Quasi-finite does not imply finite

Take the field $k$, the polynomial ring $R=k[t]$ with indeterminate $t$ [L1], and $B=R_t=\{1,t,t^2,\dots {\}}^{-1}R=k[t,t^{-1}]$ [L5]. The ring $R$ is a domain with $0\mathrm{\ne }1$ [L2], and $B$ is a localisation of it. Let $S=R\times B$ be the product ring [L4], a commutative unital ring, and let $\phi :R\to S$, $r\mapsto (r,r)$ be the diagonal map: it is a unital ring homomorphism because the operations of $S$ are componentwise [L4], and it is injective, since $(r,r)=0$ forces $r=0$ by componentwise equality [L4]. Thus $S$ is a commutative $R$-algebra with $S\mathrm{\ne }0$.

We record the prime ideals of a product ring. Let $A$ and $C$ be commutative rings and put $e=(1,0)$, $f=(0,1)$ in $A\times C$; then $e+f=1$ and $ef=0$, and for $(a,b)\in A\times C$ the componentwise operations [L4] give $e(a,b)=(a,0)$ and $f(a,b)=(0,b)$. Let $P$ be a prime ideal of $A\times C$. Since $ef=0\in P$, [L31] gives $e\in P$ or $f\in P$; not both, because then $1=e+f\in P$ would contradict the properness of a prime ideal [L31]. Suppose $f\in P$ and put $I=\{a\in A:(a,0)\in P\}$; this is an ideal of $A$: if $(a,0),(a^{\mathrm{\prime }},0)\in P$ then $(a+a^{\mathrm{\prime }},0)=(a,0)+(a^{\mathrm{\prime }},0)\in P$, and if $r\in A$ and $(a,0)\in P$ then $(ra,0)=r(a,0)\in P$, because $P$ is an additive subgroup of $A\times C$ closed under multiplication by elements of $A\times C$ [L4, L32]. For $(a,b)\in P$ the identity $(0,b)=f(a,b)$ shows $(0,b)\in P$, hence $(a,0)=(a,b)-(0,b)\in P$; conversely, if $(a,0)\in P$ then $(a,b)=(a,0)+f(0,b)$ is a sum of two elements of $P$. Hence $P=I\times C$, and $I$ is prime: it is proper, since $(1,0)\in P$ would give $1=(1,0)+(0,1)\in P$; and if $rs\in I$ for $r,s\in A$, then $(r,0)(s,0)=(rs,0)\in P$, so $r\in I$ or $s\in I$ [L31]. Symmetrically, if $e\in P$ then $P=A\times J$ with $J=\{b\in C:(0,b)\in P\}$ a prime ideal of $C$. Conversely, if $I\subseteq A$ is a prime ideal then $I\times C$ is a prime ideal of $A\times C$: it is proper because $1\mathrm{\notin }I$, and $(a,b)(a^{\mathrm{\prime }},b^{\mathrm{\prime }})=(a{a}^{\mathrm{\prime }},b{b}^{\mathrm{\prime }})\in I\times C$ forces $a{a}^{\mathrm{\prime }}\in I$, hence $a\in I$ or $a^{\mathrm{\prime }}\in I$, that is $(a,b)\in I\times C$ or $(a^{\mathrm{\prime }},b^{\mathrm{\prime }})\in I\times C$ [L4, L31]; symmetrically $A\times J$ is prime for a prime ideal $J\subseteq C$. Hence the prime ideals of a product ring $A\times C$ are exactly the ideals $I\times C$ with $I$ prime in $A$ and the ideals $A\times J$ with $J$ prime in $C$.

