The Stacks project

Lemma 15.36.2. Let $\varphi : R \to S$ be a ring map.

  1. If $R \to S$ is formally smooth in the sense of Algebra, Definition 10.136.1, 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.

  2. 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

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

    2. $\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 10.136.1. 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 15.36.1. Continuity of $S \to A/J$ means that $\mathfrak n^ k$ maps to zero in $A/J$ for some $k \geq 1$, see Lemma 15.35.2. 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 15.35.2. 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 15.36.1 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. $\square$


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