Coprime polynomial factorisations lift after an etale localisation

The universal coefficient algebra and the point. Put $A=R[b_1,\dots ,b_r,c_1,\dots ,c_s]$, $g_u=T^r+{\sum }_{i=1}^r{b}_i{T}^{r-i}$, $h_u=T^s+{\sum }_{j=1}^s{c}_j{T}^{s-j}$ and let ${\varphi }_1,\dots ,{\varphi }_n\in A$ be the coefficients of $g_u{h}_u-f$ in the basis $T^{n-1},\dots ,T,1$, so that $R^{\circ }=A\mathrm{/}({\varphi }_1,\dots ,{\varphi }_n)$ is a finitely presented $R$-algebra [F4] and $f=g_u{h}_u$ holds in $R^{\circ }[T]$ by construction. Substituting for $b_i,c_j$ the coefficients of $\bar{g},\bar{h}$ defines an $R$-algebra map $R^{\circ }\to \kappa (\mathfrak{p})$ because $\bar{g}\bar{h}=\bar{f}$ makes all ${\varphi }_k$ vanish; let ${\mathfrak{p}}_0$ be its kernel. The composite $R\to R^{\circ }\to \kappa (\mathfrak{p})$ is the canonical map, so ${\mathfrak{p}}_0$ lies over $\mathfrak{p}$, and $R^{\circ }\mathrm{/}{\mathfrak{p}}_0\to \kappa (\mathfrak{p})$ is injective by definition of the kernel and its image contains $R\mathrm{/}\mathfrak{p}$; because it lies in $\kappa (\mathfrak{p})=\mathrm{Frac}(R\mathrm{/}\mathfrak{p})$, the fraction field of $R^{\circ }\mathrm{/}{\mathfrak{p}}_0$ is exactly $\kappa ({\mathfrak{p}}_0)=\kappa (\mathfrak{p})$.

The Jacobian is a Sylvester matrix. Writing the coefficient vector of a polynomial of degree $\le n-1$ in the basis $T^{n-1},\dots ,T,1$, the $n\times n$ Jacobian matrix $J$ with entries $\mathrm{\partial }{\varphi }_k\mathrm{/}\mathrm{\partial }{b}_i$ and $\mathrm{\partial }{\varphi }_k\mathrm{/}\mathrm{\partial }{c}_j$ is the matrix of the $R$-linear map $(\delta b,\delta c)\mapsto (\delta g)h_u+g_u(\delta h)$, where $\delta g={\sum }_i\delta b_i{T}^{r-i}$ ranges over polynomials of degree $\le r-1$ and $\delta h={\sum }_j\delta c_j{T}^{s-j}$ over polynomials of degree $\le s-1$: this is the product rule applied to the coefficient functionals ${\varphi }_k$ of $g_u{h}_u$, the term $f$ contributing no derivatives. Reducing modulo ${\mathfrak{p}}_0$ gives the matrix $\bar{J}$ over the field $\kappa ({\mathfrak{p}}_0)=\kappa (\mathfrak{p})$ of the map $(\delta g,\delta h)\mapsto (\delta g)\bar{h}+\bar{g}(\delta h)$.

Invertibility of the reduced Jacobian. Let $\delta g,\delta h$ over $\kappa (\mathfrak{p})$ satisfy $(\delta g)\bar{h}+\bar{g}(\delta h)=0$ with $\mathrm{deg}\delta g\le r-1$ and $\mathrm{deg}\delta h\le s-1$. Using the Bezout identity $\bar{a}\bar{g}+\bar{b}\bar{h}=1$, one has $\delta h=\delta h(\bar{a}\bar{g}+\bar{b}\bar{h})=\bar{a}(\delta h\bar{g})+\bar{b}(\delta h\bar{h})=\bar{h}(\bar{b}\delta h-\bar{a}\delta g)$, so $\bar{h}$ divides $\delta h$; since $\mathrm{deg}\delta h<s=\mathrm{deg}\bar{h}$ and $\bar{h}\mathrm{\ne }0$, this forces $\delta h=0$. Then $\bar{g}(\delta h)=0$ gives $(\delta g)\bar{h}=0$, and since $\kappa (\mathfrak{p})[T]$ is a domain with $\bar{h}\mathrm{\ne }0$ we get $\delta g=0$. Hence $\bar{J}$ has zero kernel between spaces of dimension $n$, so it is invertible and $\det (\bar{J})\mathrm{\ne }0$ by [F5]. Therefore $d:=\det J\in R^{\circ }$ has nonzero image in $\kappa ({\mathfrak{p}}_0)=\kappa (\mathfrak{p})$ and $d\mathrm{\notin }{\mathfrak{p}}_0$.

