There is a version of formal smoothness which applies to homomorphisms of topological rings.

Definition 15.37.1. Let $R \to S$ be a homomorphism of topological rings with $R$ and $S$ linearly topologized. We say $S$ is *formally smooth over $R$* if for every commutative solid diagram

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

of homomorphisms of topological rings where $A$ is a discrete ring and $J \subset A$ is an ideal of square zero, a dotted arrow exists which makes the diagram commute.

We will mostly use this notion when given ideals $\mathfrak m \subset R$ and $\mathfrak n \subset S$ and we endow $R$ with the $\mathfrak m$-adic topology and $S$ with the $\mathfrak n$-adic topology. Continuity of $\varphi : R \to S$ holds if and only if $\varphi (\mathfrak m^ m) \subset \mathfrak n$ for some $m \geq 1$, see Lemma 15.36.2. It turns out that in this case only the topology on $S$ is relevant.

Lemma 15.37.2. Let $\varphi : R \to S$ be a ring map.

If $R \to S$ is formally smooth in the sense of Algebra, Definition 10.138.1, then $R \to S$ is formally smooth for any linear topology on $R$ and any pre-adic topology on $S$ such that $R \to S$ is continuous.

Let $\mathfrak n \subset S$ and $\mathfrak m \subset R$ ideals such that $\varphi $ is continuous for the $\mathfrak m$-adic topology on $R$ and the $\mathfrak n$-adic topology on $S$. Then the following are equivalent

$\varphi $ is formally smooth for the $\mathfrak m$-adic topology on $R$ and the $\mathfrak n$-adic topology on $S$, and

$\varphi $ is formally smooth for the discrete topology on $R$ and the $\mathfrak n$-adic topology on $S$.

**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 $m \geq 1$. 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.
$\square$

Definition 15.37.3. Let $R \to S$ be a ring map. Let $\mathfrak n \subset S$ be an ideal. If the equivalent conditions (2)(a) and (2)(b) of Lemma 15.37.2 hold, then we say $R \to S$ is *formally smooth for the $\mathfrak n$-adic topology*.

This property is inherited by the completions.

Lemma 15.37.4. Let $(R, \mathfrak m)$ and $(S, \mathfrak n)$ be rings endowed with finitely generated ideals. Endow $R$ and $S$ with the $\mathfrak m$-adic and $\mathfrak n$-adic topologies. Let $R \to S$ be a homomorphism of topological rings. The following are equivalent

$R \to S$ is formally smooth for the $\mathfrak n$-adic topology,

$R \to S^\wedge $ is formally smooth for the $\mathfrak n^\wedge $-adic topology,

$R^\wedge \to S^\wedge $ is formally smooth for the $\mathfrak n^\wedge $-adic topology.

Here $R^\wedge $ and $S^\wedge $ are the $\mathfrak m$-adic and $\mathfrak n$-adic completions of $R$ and $S$.

**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.
$\square$

The advantage of working with adic rings is that one gets a stronger lifting property.

Lemma 15.37.5. 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 15.36.2. Hence $\varphi (\mathfrak n^ r) \subset J + I$ hence $\varphi (\mathfrak n^{rt}) \subset I$ as desired.
$\square$

Lemma 15.37.6. Let $R \to S$ be a ring map. Let $\mathfrak n \subset \mathfrak n' \subset S$ be ideals. If $R \to S$ is formally smooth for the $\mathfrak n$-adic topology, then $R \to S$ is formally smooth for the $\mathfrak n'$-adic topology.

**Proof.**
Omitted.
$\square$

Lemma 15.37.7. A composition of formally smooth continuous homomorphisms of linearly topologized rings is formally smooth.

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

Lemma 15.37.8. Let $R$, $S$ be rings. Let $\mathfrak n \subset S$ be an ideal. Let $R \to S$ be formally smooth for the $\mathfrak n$-adic topology. Let $R \to R'$ be any ring map. Then $R' \to S' = S \otimes _ R R'$ is formally smooth in the $\mathfrak n' = \mathfrak nS'$-adic topology.

**Proof.**
Let a solid diagram

\[ \xymatrix{ S \ar[r] \ar@{-->}[rrd] & S' \ar[r] \ar@{-->}[rd] & A/J \\ R \ar[u] \ar[r] & R' \ar[r] \ar[u] & A \ar[u] } \]

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.
$\square$

We have seen descent for formal smoothness along faithfully flat ring maps in Algebra, Lemma 10.138.16. Something similar holds in the current setting of topological rings. However, here we just prove the following very simple and easy to prove version which is already quite useful.

Lemma 15.37.9. Let $R$, $S$ be rings. Let $\mathfrak n \subset S$ be an ideal. Let $R \to R'$ be a ring map. Set $S' = S \otimes _ R R'$ and $\mathfrak n' = \mathfrak nS$. If

the map $R \to R'$ embeds $R$ as a direct summand of $R'$ as an $R$-module, and

$R' \to S'$ is formally smooth for the $\mathfrak n'$-adic topology,

then $R \to S$ is formally smooth in the $\mathfrak n$-adic topology.

**Proof.**
Let a solid diagram

\[ \xymatrix{ S \ar[r] & A/J \\ R \ar[u] \ar[r] & A \ar[u] } \]

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

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

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

\[ J'' = \mathop{\mathrm{Im}}(J \otimes _ R C \to A \otimes _ R C) \subset J' \subset A'. \]

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.
$\square$

## Comments (2)

Comment #6453 by Yijin Wang on

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