Étale maps induce completion isomorphisms at equal-residue points

Flatness and the fibre local ring. By [F1], $\phi$ is flat at $x^{\mathrm{\prime }}$. Put $s=\phi (x^{\mathrm{\prime }})$. The local ring of the fibre at $x^{\mathrm{\prime }}$ is ${\mathcal{O}}_{X^{\mathrm{\prime }},x^{\mathrm{\prime }}}\mathrm{/}{\mathfrak{m}}_s{\mathcal{O}}_{X^{\mathrm{\prime }},x^{\mathrm{\prime }}}$ by [F2]; it is regular because the geometric fibre is regular and has dimension zero by [F1], so it is a field by [F2]. Therefore ${\mathfrak{m}}_x{\mathcal{O}}_{X^{\mathrm{\prime }},x^{\mathrm{\prime }}}={\mathfrak{m}}_s{\mathcal{O}}_{X^{\mathrm{\prime }},x^{\mathrm{\prime }}}$ is a maximal ideal of ${\mathcal{O}}_{X^{\mathrm{\prime }},x^{\mathrm{\prime }}}$, hence equals its unique maximal ideal ${\mathfrak{m}}_{x^{\mathrm{\prime }}}$ by [F3]. This proves (1) and (2).

Forward filtration detection. If $I\subseteq {\mathfrak{m}}_x^n$, extension of ideals and step 1.1 give $I{\mathcal{O}}_{X^{\mathrm{\prime }},x^{\mathrm{\prime }}}\subseteq {\mathfrak{m}}_x^n{\mathcal{O}}_{X^{\mathrm{\prime }},x^{\mathrm{\prime }}}=({\mathfrak{m}}_x{\mathcal{O}}_{X^{\mathrm{\prime }},x^{\mathrm{\prime }}}{)}^n={\mathfrak{m}}_{x^{\mathrm{\prime }}}^n\mathrm{.}$ Thus the forward implication in (3) holds.

Reverse filtration detection. Put $A={\mathcal{O}}_{X,x}$, $B={\mathcal{O}}_{X^{\mathrm{\prime }},x^{\mathrm{\prime }}}$, $\mathfrak{m}={\mathfrak{m}}_x$, and $\mathfrak{n}={\mathfrak{m}}_{x^{\mathrm{\prime }}}$. Assume $IB\subseteq {\mathfrak{n}}^N$. If some $a\in I$ were not in ${\mathfrak{m}}^N$, choose the largest $k<N$ with $a\in {\mathfrak{m}}^k$. Its class in the finite-dimensional $\kappa (x)$-vector space ${\mathfrak{m}}^k\mathrm{/}{\mathfrak{m}}^{k+1}$ is nonzero. Flatness in step 1.1 identifies $({\mathfrak{m}}^k\mathrm{/}{\mathfrak{m}}^{k+1}){\otimes }_{\kappa (x)}\kappa (x^{\mathrm{\prime }})$ with ${\mathfrak{n}}^k\mathrm{/}{\mathfrak{n}}^{k+1}$; extension of scalars along a field extension is injective on a finite-dimensional vector space, so the class of $a$ remains nonzero there. But $a\in IB\subseteq {\mathfrak{n}}^N\subseteq {\mathfrak{n}}^{k+1}$, a contradiction. Thus every $a\in I$ lies in ${\mathfrak{m}}^N$, proving the reverse implication in (3).

Completion when residue fields agree. Assume $\kappa (x)\to \kappa (x^{\mathrm{\prime }})$ is an isomorphism. By step 1.1, ${\mathfrak{m}}_x{\mathcal{O}}_{X^{\mathrm{\prime }},x^{\mathrm{\prime }}}={\mathfrak{m}}_{x^{\mathrm{\prime }}}$, so the induced map on residue fields and the degree-zero associated graded pieces is an isomorphism. For every $k\ge 0$, flatness from [F1] identifies ${\mathfrak{m}}_x^k\mathrm{/}{\mathfrak{m}}_x^{k+1}{\otimes }_{\kappa (x)}\kappa (x^{\mathrm{\prime }})\cong {\mathfrak{m}}_{x^{\mathrm{\prime }}}^k\mathrm{/}{\mathfrak{m}}_{x^{\mathrm{\prime }}}^{k+1};$ since the residue fields agree, the map of associated graded pieces is an isomorphism in every degree. The exact sequences $0\to {\mathfrak{m}}_x^{q-1}\mathrm{/}{\mathfrak{m}}_x^q\to {\mathcal{O}}_{X,x}\mathrm{/}{\mathfrak{m}}_x^q\to {\mathcal{O}}_{X,x}\mathrm{/}{\mathfrak{m}}_x^{q-1}\to 0$ and their analogues for ${\mathcal{O}}_{X^{\mathrm{\prime }},x^{\mathrm{\prime }}}$ show by induction on $q$ that the induced map on each finite quotient by the $q$th power is an isomorphism. Taking inverse limits using [F4] proves the asserted isomorphism of completions.