The Stacks project

Proof. Assume $R \to S$ is formally smooth in the sense of Algebra, Definition 10.138.1. If $S$ has a pre-adic topology, then there exists an ideal $\mathfrak n \subset S$ such that $S$ has the $\mathfrak n$-adic topology. Suppose given a solid commutative diagram as in Definition 15.37.1. Continuity of $S \to A/J$ means that $\mathfrak n^ k$ maps to zero in $A/J$ for some $k \geq 1$, see Lemma 15.36.2. We obtain a ring map $\psi : S \to A$ from the assumed formal smoothness of $S$ over $R$. Then $\psi (\mathfrak n^ k) \subset J$ hence $\psi (\mathfrak n^{2k}) = 0$ as $J^2 = 0$. Hence $\psi $ is continuous by Lemma 15.36.2. This proves (1).

The proof of (2)(b) $\Rightarrow $ (2)(a) is the same as the proof of (1). Assume (2)(a). Suppose given a solid commutative diagram as in Definition 15.37.1 where we use the discrete topology on $R$. Since $\varphi $ is continuous we see that $\varphi (\mathfrak m^ n) \subset \mathfrak n$ for some $n \geq 1$. As $S \to A/J$ is continuous we see that $\mathfrak n^ k$ maps to zero in $A/J$ for some $k \geq 1$. Hence $\mathfrak m^{nk}$ maps into $J$ under the map $R \to A$. Thus $\mathfrak m^{2nk}$ maps to zero in $A$ and we see that $R \to A$ is continuous in the $\mathfrak m$-adic topology. Thus (2)(a) gives a dotted arrow as desired. $\square$

Proof. The assumption that $\mathfrak m$ is finitely generated implies that $R^\wedge $ is $\mathfrak mR^\wedge $-adically complete, that $\mathfrak mR^\wedge = \mathfrak m^\wedge $ and that $R^\wedge /\mathfrak m^ nR^\wedge = R/\mathfrak m^ n$, see Algebra, Lemma 10.96.3 and its proof. Similarly for $(S, \mathfrak n)$. Thus it is clear that diagrams as in Definition 15.37.1 for the cases (1), (2), and (3) are in 1-to-1 correspondence. $\square$

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 $A_{n, m} = A/J^ n + I^ m$:

Note that each of the commutative squares defines a surjection

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 $\psi (\mathfrak n^ r) \subset J + I/J$ for some $r \geq 1$ by our assumption that $\psi $ is a morphism of topological rings, see Lemma 15.36.2. Hence $\varphi (\mathfrak n^ r) \subset J + I$ hence $\varphi (\mathfrak n^{rt}) \subset I$ as desired. $\square$

Proof. Omitted. $\square$

Proof. Omitted. (Hint: This is completely formal, and follows from considering a suitable diagram.) $\square$

Proof. Let a solid diagram

as in Definition 15.37.1 be given. Then the composition $S \to S' \to A/J$ is continuous. By assumption the longer dotted arrow exists. By the universal property of tensor product we obtain the shorter dotted arrow. $\square$

Proof. Let a solid diagram

as in Definition 15.37.1 be given. Set $A' = A \otimes _ R R'$ and $J' = \mathop{\mathrm{Im}}(J \otimes _ R R' \to A')$. The base change of the diagram above is the diagram

with continuous arrows. By condition (2) we obtain the dotted arrow $\psi ' : S' \to A'$. Using condition (1) choose a direct summand decomposition $R' = R \oplus C$ as $R$-modules. (Warning: $C$ isn't an ideal in $R'$.) Then $A' = A \oplus A \otimes _ R C$. Set

Then $J' = J \oplus J''$ as $A$-modules. The image of the composition $\psi : S \to A'$ of $\psi '$ with $S \to S'$ is contained in $A + J' = A \oplus J''$. However, in the ring $A + J' = A \oplus J''$ the $A$-submodule $J''$ is an ideal! (Use that $J^2 = 0$.) Hence the composition $S \to A + J' \to (A + J')/J'' = A$ is the arrow we were looking for. $\square$