Total fractions split over the branches of a reduced hypersurface

Write $D_i:=\mathcal{O}\mathrm{/}(q_i)$ for each $i$ and $R:={\prod }_{i=1}^r{D}_i$. Write $a^{(i)}$ for the $i$-th component in $D_i$ of a class $a\in A$. The quotient map $\psi :A\to R$, $h+(f)\mapsto (h+(q_1),\dots ,h+(q_r))$, is well defined by [F2] and [F3], and its kernel is $({\bigcap }_i(q_i))\mathrm{/}(f)$. By [F2] and [F3], an element $h$ lies in every $(q_i)$ exactly when each $q_i$ divides $h$, and since the $q_i$ are pairwise nonassociate irreducibles this happens exactly when $q_1\cdots q_r=f$ divides $h$; hence ${\bigcap }_i(q_i)=(f)$ and $\psi$ is injective. Each $D_i$ is a domain by [F3], so the fraction fields $K_i:=\mathrm{Frac}(D_i)$ exist.

For each $i$ let ${\sigma }_i\in A$ be the class of ${\prod }_{j\mathrm{\ne }i}{q}_j$. Its $i$-th component ${\sigma }_i^{(i)}$ is nonzero in $D_i$ — it is a product of the nonzero classes of the $q_j$, $j\mathrm{\ne }i$, in the domain $D_i$ by [F3] and [F4] — and its $j$-th component vanishes for every $j\mathrm{\ne }i$. The sum $s:={\sigma }_1+\cdots +{\sigma }_r$ therefore has $s^{(i)}={\sigma }_i^{(i)}\mathrm{\ne }0$ for every $i$. An element $x\in A$ has all components nonzero in $R$ if and only if it is a nonzerodivisor: if $x^{(i)}=0$ for some $i$, then $x{\sigma }_i=0$ with ${\sigma }_i\mathrm{\ne }0$, so $x$ is a zerodivisor; conversely, if $x^{(i)}\mathrm{\ne }0$ for all $i$ and $xy=0$ for some $y\in A$, then $x^{(i)}{y}^{(i)}=0$ in the domain $D_i$ gives $y^{(i)}=0$ for every $i$, hence $y=0$. Therefore $S=\{x\in A:x^{(i)}\mathrm{\ne }0\text{ for all }i\}$, and in particular $s\in S$.

In $Q(A)$ the element $s$ is invertible, and the elements $e_i:={\sigma }_i\mathrm{/}s$ are orthogonal idempotents with ${\sum }_i{e}_i=1$: componentwise in $R$ one has ${\sigma }_i^2={\sigma }_{i}s$ and ${\sigma }_i{\sigma }_j=0$ for $i\mathrm{\ne }j$, while ${\sum }_i{\sigma }_i=s$, so $e_i^2=e_i$, $e_i{e}_j=0$ and ${\sum }_i{e}_i=1$ in $Q(A)$ by the arithmetic of the localisation.

Define $\mathrm{\Phi }:Q(A)\to R^{\mathrm{\prime }}:={\prod }_{i=1}^r{K}_i$ by $\mathrm{\Phi }(a\mathrm{/}s):=(a^{(1)}\mathrm{/}{s}^{(1)},\dots ,a^{(r)}\mathrm{/}{s}^{(r)})$. This is well defined: if $a\mathrm{/}s=a^{\mathrm{\prime }}\mathrm{/}{s}^{\mathrm{\prime }}$ in $Q(A)$, then $u(a{s}^{\mathrm{\prime }}-a^{\mathrm{\prime }}s)=0$ in $A$ for some $u\in S$, and applying the injective map of step 1.1 componentwise gives $u^{(i)}(a^{(i)}{s}^{\mathrm{\prime }(i)}-a^{\mathrm{\prime }(i)}{s}^{(i)})=0$ in the domain $D_i$ with $u^{(i)}\mathrm{\ne }0$, hence $a^{(i)}\mathrm{/}{s}^{(i)}=a^{\mathrm{\prime }(i)}\mathrm{/}{s}^{\mathrm{\prime }(i)}$ in $K_i$. The map $\mathrm{\Phi }$ is a ring homomorphism, and it is injective: if $\mathrm{\Phi }(a\mathrm{/}s)=0$, then $a^{(i)}=0$ for every $i$, so $a=0$ by injectivity of $\psi$, and $a\mathrm{/}s=0$ in the localisation.

$\mathrm{\Phi }$ is surjective. Let $(y_1,\dots ,y_r)\in R^{\mathrm{\prime }}$, and for each $i$ write $y_i=X_i\mathrm{/}{D}_i$ with $X_i,D_i\in D_i$ and $D_i\mathrm{\ne }0$; since $\psi$ is surjective onto $D_i$, choose lifts $X,D\in A$ of $X_i$ and $D_i$. Put $u_i:=D{\sigma }_i+(s-{\sigma }_i)\in A$ and $z_i:=X{\sigma }_i\mathrm{/}{u}_i\in Q(A)$. The element $u_i$ has all components nonzero: $u_i^{(i)}=D^{(i)}{\sigma }_i^{(i)}\mathrm{\ne }0$ in the domain $D_i$, and for $j\mathrm{\ne }i$ one has $u_i^{(j)}=(s-{\sigma }_i{)}^{(j)}={\sigma }_j^{(j)}\mathrm{\ne }0$; hence $u_i\in S$ by step 2.1 and $z_i$ is a legitimate fraction. Moreover $\mathrm{\Phi }(z_i)$ has $i$-th component $\left(X^{(i)}{\sigma }_i^{(i)}\right)/\left(D^{(i)}{\sigma }_i^{(i)}\right)=X_i\mathrm{/}{D}_i=y_i$ and vanishes in every component $j\mathrm{\ne }i$, because the numerator $X{\sigma }_i$ has vanishing $j$-th component. Therefore $\mathrm{\Phi }\left({\sum }_i{z}_i\right)=(y_1,\dots ,y_r)$, so $\mathrm{\Phi }$ is surjective.

Together with step 3.2 this makes $\mathrm{\Phi }$ an isomorphism $Q(A)\cong {\prod }_i{K}_i={\prod }_i\mathrm{Frac}(\mathcal{O}\mathrm{/}(q_i))$. For $r=1$ we have $f=q_1$, $A=\mathcal{O}\mathrm{/}(q_1)$ is a domain by [F3], $R=D_1$, and $\mathrm{\Phi }$ identifies $Q(A)$ with its fraction field, the ordinary case of the total quotient ring.