The Stacks project

86.23 The completion functor

In this section we consider the following situation. First we fix a base scheme $S$. All rings, topological rings, schemes, algebraic spaces, and formal algebraic spaces and morphisms between these will be over $S$. Next, we fix an algebraic space $X$ and a closed subset $T \subset |X|$. We denote $U \subset X$ be the open subspace with $|U| = |X| \setminus T$. Picture

\[ U \to X \quad |X| = |U| \amalg T \]

In this situation, given an algebraic space $X'$ over $X$, i.e., an algebraic space $X'$ endowed with a morphism $f : X' \to X$, then we denote $T' \subset |X'|$ the inverse image of $T$ and we let $U' \subset X'$ be the open subspace with $|U'| = |X'| \setminus T'$. Picture

\[ U' = f^{-1}U \quad \quad \vcenter { \xymatrix{ U' \ar[d] \ar[r] & X' \ar[d]_ f \\ U \ar[r] & X } } \quad \quad \vcenter { \xymatrix{ |U'| \ar[r] \ar[d] & |X'| \ar[d]^{|f|} & T' \ar[l] \ar[d] \\ |U| \ar[r] & |X| & T \ar[l] } } \quad \quad T' = |f|^{-1}T \]

We will relate properties of $f$ to properties of the induced morphism

\[ f_{/T} : X'_{/T'} \longrightarrow X_{/T} \]

of formal completions. As indicated in the displayed formula, we will denote this morphism $f_{/T}$. We have already seen that $f_{/T}$ is representable by algebraic spaces in Formal Spaces, Lemma 85.10.4. In fact, as the proof of that lemma shows, the diagram

\[ \xymatrix{ X'_{/T'} \ar[d]_{f_{/T}} \ar[r] & X' \ar[d]^ f \\ X_{/T} \ar[r] & X } \]

is cartesian. Please keep this fact in mind whilst reading the lemmas stated and proved below.

Lemma 86.23.1. In the situation above. If $f$ is locally of finite type, then $f_{/T}$ is locally of finite type.

Proof. (Finite type morphisms of formal algebraic spaces are discussed in Formal Spaces, Section 85.20.) Namely, suppose that $Z \to X$ is a morphism from a scheme into $X$ such that $|Z|$ maps into $T$. From the cartesian square above we see that $Z \times _ X X'$ is an algebraic space representing $Z \times _{X_{/T}} X'_{/T'}$. Since $Z \times _ X X' \to Z$ is locally of finite type by Morphisms of Spaces, Lemma 65.23.3 we conclude. $\square$

Lemma 86.23.2. In the situation above. If $f$ is étale, then $f_{/T}$ is étale.

Proof. By the same argument as in the proof of Lemma 86.23.1 this follows from Morphisms of Spaces, Lemma 65.39.4. $\square$

Lemma 86.23.3. In the situation above. If $f$ is a closed immersion, then $f_{/T}$ is a closed immersion.

Proof. (Closed immersions of formal algebraic spaces are discussed in Formal Spaces, Section 85.23.) By the same argument as in the proof of Lemma 86.23.1 this follows from Spaces, Lemma 63.12.3. $\square$

Lemma 86.23.4. In the situation above. If $f$ is proper, then $f_{/T}$ is proper.

Proof. (Proper morphisms of formal algebraic spaces are discussed in Formal Spaces, Section 85.27.) By the same argument as in the proof of Lemma 86.23.1 this follows from Morphisms of Spaces, Lemma 65.40.3. $\square$

Lemma 86.23.5. In the situation above. If $f$ is quasi-compact, then $f_{/T}$ is quasi-compact.

Proof. (Quasi-compact morphisms of formal algebraic spaces are discussed in Formal Spaces, Section 85.13.) We have to show that $(X'_{/T'})_{red} \to (X_{/T})_{red}$ is a quasi-compact morphism of algebraic spaces. By Formal Spaces, Lemma 85.10.5 this is the morphism $Z' \to Z$ where $Z' \subset X'$, resp. $Z \subset X$ is the reduced induced algebraic space structure on $T'$, resp. $T$. It follows that $Z' \to f^{-1}Z = Z \times _ X X'$ is a thickening (a closed immersion defining an isomorphism on underlying topological spaces). Since $Z \times _ X X' \to Z$ is quasi-compact as a base change of $f$ (Morphisms of Spaces, Lemma 65.8.4) we conclude that $Z' \to Z$ is too by More on Morphisms of Spaces, Lemma 74.10.1. $\square$

