Completion of a Noetherian local ring is local with the same residue field

Since $R$ is local, its unique maximal ideal $\mathfrak{m}$ equals $J(R)$. Hence [L2] applies and shows that $R\to \widehat{R}$ is faithfully flat.

By [L1], the ring $\widehat{R}$ is Noetherian. By [L3], $\widehat{R}\mathrm{/}\mathfrak{m}\widehat{R}\cong R\mathrm{/}\mathfrak{m},$ and the right-hand side is a field because $R$ is local. Thus $\mathfrak{m}\widehat{R}$ is a maximal ideal of $\widehat{R}$.

For each $n\ge 1$, part 3 of [L3] gives $\widehat{R}\mathrm{/}{\mathfrak{m}}^n\widehat{R}\cong R\mathrm{/}{\mathfrak{m}}^n\mathrm{.}$ Therefore the canonical map identifies with the identity of So $\widehat{R}$ is complete for the $\mathfrak{m}\widehat{R}$-adic topology.

Let $x\in \widehat{R}$ with $x\mathrm{\notin }\mathfrak{m}\widehat{R}$. Its residue class in $\widehat{R}\mathrm{/}\mathfrak{m}\widehat{R}$ is then nonzero, hence a unit. Choose $y\in \widehat{R}$ with $xy\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}\mathfrak{m}\widehat{R})\mathrm{.}$ By [L4] and step 1.3, the element $xy$ is a unit, hence $x$ is a unit. Therefore every nonunit lies in $\mathfrak{m}\widehat{R}$, so $\mathfrak{m}\widehat{R}$ is the unique maximal ideal of $\widehat{R}$.

Step 1.2 proves the residue-field isomorphism, and steps 1.1, 1.3, and 2.1 prove that $\widehat{R}$ is Noetherian local with maximal ideal $\mathfrak{m}\widehat{R}$ and that $R\to \widehat{R}$ is faithfully flat.