Put $u:=(0,t^{-1})\in S$, $e:=(1,0)\in S$ and $\bar{t}:=\phi (t)=(t,t)$. Multiplying, $\bar{t}u=(t,t)(0,t^{-1})=(0,1)=1_S-e$ [L4]. We claim $S=R[u]$. First, every element of $B=R_t$ is of the form ${\sum }_{i=0}^n{a}_i{t}^{-i}$ with $a_i\in R$: an element of $R_t$ is $r\mathrm{/}{t}^n$ [L5], and writing $r={\sum }_{j=0}^d{c}_j{t}^j$ in $R=k[t]$ [L1] gives $r\mathrm{/}{t}^n={\sum }_{j<n}{c}_j{t}^{j-n}+{\sum }_{j\ge n}{c}_j{t}^{j-n}$, where the second sum lies in $R$ and the first is ${\sum }_{i=1}^n{c}_{n-i}{t}^{-i}$ [L1]. Second, for $g=a_0+{\sum }_{i=1}^n{a}_i{t}^{-i}\in B$ and $f\in R$ one has $(f,g)=(f-a_0)e+\phi (a_0)+{\sum }_{i=1}^n\phi (a_i)u^i$: indeed $u^i=(0,t^{-i})$ and multiplication by $\phi (a_i)=(a_i,a_i)$ is componentwise [L4], while $\phi$ is the structure map [step 1.1]. Hence every element of $S$ is a polynomial in $u$ with coefficients in the image of $R$, so $S=R[u]=R[e,u]$ because $e=1_S-\bar{t}u$ is itself a polynomial in $u$; therefore $R\to S$ is of finite type [L7]. This proves part 1.

The element $(1,t)\in S$ has inverse $(1,t^{-1})$ because $t^{-1}\in B$ is the inverse of $t\in B$ [L5] and units multiply componentwise [L4]: $(1,t)(1,t^{-1})=(1,1)=1_S$. In particular $t$ is a unit of $B$, so $B_t=B$.

We compute the residue fields of $R=k[t]$. By [L16] with $\mathfrak{p}=(t)$ we have $\kappa ((t))\cong \mathrm{Frac}(R\mathrm{/}(t))$, the residue field being defined by that isomorphism. The evaluation map ${\mathrm{ev}}_{k,0}:R=k[t]\to k$, ${\sum }_i{a}_i{t}^i\mapsto a_0$, is the ring homomorphism of [L17] for $\phi ={\mathrm{id}}_k$ and $s=0$; it is surjective, since it is the identity on constants. Its kernel is exactly the principal ideal $(t)$: for $f={\sum }_i{a}_i{t}^i$ the division of [L18] by the monic polynomial $g=t=x$ gives unique $q,r$ with $f=qt+r$ and $r=0$ or $\mathrm{deg}r<1$, that is $r=c$ a constant, and evaluating gives $c=f(0)$; hence $f(0)=0$ if and only if $r=0$, that is $t\mid f$. So [L19] gives $R\mathrm{/}(t)\cong k$, and then $\kappa ((t))\cong \mathrm{Frac}(k)=k$ by [L13] and [L3]: the field of fractions of a field is the field itself, because all its nonzero elements are already units. Moreover the image of $t$ in $\kappa ((t))$ is zero, since $t\in (t)$ and $R\to R\mathrm{/}(t)\to \mathrm{Frac}(R\mathrm{/}(t))$ is the canonical composite.

The underlying additive group of $S=R\times B$ is the direct sum $R\oplus B$, the $R$-module structure being componentwise [L4]. Let $\mathfrak{p}\in \mathrm{Spec}(R)$ and put $\kappa :=\kappa (\mathfrak{p})$. Tensoring the decomposition with $\kappa$ over $R$ and applying [L20] and [L21] gives $S{\otimes }_R\kappa \cong (R{\otimes }_R\kappa )\oplus (B{\otimes }_R\kappa )\cong \kappa \oplus (B{\otimes }_R\kappa )$, and applying [L22] to the multiplicative subset $M=\{1,t,t^2,\dots \}\subseteq R$ with $A=R$ and $C=\kappa$ gives $B{\otimes }_R\kappa =R_t{\otimes }_R\kappa \cong {\kappa }_t$, the principal localisation of the field $\kappa$ at the image of $t$ [L5]. Now take $\mathfrak{p}=(t)$: the element $t$ maps to $0\in \kappa ((t))$ by step 2.3, and inverting the zero element of a ring gives the zero ring [L6], so $\kappa ((t){)}_t=0$ and $S{\otimes }_R\kappa ((t))\cong \kappa ((t))\oplus 0\cong k$ by step 2.3.

