the open set $U$ are the ones constructed in steps 3.2 and 4.2, and first record that the images of $g_1,\dots ,g_n$ generate the unit ideal of $T_g$. By [L9] applied to the principal localisation $T\to T_g$ the spectrum $\mathrm{Spec}(T_g)$ is identified with $D_T(g)$, and under this identification the subset $D_{T_g}(g_i\mathrm{/}1)$ corresponds to $D_T(g)\cap D_T(g_i)$: a prime $\mathfrak{p}\in D_T(g)$ contains $g_i$ exactly when the corresponding prime of $T_g$ contains $g_i\mathrm{/}1$ by [L13] and [L14]. Hence the inclusion $D_T(g)\subseteq {\bigcup }_i{D}_T(g_i)$ gives $\mathrm{Spec}(T_g)={\bigcup }_i{D}_{T_g}(g_i\mathrm{/}1)$, and [L8] applied to the ring $T_g$ and the finite family $g_i\mathrm{/}1$ shows that the ideal generated by the images of $g_1,\dots ,g_n$ in $T_g$ is the unit ideal, say $1={\sum }_i{a}_i(g_i\mathrm{/}1)$ with $a_i\in T_g$.

The map $\phi$ is injective and has image $U$. For injectivity, let ${\mathfrak{q}}_1,{\mathfrak{q}}_2\in \mathrm{Spec}(S)$ with $\phi ({\mathfrak{q}}_1)=\phi ({\mathfrak{q}}_2)=:\mathfrak{p}\in U$; choose $i$ with $\mathfrak{p}\in D_T(g_i)$. Since $g_i\in T$ and $g_i\mathrm{\notin }\mathfrak{p}={\mathfrak{q}}_j\cap T$, we get $g_i\mathrm{\notin }{\mathfrak{q}}_j$ for $j=1,2$, so both primes lie in $D_S(g_i)$, where $\phi$ is injective by step 4.3. For surjectivity onto $U$, let $\mathfrak{p}\in U$ and choose $i$ with $\mathfrak{p}\in D_T(g_i)$; the prime $\mathfrak{p}$ corresponds under [L9] to a prime of $T_{g_i}$, which we transport across the isomorphism $T_{g_i}\cong S_{g_i}$ of step 3.2 to a prime of $S_{g_i}$, and its contraction to $S$ by [L9] is a prime $\mathfrak{q}\in D_S(g_i)$ with $\phi (\mathfrak{q})=\mathfrak{p}$ by step 4.3.

The map $T_g\to S_g$ induced by the inclusion is an isomorphism. It is injective: $T\to S$ is injective with $S$ viewed as a $T$-module, so [L15] applies to the multiplicative subset $\{1,g,g^2,\dots \}$ of $T$. For surjectivity let $s\in S_g$. By step 4.4 the localisation $(T_g{)}_{g_i}\to (S_g{)}_{g_i}$ is an isomorphism for every $i$, so for each $i$ there are $m_i\ge 0$ and $t_i\in T_g$ with $g_i^{m_i}s=t_i$. Put $M:=m_1+\cdots +m_n$ and expand $1=({\sum }_i{a}_i(g_i\mathrm{/}1){)}^M$ from step 5.1: every monomial in the expansion has the form $(\text{product of }a\text{’s})\cdot {\prod }_i{g}_i^{{\nu }_i}$ with ${\nu }_1+\cdots +{\nu }_n=M$, hence some ${\nu }_i\ge m_i$ and the monomial is divisible by $g_i^{m_i}$; consequently $1\in (g_1^{m_1},\dots ,g_n^{m_n})$ in $T_g$, say $1={\sum }_i{c}_i{g}_i^{m_i}$ with $c_i\in T_g$. Then $s={\sum }_i{c}_i{g}_i^{m_i}s={\sum }_i{c}_i{t}_i$ lies in $T_g$. Hence $T_g\to S_g$ is bijective, and being a ring homomorphism induced by the inclusion it is an isomorphism.

The restricted map $\phi :\mathrm{Spec}(S)\to U$ is a homeomorphism. It is continuous by step 4.2 and bijective by step 5.2. To see that it is open, let $W\subseteq \mathrm{Spec}(S)$ be open and write $W={\bigcup }_i(W\cap D_S(g_i))$ using that the $D_S(g_i)$ cover $\mathrm{Spec}(S)$ by step 3.1; each $W\cap D_S(g_i)$ is open in $D_S(g_i)$ and therefore has image $\phi (W\cap D_S(g_i))$ open in $D_T(g_i)$ by step 4.3, hence open in $U$ because $D_T(g_i)\subseteq U$ is open in $\mathrm{Spec}(T)$. Thus $\phi (W)={\bigcup }_i\phi (W\cap D_S(g_i))$ is a union of subsets open in $U$ and is open in $U$. A continuous, open bijection onto $U$ is a homeomorphism.

Part 1 is now established by the objects constructed above: $T$ is a finite $R$-subalgebra of $S^{\mathrm{\prime }}$ by step 4.1, the set $U=D_T(g_1)\cup \cdots \cup D_T(g_n)$ is open by step 4.2, the contraction map $\phi :\mathrm{Spec}(S)\to \mathrm{Spec}(T)$ is a homeomorphism onto $U$ by step 6.2, and $T_{g_i}\cong S_{g_i}$ for every $i$ by step 3.2. Part 2 is step 6.1.

The Axiom of Choice was used in step 2.1 through [L3], in step 3.1 through the compactness of $\mathrm{Spec}(S)$ [L7] and the cover-to-unit-ideal statement [L8], and in step 5.1 through [L8] again. Every other selection was finite: the generators $s_i$ of step 1.1, the finitely many $g_i$ of step 3.1, the generators $z_{ij}$ and numerators $y_{ij}$ of step 3.2, the index $i$ in steps 5.2 and 5.1 and the exponents $m_i$ of step 6.1. ∎