Semilocal completion decomposes into completed local factors

Distinct maximal ideals are pairwise comaximal. If $I+J=R$, expanding $(a+b{)}^{2n-1}=1$ for $a\in I$, $b\in J$, and $a+b=1$ shows that $I^n+J^n=R$; hence the powers ${\mathfrak{m}}_i^n$ are again pairwise comaximal. Applying [L2] first to the ${\mathfrak{m}}_i$ and then to their powers gives, for every $n\ge 1$, ${\mathfrak{m}}^n={\left({\prod }_{i=1}^t{\mathfrak{m}}_i\right)}^n={\prod }_{i=1}^t{\mathfrak{m}}_i^n={\bigcap }_{i=1}^t{\mathfrak{m}}_i^n$ and $R\mathrm{/}{\mathfrak{m}}^n\cong {\prod }_{i=1}^{t}R\mathrm{/}{\mathfrak{m}}_i^n\mathrm{.}$

Localizing $R\mathrm{/}{\mathfrak{m}}_i^n$ at $R\setminus {\mathfrak{m}}_i$ changes nothing. Indeed, if $s\mathrm{\notin }{\mathfrak{m}}_i$, maximality gives $a\in R$ and $u\in {\mathfrak{m}}_i$ with $as+u=1$, and $as(1+u+\cdots +u^{n-1})=1-u^n\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}{\mathfrak{m}}_i^n)\mathrm{.}$ Thus every such $s$ is already a unit modulo ${\mathfrak{m}}_i^n$, and $R\mathrm{/}{\mathfrak{m}}_i^n\cong R_{{\mathfrak{m}}_i}\mathrm{/}{\mathfrak{m}}_i^n{R}_{{\mathfrak{m}}_i}\mathrm{.}$ Taking inverse limits and using [L1] yields

A compatible tuple in the inverse limit of the finite products in step 2.1 is exactly a choice, for each $i$, of a compatible tuple in the $i$th quotient tower. Therefore inverse limit commutes with this finite product, and This is the claimed decomposition.