The punctured affine line as an open finite factorization

Take $R=k[t]$ and $S=R_t=k[t,t^{-1}]$, the principal localisation of $R$ at $t$ as in [L3], and note $S\mathrm{\ne }0$ since $R$ is a domain by [L13] and a localisation of a nonzero ring at a nonzerodivisor is nonzero: if $r\mathrm{/}{t}^n=0$ then $t^{m}r=0$ for some $m\ge 0$ by [L12], whence $r=0$. The localisation map $R\to S$ is injective by the same computation, so we may view $R\subseteq S$. Every element of $S$ is of the form $r\mathrm{/}{t}^n$, that is $r\cdot (1\mathrm{/}t{)}^n$, so $S=R[1\mathrm{/}t]$ is generated as an $R$-algebra by the single element $1\mathrm{/}t$; hence $R\to S$ is of finite type by [L2].

We compute the fibres. Let $\mathfrak{p}\in \mathrm{Spec}(R)$ and apply [L5] with $A=R$, $C=\kappa (\mathfrak{p})$ and the multiplicative subset $M=\{1,t,t^2,\dots \}$ of $R$, whose image in $\kappa (\mathfrak{p})$ is generated by the image of $t$: $S{\otimes }_R\kappa (\mathfrak{p})=(M^{-1}R){\otimes }_R\kappa (\mathfrak{p})\cong {\bar{M}}^{-1}\kappa (\mathfrak{p})=\kappa (\mathfrak{p}{)}_t\mathrm{.}$ By [L7] the field $\kappa (\mathfrak{p})$ is $\mathrm{Frac}(R\mathrm{/}\mathfrak{p})$, so the image of $t$ in $\kappa (\mathfrak{p})$ is zero exactly when $t\in \mathfrak{p}$. If $t\in \mathfrak{p}$, then inverting the zero element of a ring gives the zero ring, so $\kappa (\mathfrak{p}{)}_t=0$ and therefore $S{\otimes }_R\kappa (\mathfrak{p})=0$; if $t\mathrm{\notin }\mathfrak{p}$, then the image of $t$ in the field $\kappa (\mathfrak{p})$ is a nonzero element, hence a unit, and then every fraction $r\mathrm{/}{t}^n=r(t^{-1}{)}^n$ already lies in $\kappa (\mathfrak{p})$, so $\kappa (\mathfrak{p}{)}_t=\kappa (\mathfrak{p})$.

For part 2 put $T:=R$, viewed as a subring of $S$ by step 1.1. Then $T$ is module-finite over $R$ because it is generated as an $R$-module by the single element $1$, hence a finite $R$-algebra in the sense of [L2]. Every element $r\in R$ is integral over $R$, being a root of the monic polynomial $X-r\in R[X]$ by [L10]; so $R\subseteq {\mathrm{Int}}_R(S)=S^{\mathrm{\prime }}$ by [L11], and $T=R$ is a finite $R$-subalgebra of $S^{\mathrm{\prime }}$.

By the fibre form of [L1] the fibre of $\mathrm{Spec}(S)\to \mathrm{Spec}(R)$ over $\mathfrak{p}$ is $\mathrm{Spec}(S{\otimes }_R\kappa (\mathfrak{p}))$. Hence by step 2.1 the fibre over the prime $(t)$ is $\mathrm{Spec}(0)=\mathrm{\varnothing }$, in particular $S{\otimes }_R\kappa ((t))=0$ and no prime of $S$ lies over $(t)$, while over every $\mathfrak{p}$ with $t\mathrm{\notin }\mathfrak{p}$ the fibre is $\mathrm{Spec}(\kappa (\mathfrak{p}))$, a single point. This matches [L8], which shows that the primes of $S=R_t$ are exactly the primes of $R$ not containing $t$.

With $g:=t\in T$ we have $T_t=R_t=S$ by [L3] and $S_t=S$, since $t$ is already invertible in $S=R_t$; and $g\mathrm{\notin }\mathfrak{q}$ for every $\mathfrak{q}\in \mathrm{Spec}(S)$ by [L8]. The contraction map $\mathrm{Spec}(S)\to \mathrm{Spec}(T)=\mathrm{Spec}(R)$ is a homeomorphism onto $D_R(t)$ by [L4] applied to $f=t$, which is the open set $U=D_T(t)$; this is exactly the configuration of [L9] with $n=1$, $g_1=t$ and a single principal open.

We compute the local fibres and quasi-finiteness. Let $\mathfrak{q}\in \mathrm{Spec}(S)$ and let $\mathfrak{p}=\mathfrak{q}\cap R$ be its contraction; by [L8] we have $t\mathrm{\notin }\mathfrak{p}$. Localizing $S=R_t$ at $\mathfrak{q}$ is the same as localizing $R$ at the image of $R\setminus \mathfrak{p}$, because $t\in R\setminus \mathfrak{p}$ and therefore the multiplicative set generated by $t$ and $R\setminus \mathfrak{p}$ is just $R\setminus \mathfrak{p}$, by [L6]. Every $s=r\mathrm{/}{t}^n\in S$ outside $\mathfrak{q}$ has $r\mathrm{\notin }\mathfrak{p}$, so it becomes a unit after localizing by $R\setminus \mathfrak{p}$; conversely every element of $R\setminus \mathfrak{p}$ maps outside $\mathfrak{q}$. Thus the two localizations have the same universal property and $S_{\mathfrak{q}}=(R_t{)}_{\mathfrak{q}}\cong R_{\mathfrak{p}},$ an isomorphism of $R$-algebras carrying the extension of $\mathfrak{p}$ to the extension of $\mathfrak{p}$. It follows that $S_{\mathfrak{q}}\mathrm{/}\mathfrak{p}{S}_{\mathfrak{q}}\cong R_{\mathfrak{p}}\mathrm{/}\mathfrak{p}{R}_{\mathfrak{p}}\cong \kappa (\mathfrak{p})$ by [L7]. This is finite over the field $\kappa (\mathfrak{p})$, of dimension one; by [L1] the map $R\to S$ is quasi-finite at $\mathfrak{q}$, and since $\mathfrak{q}$ was arbitrary and $R\to S$ is of finite type by step 1.1, the map is quasi-finite.

Part 1 is proved by steps 1.1, 2.1 and 4.1: the map is finite type and quasi-finite, the fibre over $(t)$ is empty with $S{\otimes }_R\kappa ((t))=0$, over every other prime the fibre ring is $\kappa (\mathfrak{p})$, and the local fibre at the prime above $\mathfrak{p}$ is that same field.

The Axiom of Choice was recorded in the Statement for the same reason it appears in [L9], namely the general factorization theorem invoked in step 3.2; the computations of steps 1.1, 2.1, 4.1, 2.2 and 3.2 manipulate finitely many explicit elements ($t$, $1\mathrm{/}t$, the fractions $r\mathrm{/}{t}^n$) and no family of nonempty sets is selected anywhere. This proves both parts. ∎