Suppose, for contradiction, that $S$ is module-finite over $R$, that is, finitely generated as an $R$-module [L7]. By step 2.1 we have $S=R[u]$, so $R[u]$ is a finitely generated $R$-module, and the image $\phi (R)$ of $R$ in $S$ is nonzero by step 1.1; applying the equivalence of [L15] with $A=\phi (R)\cong R$ and $b=u$ — the module structure being the one transported along $\phi$ [L7] — we conclude that $u$ is integral over $\phi (R)$. By [L10] and [L11] there are $n\ge 1$ and $a_0,\dots ,a_{n-1}\in R$ with $u^n+\phi (a_{n-1})u^{n-1}+\cdots +\phi (a_0)=0$ in $S$.

We compute the relative integral closure $S^{\mathrm{\prime }}={\mathrm{Int}}_R(S)$ [L11]. First let $(f,g)\in S$ be integral over $\phi (R)$. The second projection ${\pi }_2:S=R\times B\to B$, $(a,b)\mapsto b$, is a unital ring homomorphism — the operations are componentwise [L4] — and it carries $\phi (a)=(a,a)$ to $a$, so applying it to a monic equation for $(f,g)$ over $\phi (R)$ produces a monic equation for $g$ over $R$ [L10, L11]; that is, $g$ is integral over $R$. Now $B=R_t\subseteq \mathrm{Frac}(R)$: every element of $B$ is a fraction $r\mathrm{/}{t}^n$ with $r\in R$ and $t^n\mathrm{\ne }0$ [L5, L13]. The field $k$ is an integrally closed domain, since $\mathrm{Frac}(k)=k$ by [L3, L13] and every element of $k$ lies in $k$ [L12]; hence $R=k[t]$ is an integrally closed domain by [L14]. So the element $g\in \mathrm{Frac}(R)$, being integral over $R$, lies in $R$. The first coordinate $f$ lies in $R$ by the definition of $S$ [L4]. Conversely, let $f,g\in R$; then $(f,g)=\phi (g)+(f-g)e$, where $\phi (g)=(g,g)$ lies in the image of $R$ and hence in $S^{\mathrm{\prime }}$ [L11], while $e=(1,0)$ is integral over $\phi (R)$ because $e^2-e=0$ [L4, L10], and $(f-g)e$ is a product of two elements of the subring $S^{\mathrm{\prime }}$ [L11]. Hence $(f,g)\in S^{\mathrm{\prime }}$, and $S^{\mathrm{\prime }}=R\times R=\{\phi (g)+(h,0):g,h\in R\}$.

Now let $\mathfrak{p}\in \mathrm{Spec}(R)$ with $t\mathrm{\notin }\mathfrak{p}$ and again put $\kappa =\kappa (\mathfrak{p})\cong \mathrm{Frac}(R\mathrm{/}\mathfrak{p})$ [L16]. The image of $t$ in $R\mathrm{/}\mathfrak{p}$ is nonzero, because $R\mathrm{/}\mathfrak{p}$ is a domain [L2] and $t\mathrm{\notin }\mathfrak{p}$, so its image in the field $\kappa$ is a nonzero element, hence a unit [L3]. Therefore every fraction $r\mathrm{/}{t}^n$ of ${\kappa }_t$ already lies in $\kappa$ [L5], so the localisation map $\kappa \to {\kappa }_t$ is surjective as well as injective, that is ${\kappa }_t=\kappa$; with step 3.1 this gives $S{\otimes }_R\kappa \cong \kappa \oplus \kappa \cong \kappa \times \kappa$ as rings, a product of two copies of the residue field. This proves part 2.

