Lemma 88.23.8. In the situation above. If $X$ is locally Noetherian, $f$ is locally of finite type, and $U' \to U$ is smooth, then $f_{/T}$ is rig-smooth.

Proof. The strategy of the proof is this: reduce to the case where $X$ and $X'$ are affine, translate the affine case into algebra, and finally apply Lemma 88.4.3. We urge the reader to skip the details.

Choose a surjective étale morphism $W \to X$ with $W = \coprod W_ i$ a disjoint union of affine schemes, see Properties of Spaces, Lemma 66.6.1. For each $i$ choose a surjective étale morphism $W'_ i \to W_ i \times _ X X'$ where $W'_ i = \coprod W'_{ij}$ is a disjoint union of affines. In particular $\coprod W'_{ij} \to X'$ is surjective and étale. Denote $f_{ij} : W_{ij} \to W_ i$ the given morphism. Denote $T_ i \subset W_ i$ and $T'_{ij} \subset W_{ij}$ the inverse images of $T$. Since taking the completion along the inverse image of $T$ produces cartesian diagrams (see above) we have $(W_ i)_{/T_ i} = W_ i \times _ X X_{/T}$ and similarly $(W'_{ij})_{/T'_{ij}} = W'_{ij} \times _{X'} X'_{/T'}$. Moreover, recall that $(W_ i)_{/T_ i}$ and $(W'_{ij})_{/T'_{ij}}$ are affine formal algebraic spaces. Hence $\{ W'_{ij})_{/T'_{ij}} \to X'_{/T'}\}$ is a covering as in Formal Spaces, Definition 87.11.1. By Lemma 88.18.2 we see that it suffices to prove that

$(W'_{ij})_{/T'_{ij}} \longrightarrow (W_ i)_{/T_ i}$

is rig-smooth. Observe that $W'_{ij} \to W_ i$ is locally of finite type and induces a smooth morphism $W'_{ij} \setminus T'_{ij} \to W_ i \setminus T_ i$ (as this is true for $f$ and these properties of morphisms are étale local on the source and target). Observe that $W_ i$ is locally Noetherian (as $X$ is locally Noetherian and this property is étale local on the algebraic space). Hence it suffices to prove the lemma when $X$ and $X'$ are affine schemes.

Assume $X = \mathop{\mathrm{Spec}}(A)$ and $X' = \mathop{\mathrm{Spec}}(A')$ are affine schemes. Since $X$ is Noetherian, we see that $A$ is Noetherian. The morphism $f$ is given by a ring map $A \to A'$ of finite type. Let $I \subset A$ be an ideal cutting out $T$. Then $IA'$ cuts out $T'$. Also $\mathop{\mathrm{Spec}}(A') \to \mathop{\mathrm{Spec}}(A)$ is smooth over $\mathop{\mathrm{Spec}}(A) \setminus T$. Let $A^\wedge$ and $(A')^\wedge$ be the $I$-adic completions. We have $X_{/T} = \text{Spf}(A^\wedge )$ and $X'_{/T'} = \text{Spf}((A')^\wedge )$, see proof of Formal Spaces, Lemma 87.20.8. By Lemma 88.4.3 we see that $(A')^\wedge$ is rig-smooth over $(A. I)$ which in turn means that $A^\wedge \to (A')^\wedge$ is rig-smooth which finally implies that $X'_{/T'} \to X_{/T}$ is rig smooth by Lemma 88.18.2. $\square$

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