Lemma 15.36.5. Let $R \to S$ be a ring map. Let $\mathfrak n$ be an ideal of $S$. Assume that $R \to S$ is formally smooth in the $\mathfrak n$-adic topology. Consider a solid commutative diagram

$\xymatrix{ S \ar[r]_\psi \ar@{-->}[rd] & A/J \\ R \ar[r] \ar[u] & A \ar[u] }$

of homomorphisms of topological rings where $A$ is adic and $A/J$ is the quotient (as topological ring) of $A$ by a closed ideal $J \subset A$ such that $J^ t$ is contained in an ideal of definition of $A$ for some $t \geq 1$. Then there exists a dotted arrow in the category of topological rings which makes the diagram commute.

Proof. Let $I \subset A$ be an ideal of definition so that $I \supset J^ t$ for some $n$. Then $A = \mathop{\mathrm{lim}}\nolimits A/I^ n$ and $A/J = \mathop{\mathrm{lim}}\nolimits A/J + I^ n$ because $J$ is assumed closed. Consider the following diagram of discrete $R$ algebras $A_{n, m} = A/J^ n + I^ m$:

$\xymatrix{ A/J^3 + I^3 \ar[r] \ar[d] & A/J^2 + I^3 \ar[r] \ar[d] & A/J + I^3 \ar[d] \\ A/J^3 + I^2 \ar[r] \ar[d] & A/J^2 + I^2 \ar[r] \ar[d] & A/J + I^2 \ar[d] \\ A/J^3 + I \ar[r] & A/J^2 + I \ar[r] & A/J + I }$

Note that each of the commutative squares defines a surjection

$A_{n + 1, m + 1} \longrightarrow A_{n + 1, m} \times _{A_{n, m}} A_{n, m + 1}$

of $R$-algebras whose kernel has square zero. We will inductively construct $R$-algebra maps $\varphi _{n, m} : S \to A_{n, m}$. Namely, we have the maps $\varphi _{1, m} = \psi \bmod J + I^ m$. Note that each of these maps is continuous as $\psi$ is. We can inductively choose the maps $\varphi _{n, 1}$ by starting with our choice of $\varphi _{1, 1}$ and lifting up, using the formal smoothness of $S$ over $R$, along the right column of the diagram above. We construct the remaining maps $\varphi _{n, m}$ by induction on $n + m$. Namely, we choose $\varphi _{n + 1, m + 1}$ by lifting the pair $(\varphi _{n + 1, m}, \varphi _{n, m + 1})$ along the displayed surjection above (again using the formal smoothness of $S$ over $R$). In this way all of the maps $\varphi _{n, m}$ are compatible with the transition maps of the system. As $J^ t \subset I$ we see that for example $\varphi _ n = \varphi _{nt, n} \bmod I^ n$ induces a map $S \to A/I^ n$. Taking the limit $\varphi = \mathop{\mathrm{lim}}\nolimits \varphi _ n$ we obtain a map $S \to A = \mathop{\mathrm{lim}}\nolimits A/I^ n$. The composition into $A/J$ agrees with $\psi$ as we have seen that $A/J = \mathop{\mathrm{lim}}\nolimits A/J + I^ n$. Finally we show that $\varphi$ is continuous. Namely, we know that $\psi (\mathfrak n^ r) \subset J + I^ r/J$ for some $r$ by our assumption that $\psi$ is a morphism of topological rings, see Lemma 15.35.2. Hence $\varphi (\mathfrak n^ r) \subset J + I$ hence $\varphi (\mathfrak n^{rt}) \subset I$ as desired. $\square$

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