We verify quasi-finiteness at a prime in the case $\mathfrak{p}=(t)$. Let $\mathfrak{q}\in \mathrm{Spec}(S)$ with $\mathfrak{q}\cap R=(t)$. By [L23] the quotient $S_{\mathfrak{q}}\mathrm{/}\mathfrak{p}{S}_{\mathfrak{q}}$ is the local ring of the fibre $\mathrm{Spec}(S{\otimes }_R\kappa (\mathfrak{p}))$ at the prime $\bar{\mathfrak{q}}$ determined by $\mathfrak{q}$, and by step 3.1 the fibre ring is the field $k$. A prime ideal of $k$ is proper, and no unit of a ring lies in a prime ideal — if $u\in P$ were a unit then $1=u{u}^{-1}\in P$ — [L31, L32], while every nonzero element of the field $k$ is a unit [L3]; hence every prime ideal of $k$ is $(0)$, and since the fibre has the prime $\bar{\mathfrak{q}}$, that prime is $(0)$. The local ring of the fibre there is therefore the localisation $k_{(0)}=(k\setminus \{0\}{)}^{-1}k$ at the prime $(0)$, and every class $a\mathrm{/}s$ of that localisation equals $a\cdot s^{-1}\mathrm{/}1$, because $s\mathrm{\ne }0$ is a unit of $k$ [L3, L6]; so the localisation map $k\to k_{(0)}$ is bijective and Hence $S_{\mathfrak{q}}\mathrm{/}\mathfrak{p}{S}_{\mathfrak{q}}\cong k=\kappa ((t))$ — the equality of fields being step 2.3 — a one-dimensional vector space over $\kappa (\mathfrak{p})$, hence a finite $\kappa (\mathfrak{p})$-module [L23].

We compute that equation in the two coordinates of $S=R\times B$ [L4]. Since $u^i=(0,t^{-i})$ and $\phi (a_i)=(a_i,a_i)$, the second coordinate of the equation is $t^{-n}+a_{n-1}{t}^{-(n-1)}+\cdots +a_1{t}^{-1}+a_0=0$ in $B$. Multiplying this equation by $t^n\in B$ gives $1+a_{n-1}t+\cdots +a_1{t}^{n-1}+a_0{t}^n=0$ in $B$. All terms of this equation lie in the subring $R\subseteq B$, and $R\to B=R_t$ is injective: if $r\mathrm{/}1=0$ in $B$, then $t^{m}r=0$ in the domain $R$ for some $m$, so $r=0$, by the fraction criterion [L6, L2]. Hence the equation holds in $R$, and $1=-t(a_{n-1}+a_{n-2}t+\cdots +a_1{t}^{n-2}+a_0{t}^{n-1})$, that is $1\in tR$.

Put $T:=S^{\mathrm{\prime }}=R\times R$. As an $R$-module, $T$ is generated by $(1,0)$ and $(0,1)$: every $(g,h)\in R\times R$ is $(g,h)=g\cdot (1,0)+h\cdot (0,1)$ with the componentwise action [L4]. Hence $T$ is a finite $R$-algebra [L7], in particular a finite $R$-subalgebra of $S^{\mathrm{\prime }}$ in the sense of [L29].

We verify quasi-finiteness at a prime in the case $t\mathrm{\notin }\mathfrak{p}$, where $\mathfrak{p}=\mathfrak{q}\cap R$. By step 4.1 the fibre ring is $K\times K$ with $K:=\kappa (\mathfrak{p})$ a field [L3], and by step 1.2, applied to the product ring $K\times K$, its prime ideals are exactly $K\times 0$ and $0\times K$. Let $P:=K\times 0$; the multiplicative set of the localisation at $P$ is the complement $M=\{(a,b):b\mathrm{\ne }0\}$. In $M^{-1}(K\times K)$ the element $(0,1)\in M$ is a unit [L6] and $(0,1)(1,0)=0$ [L4], so $(1,0)\mathrm{/}1=0$; consequently the map $\psi :K\to M^{-1}(K\times K)$, $a\mapsto (0,a)\mathrm{/}1$, is a well-defined unital ring homomorphism [L6], injective because $(0,a)\mathrm{/}1=0$ means $(c,d)(0,a)=(0,da)=0$ for some $(c,d)\in M$, whence $a=0$ as $d\mathrm{\ne }0$ and $K$ is a field [L3, L6], and surjective because every class in the localisation equals $0\mathrm{/}1$ or $(0,b)\mathrm{/}(c,d)=(0,b\mathrm{/}d)\mathrm{/}1$ for $d\mathrm{\ne }0$, again by the fraction criterion [L6]. Hence the local ring of the fibre at the prime over $P$ is $K$, and symmetrically the one at the prime over $0\times K$ is $K$; both are one-dimensional over $\kappa (\mathfrak{p})$. By [L23] the map $R\to S$ is quasi-finite at every $\mathfrak{q}\in \mathrm{Spec}(S)$.