Remark 86.23.6. In the situation above consider the diagonal morphisms $\Delta _ f : X' \to X' \times _ X X'$ and $\Delta _{f_{/T}} : X'_{/T'} \to X'_{/T'} \times _{X_{/T}} X'_{/T'}$. It is easy to see that

\[ X'_{/T'} \times _{X_{/T}} X'_{/T'} = (X' \times _ X X')_{/T''} \]

as subfunctors of $X' \times _ X X'$ where $T'' \subset |X' \times _ X X'|$ is the inverse image of $T$. Hence we see that $\Delta _{f_{/T}} = (\Delta _ f)_{/T''}$. We will use this below to show that properties of $\Delta _ f$ are inherited by $\Delta _{f_{/T}}$.

Lemma 86.23.7. In the situation above. If $f$ is (quasi-)separated, then $f_{/T}$ is too.

Proof. (Separation conditions on morphisms of formal algebraic spaces are discussed in Formal Spaces, Section 85.26.) We have to show that if $\Delta _ f$ is quasi-compact, resp. a closed immersion, then the same is true for $\Delta _{f_{/T}}$. This follows from the discussion in Remark 86.23.6 and Lemmas 86.23.5 and 86.23.3. $\square$

Lemma 86.23.8. 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 86.4.3. 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 64.6.1. 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 85.7.1. By Lemma 86.18.2 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 85.16.6. By Lemma 86.4.3 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 86.18.2. $\square$

Lemma 86.23.9. In the situation above. If $X$ is locally Noetherian, $f$ is locally of finite type, and $U' \to U$ is étale, then $f_{/T}$ is rig-étale.

Lemma 86.23.10. In the situation above. If $X$ is locally Noetherian, $f$ is proper, and $U' \to U$ is surjective, then $f_{/T}$ is rig-surjective.

Proof. (The statement makes sense by Lemma 86.23.1 and Formal Spaces, Lemma 85.16.6.) Let $R$ be a complete discrete valuation ring with fraction field $K$. Let $p : \text{Spf}(R) \to X_{/T}$ be an adic morphism of formal algebraic spaces. By Formal Spaces, Lemma 85.29.4 the composition $\text{Spf}(R) \to X_{/T} \to X$ corresponds to a morphism $q : \mathop{\mathrm{Spec}}(R) \to X$ which maps $\mathop{\mathrm{Spec}}(K)$ into $U$. Since $U' \to U$ is proper and surjective we see that $\mathop{\mathrm{Spec}}(K) \times _ U U'$ is nonempty and proper over $K$. Hence we can choose a field extension $K'/K$ and a commutative diagram

\[ \xymatrix{ \mathop{\mathrm{Spec}}(K') \ar[r] \ar[d] & U' \ar[r] \ar[d] & X' \ar[d] \\ \mathop{\mathrm{Spec}}(K) \ar[r] & U \ar[r] & X } \]

Let $R' \subset K'$ be a discrete valuation ring dominating $R$ with fraction field $K'$, see Algebra, Lemma 10.119.13. Since $\mathop{\mathrm{Spec}}(K) \to X$ extends to $\mathop{\mathrm{Spec}}(R) \to X$ we see by the valuative criterion of properness (Morphisms of Spaces, Lemma 65.44.1) that we can extend our $K'$-point of $U'$ to a morphism $\mathop{\mathrm{Spec}}(R') \to X'$ over $\mathop{\mathrm{Spec}}(R) \to X$. It follows that the inverse image of $T'$ in $\mathop{\mathrm{Spec}}(R')$ is the closed point and we find an adic morphism $\text{Spf}((R')^\wedge ) \to X'_{/T'}$ lifting $p$ as desired (note that $(R')^\wedge $ is a complete discrete valuation ring by More on Algebra, Lemma 15.43.5). $\square$

Lemma 86.23.11. In the situation above. If $X$ is locally Noetherian, $f$ is separated and locally of finite type, and $U' \to U$ is a monomorphism, then $\Delta _{f_{/T}}$ is rig-surjective.

Proof. The diagonal $\Delta _ f : X' \to X' \times _ X X'$ is a closed immersion and the restriction $U' \to U' \times _ U U'$ of $\Delta _ f$ is surjective. Hence the lemma follows from the discussion in Remark 86.23.6 and Lemma 86.23.10. $\square$


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0GDB. Beware of the difference between the letter 'O' and the digit '0'.