The Stacks project

Proof. We replace $S$ by $\mathop{\mathrm{Spec}}(\mathcal{O}_{S, s})$ and $T$ by the base change to $\mathop{\mathrm{Spec}}(\mathcal{O}_{S, s})$. Then $T$ is locally Noetherian and hence $\mathcal{O}_{T, t}$ is Noetherian. Set $A = \mathcal{O}_{S, s}^\wedge $, $\mathfrak m = \mathfrak m_ A$, and $B = \mathcal{O}_{T, t}^\wedge $. By More on Algebra, Lemma 15.43.8 we see that $A \to B$ is flat. Since $\mathcal{O}_{T, t}/\mathfrak m_ s \mathcal{O}_{T, t} = \mathcal{O}_{T_ s, t}$ we see that $B/\mathfrak m B = \mathcal{O}_{T_ s, t}^\wedge $. By assumption (4) and Lemma 53.19.11 we conclude there exist $\overline{u}, \overline{v} \in B/\mathfrak m B$ such that the map

is surjective with kernel $(xy)$.

Assume we have $n \geq 1$ and $u, v \in B$ mapping to $\overline{u}, \overline{v}$ such that

for some $h \in A$ and $\delta \in \mathfrak m^ nB$. We claim that there exist $u', v' \in B$ with $u - u', v - v' \in \mathfrak m^ n B$ such that

for some $h' \in A$ and $\delta ' \in \mathfrak m^{n + 1}B$. To see this, write $\delta = \sum f_ i b_ i$ with $f_ i \in \mathfrak m^ n$ and $b_ i \in B$. Then write $b_ i = a_ i + u b_{i, 1} + v b_{i, 2} + \delta _ i$ with $a_ i \in A$, $b_{i, 1}, b_{i, 2} \in B$ and $\delta _ i \in \mathfrak m B$. This is possible because the residue field of $B$ agrees with the residue field of $A$ and the images of $u$ and $v$ in $B/\mathfrak m B$ generate the maximal ideal. Then we set

and we obtain

for some $c_{i, j} \in B$. Thus we get a formula as above with $h' = h + \sum a_ if_ i$ and $\delta ' = \sum f_ i \delta _ i + \sum c_{ij}f_ if_ j$.

Arguing by induction and starting with any lifts $u_1, v_1 \in B$ of $\overline{u}, \overline{v}$ the result of the previous paragraph shows that we find a sequence of elements $u_ n, v_ n \in B$ and $h_ n \in A$ such that $u_ n - u_{n + 1} \in \mathfrak m^ n B$, $v_ n - v_{n + 1} \in \mathfrak m^ n B$, $h_ n - h_{n + 1} \in \mathfrak m^ n$, and such that $u_ n v_ n - h_ n \in \mathfrak m^ n B$. Since $A$ and $B$ are complete we can set $u_\infty = \mathop{\mathrm{lim}}\nolimits u_ n$, $v_\infty = \mathop{\mathrm{lim}}\nolimits v_ n$, and $h_\infty = \mathop{\mathrm{lim}}\nolimits h_ n$, and then we obtain $u_\infty v_\infty = h_\infty $ in $B$. Thus we have an $A$-algebra map

sending $x$ to $u_\infty $ and $v$ to $v_\infty $. This is a map of flat $A$-algebras which is an isomorphism after dividing by $\mathfrak m$. It is surjective modulo $\mathfrak m$ and hence surjective by completeness and Algebra, Lemma 10.96.1. Then we can apply Algebra, Lemma 10.99.1 to conclude it is an isomorphism. $\square$