We show $1\mathrm{\notin }tR$. If $1=th$ for some $h\in R$, then $h\mathrm{\ne }0$ since $1\mathrm{\ne }0$ in the domain $R$ [L2], and $\mathrm{deg}(t)=1$ with $\mathrm{lc}(t)=1$ while $\mathrm{deg}(1)=0$ by [L8]. By [L9] the coefficient of $x^{1+\mathrm{deg}h}$ in $th$ is $\mathrm{lc}(t)\mathrm{lc}(h)=\mathrm{lc}(h)\mathrm{\ne }0$, so $\mathrm{deg}(th)\ge 1+\mathrm{deg}h\ge 1$ and $th\mathrm{\ne }0$; but $th=1$ has degree $0$, a contradiction. Hence $1\mathrm{\notin }tR$, contradicting step 4.3, and $S$ is not module-finite over $R$. This proves part 4.

Put $g:=(1,t)\in T$. For a product ring the localisation at a product element is the product of the localisations: the map $\theta :(A\times A^{\mathrm{\prime }}{)}_{(f,f^{\mathrm{\prime }})}\to A_f\times A_{f^{\mathrm{\prime }}}^{\mathrm{\prime }}$, $(a,a^{\mathrm{\prime }})\mathrm{/}(f,f^{\mathrm{\prime }}{)}^n\mapsto (a\mathrm{/}{f}^n,a^{\mathrm{\prime }}\mathrm{/}{f}^{\mathrm{\prime }n})$, is well defined and a unital ring homomorphism by the fraction criterion [L6] and the componentwise operations [L4], it is injective because $(f,f^{\mathrm{\prime }}{)}^k(a,a^{\mathrm{\prime }})=0$ for some $k$ means $f^{k}a=0$ and $f^{\mathrm{\prime }k}{a}^{\mathrm{\prime }}=0$, and it is surjective because $(x{f}^{N-m},x^{\mathrm{\prime }}{f}^{\mathrm{\prime }N-n})\mathrm{/}(f,f^{\mathrm{\prime }}{)}^N$ maps to $(x\mathrm{/}{f}^m,x^{\mathrm{\prime }}\mathrm{/}{f}^{\mathrm{\prime }n})$ for $N=\max \{m,n\}$. Applying this with $A=A^{\mathrm{\prime }}=R$, $f=1$, $f^{\mathrm{\prime }}=t$ and [L5] gives $T_g=(R\times R{)}_{(1,t)}\cong R_1\times R_t=R\times B=S$; applying it with $A=R$, $A^{\mathrm{\prime }}=B$ and [L5] gives $S_g=(R\times B{)}_{(1,t)}\cong R_1\times B_t=R\times B=S$, since $B_t=B$ by step 2.2. Both isomorphisms are the canonical maps induced by the inclusion $T\subseteq S$, so $T_g=S_g=S$ and the inclusion $T\to S$ induces an isomorphism of principal localisations [L5].

By step 2.1 the map $R\to S$ is of finite type and by steps 4.2 and 5.1 it is quasi-finite at every prime of $S$; hence it is quasi-finite [L23], and for every $\mathfrak{q}$ with contraction $\mathfrak{p}$ the quotient $S_{\mathfrak{q}}\mathrm{/}\mathfrak{p}{S}_{\mathfrak{q}}$ is isomorphic to $\kappa (\mathfrak{p})$, as computed in steps 4.2 and 5.1. This proves part 3.