Localisation at the determinant. Put $R^{\mathrm{\prime }}=(R^{\circ }{)}_d$ and ${\mathfrak{p}}^{\mathrm{\prime }}={\mathfrak{p}}_0{R}^{\mathrm{\prime }}$, a prime of $R^{\mathrm{\prime }}$ over $\mathfrak{p}$ because $d\mathrm{\notin }{\mathfrak{p}}_0$ [F3]. In $R^{\mathrm{\prime }}$ the element $d$ is a unit, $\kappa ({\mathfrak{p}}^{\mathrm{\prime }})=\mathrm{Frac}((R^{\circ }\mathrm{/}{\mathfrak{p}}_0{)}_d)=\mathrm{Frac}(R^{\circ }\mathrm{/}{\mathfrak{p}}_0)=\kappa (\mathfrak{p})$ by step 1.1, since localizing a domain at a nonzero element does not change its fraction field, and the equation $f=g_u{h}_u$ of step 1.1 persists in $R^{\mathrm{\prime }}[T]$ with the images of $g_u,h_u$, which are monic of degrees $r,s$ because $R^{\mathrm{\prime }}\mathrm{\ne }0$ and leading coefficient $1$ remains a unit. This gives claim 2 in the form $g=g_u$, $h=h_u$ inside $R^{\mathrm{\prime }}$.

'Etaleness on the whole chart. The algebra $R^{\mathrm{\prime }}$ is the localisation at $d$ of the finitely presented $R$-algebra $R^{\circ }=A\mathrm{/}({\varphi }_1,\dots ,{\varphi }_n)$, hence is finitely presented over $R$ [F4], and the full $n\times n$ Jacobian determinant is the unit $d$ of $R^{\mathrm{\prime }}$; the presentation $R^{\mathrm{\prime }}=(R[b_1,\dots ,b_r,c_1,\dots ,c_s]\mathrm{/}({\varphi }_1,\dots ,{\varphi }_n){)}_d$ therefore has an invertible $n\times n$ minor with $m=n$ variables and $n$ equations. By the Jacobian criterion [F1] (AC) the morphism $\mathrm{Spec}{R}^{\mathrm{\prime }}\to \mathrm{Spec}R$ is smooth of relative dimension $n-n=0$ at every prime of $R^{\mathrm{\prime }}$, so it is 'etale everywhere by [F2]. This gives claim 1.

Coprimality of the lifted factors. Since $\det J=d$ is a unit of $R^{\mathrm{\prime }}$, the matrix $J$ is invertible over $R^{\mathrm{\prime }}$ with inverse $d^{-1}\mathrm{adj}(J)$ [F5]. In the identification of step 1.2 the same matrix $J$ represents the $R^{\mathrm{\prime }}$-linear map $(b,a)\mapsto ag+bh$, where $a$ runs over the polynomials of degree $\le s-1$ (coefficients the images of $c_1,\dots ,c_s$) and $b$ over the polynomials of degree $\le r-1$ (coefficients the images of $b_1,\dots ,b_r$), into the polynomials of degree $\le n-1$; invertibility of $J$ makes this map surjective, so there are $a,b\in R^{\mathrm{\prime }}[T]$ with $ag+bh=1$; in particular $g$ and $h$ are coprime in $R^{\mathrm{\prime }}[T]$.

Conclusion and choice accounting. Steps 1.1, 1.2 and 2.1 construct $R^{\mathrm{\prime }},{\mathfrak{p}}^{\mathrm{\prime }}$ with $\kappa ({\mathfrak{p}}^{\mathrm{\prime }})=\kappa (\mathfrak{p})$, step 4.1 gives the 'etaleness of claim 1 and steps 3.1 and 4.2 the factorisation with coprime monic factors of claim 2. The Axiom of Choice [F6] is assumed in the Statement and used exactly through the Jacobian criterion [F1] in step 4.1; the coefficient construction, the residue-field identification and the linear algebra of steps 1.2, 2.1 and 4.2 are choice-free. [F1, F6, step 2.1, step 3.1, step 4.1, step 4.2] $\mathrm{\square }$