By step 1.2 the primes of $S=R\times B$ are exactly the ideals $\mathfrak{p}\times B$ with $\mathfrak{p}\in \mathrm{Spec}(R)$ and the ideals $R\times \mathfrak{q}$ with $\mathfrak{q}\in \mathrm{Spec}(B)$, and the primes of $T=R\times R$ are the ideals $\mathfrak{p}\times R$ and $R\times {\mathfrak{p}}^{\mathrm{\prime }}$ with $\mathfrak{p},{\mathfrak{p}}^{\mathrm{\prime }}\in \mathrm{Spec}(R)$. For a prime $\mathfrak{q}=\mathfrak{p}\times B$ of $S$ we compute $\mathfrak{q}\cap T=(\mathfrak{p}\times B)\cap (R\times R)=\mathfrak{p}\times R$, and for $\mathfrak{q}=R\times {\mathfrak{q}}^{\mathrm{\prime }}$ we get $\mathfrak{q}\cap T=R\times ({\mathfrak{q}}^{\mathrm{\prime }}\cap R)$ by the componentwise description of ideals of a product [L4]. Hence the image of the contraction map $\mathrm{Spec}(S)\to \mathrm{Spec}(T)$ is the union of the set of primes $\mathfrak{p}\times R$, which is the entire first component of $\mathrm{Spec}(T)$, and of the set of primes $R\times {\mathfrak{p}}^{\mathrm{\prime }}$ with ${\mathfrak{p}}^{\mathrm{\prime }}$ in the image of the contraction $\mathrm{Spec}(B)\to \mathrm{Spec}(R)$. That contraction is the localisation map of $R$ at $M=\{1,t,t^2,\dots \}$, so by [L25] its image is $\{\mathfrak{p}\in \mathrm{Spec}(R):\mathfrak{p}\cap M=\mathrm{\varnothing }\}=D(t)$ [L24, L5]. Finally, a prime $\mathfrak{p}\times R$ contains $g=(1,t)$ only if $1\in \mathfrak{p}$, which is impossible [L4], while $R\times {\mathfrak{p}}^{\mathrm{\prime }}$ contains $(1,t)$ exactly when $t\in {\mathfrak{p}}^{\mathrm{\prime }}$ [L4]; so the image is precisely the open set $D_T(g)=\{\mathfrak{P}\in \mathrm{Spec}(T):g\mathrm{\notin }\mathfrak{P}\}$ [L24].

We record the topological form of step 6.2. The first projection ${\pi }_1^S:S\to R$ of $S=R\times B$ is a surjective unital ring homomorphism with kernel $0\times B$, and the second projection ${\pi }_2^S:S\to B$ is surjective with kernel $R\times 0$ [L4]; by step 1.2 the primes of $S$ are exactly the primes of the two complementary closed sets $V(0\times B)=\{\mathfrak{p}\times B:\mathfrak{p}\in \mathrm{Spec}(R)\}$ and $V(R\times 0)=\{R\times \mathfrak{q}:\mathfrak{q}\in \mathrm{Spec}(B)\}$, and the analogous statements hold for the projections ${\pi }_1^T,{\pi }_2^T$ of $T=R\times R$, whose kernels are $0\times R$ and $R\times 0$. By [L26] the four contraction maps along these projections are homeomorphisms onto $V(0\times B)$, $V(R\times 0)$, $V(0\times R)$ and $V(R\times 0)$ respectively, while by functoriality [L27] the contraction map $\phi =\mathrm{Spec}(\iota )$ along the inclusion $\iota :T\to S$ satisfies $\phi \circ \mathrm{Spec}({\pi }_1^S)=\mathrm{Spec}({\pi }_1^T)$ and $\phi \circ \mathrm{Spec}({\pi }_2^S)=\mathrm{Spec}({\iota }_B\circ {\pi }_2^T)$, where ${\pi }_2^T:T\to R$ is the second projection and ${\iota }_B:R\to B=R_t$ is the localisation map [L4, L5]. Consequently $\phi$ carries the piece $V(0\times B)$ homeomorphically onto the piece $V(0\times R)=\{\mathfrak{p}\times R\}$ of $\mathrm{Spec}(T)$, and carries the piece $V(R\times 0)$ homeomorphically onto the image of $\mathrm{Spec}({\iota }_B)$, which by [L25] is $D(t)\subseteq \mathrm{Spec}(R)$ carried into the piece $V(R\times 0)=\{R\times {\mathfrak{p}}^{\mathrm{\prime }}\}$ of $\mathrm{Spec}(T)$. The two target pieces are disjoint by step 1.2, so $\phi$ is a homeomorphism onto their union, which is the open set $D_T(g)$ of step 6.2. Therefore the data $T=S^{\mathrm{\prime }}=R\times R$, $n=1$ and $g_1=g=(1,t)$ satisfy the conclusion of part 1 of [L29]; and since $g$ is a unit of $S$ by step 2.2, it lies in no prime ideal of $S$ [L31], so the same single element $g$ witnesses the conclusion of [L30] at every prime of $S$ simultaneously.