The Stacks project

85.26 Maps out of affine formal schemes

We prove a few results that will be useful later. In the paper [Bhatt-Algebraize] the reader can find very general results of a similar nature.

Lemma 85.26.1. Let $S$ be a scheme. Let $A$ be a weakly admissible topological $S$-algebra. Let $X$ be an affine scheme over $S$. Then the natural map

\[ \mathop{Mor}\nolimits _ S(\mathop{\mathrm{Spec}}(A), X) \longrightarrow \mathop{Mor}\nolimits _ S(\text{Spf}(A), X) \]

is bijective.

Proof. If $X$ is affine, say $X = \mathop{\mathrm{Spec}}(B)$, then we see from Lemma 85.5.10 that morphisms $\text{Spf}(A) \to \mathop{\mathrm{Spec}}(B)$ correspond to continuous $S$-algebra maps $B \to A$ where $B$ has the discrete topology. These are just $S$-algebra maps, which correspond to morphisms $\mathop{\mathrm{Spec}}(A) \to \mathop{\mathrm{Spec}}(B)$. $\square$

Lemma 85.26.2. Let $S$ be a scheme. Let $A$ be a weakly admissible topological $S$-algebra such that $A/I$ is a local ring for some weak ideal of definition $I \subset A$. Let $X$ be a scheme over $S$. Then the natural map

\[ \mathop{Mor}\nolimits _ S(\mathop{\mathrm{Spec}}(A), X) \longrightarrow \mathop{Mor}\nolimits _ S(\text{Spf}(A), X) \]

is bijective.

Proof. Let $\varphi : \text{Spf}(A) \to X$ be a morphism. Since $\mathop{\mathrm{Spec}}(A/I)$ is local we see that $\varphi $ maps $\mathop{\mathrm{Spec}}(A/I)$ into an affine open $U \subset X$. However, this then implies that $\mathop{\mathrm{Spec}}(A/J)$ maps into $U$ for every ideal of definition $J$. Hence we may apply Lemma 85.26.1 to see that $\varphi $ comes from a morphism $\mathop{\mathrm{Spec}}(A) \to X$. This proves surjectivity of the map. We omit the proof of injectivity. $\square$

Lemma 85.26.3. Let $S$ be a scheme. Let $R$ be a complete local Noetherian $S$-algebra. Let $X$ be an algebraic space over $S$. Then the natural map

\[ \mathop{Mor}\nolimits _ S(\mathop{\mathrm{Spec}}(R), X) \longrightarrow \mathop{Mor}\nolimits _ S(\text{Spf}(R), X) \]

is bijective.

Proof. Let $\mathfrak m$ be the maximal ideal of $R$. We have to show that

\[ \mathop{Mor}\nolimits _ S(\mathop{\mathrm{Spec}}(R), X) \longrightarrow \mathop{\mathrm{lim}}\nolimits \mathop{Mor}\nolimits _ S(\mathop{\mathrm{Spec}}(R/\mathfrak m^ n), X) \]

is bijective for $R$ as above.

Injectivity: Let $x, x' : \mathop{\mathrm{Spec}}(R) \to X$ be two morphisms mapping to the same element in the right hand side. Consider the fibre product

\[ T = \mathop{\mathrm{Spec}}(R) \times _{(x, x'), X \times _ S X, \Delta } X \]

Then $T$ is a scheme and $T \to \mathop{\mathrm{Spec}}(R)$ is locally of finite type, monomorphism, separated, and locally quasi-finite, see Morphisms of Spaces, Lemma 65.4.1. In particular $T$ is locally Noetherian, see Morphisms, Lemma 29.15.6. Let $t \in T$ be the unique point mapping to the closed point of $\mathop{\mathrm{Spec}}(R)$ which exists as $x$ and $x'$ agree over $R/\mathfrak m$. Then $R \to \mathcal{O}_{T, t}$ is a local ring map of Noetherian rings such that $R/\mathfrak m^ n \to \mathcal{O}_{T, t}/\mathfrak m^ n\mathcal{O}_{T, t}$ is an isomorphism for all $n$ (because $x$ and $x'$ agree over $\mathop{\mathrm{Spec}}(R/\mathfrak m^ n)$ for all $n$). Since $\mathcal{O}_{T, t}$ maps injectively into its completion (see Algebra, Lemma 10.50.4) we conclude that $R = \mathcal{O}_{T, t}$. Hence $x$ and $x'$ agree over $R$.

Surjectivity: Let $(x_ n)$ be an element of the right hand side. Choose a scheme $U$ and a surjective étale morphism $U \to X$. Denote $x_0 : \mathop{\mathrm{Spec}}(k) \to X$ the morphism induced on the residue field $k = R/\mathfrak m$. The morphism of schemes $U \times _{X, x_0} \mathop{\mathrm{Spec}}(k) \to \mathop{\mathrm{Spec}}(k)$ is surjective étale. Thus $U \times _{X, x_0} \mathop{\mathrm{Spec}}(k)$ is a nonempty disjoint union of spectra of finite separable field extensions of $k$, see Morphisms, Lemma 29.35.7. Hence we can find a finite separable field extension $k \subset k'$ and a $k'$-point $u_0 : \mathop{\mathrm{Spec}}(k') \to U$ such that

\[ \xymatrix{ \mathop{\mathrm{Spec}}(k') \ar[d] \ar[r]_-{u_0} & U \ar[d] \\ \mathop{\mathrm{Spec}}(k) \ar[r]^-{x_0} & X } \]

commutes. Let $R \subset R'$ be the finite étale extension of Noetherian complete local rings which induces $k \subset k'$ on residue fields (see Algebra, Lemmas 10.152.7 and 10.152.9). Denote $x'_ n$ the restriction of $x_ n$ to $\mathop{\mathrm{Spec}}(R'/\mathfrak m^ nR')$. By More on Morphisms of Spaces, Lemma 74.16.8 we can find an element $(u'_ n) \in \mathop{\mathrm{lim}}\nolimits \mathop{Mor}\nolimits _ S(\mathop{\mathrm{Spec}}(R'/\mathfrak m^ nR'), U)$ mapping to $(x'_ n)$. By Lemma 85.26.2 the family $(u'_ n)$ comes from a unique morphism $u' : \mathop{\mathrm{Spec}}(R') \to U$. Denote $x' : \mathop{\mathrm{Spec}}(R') \to X$ the composition. Note that $R' \otimes _ R R'$ is a finite product of spectra of Noetherian complete local rings to which our current discussion applies. Hence the diagram

\[ \xymatrix{ \mathop{\mathrm{Spec}}(R' \otimes _ R R') \ar[r] \ar[d] & \mathop{\mathrm{Spec}}(R') \ar[d]^{x'} \\ \mathop{\mathrm{Spec}}(R') \ar[r]^{x'} & X } \]

is commutative by the injectivity shown above and the fact that $x'_ n$ is the restriction of $x_ n$ which is defined over $R/\mathfrak m^ n$. Since $\{ \mathop{\mathrm{Spec}}(R') \to \mathop{\mathrm{Spec}}(R)\} $ is an fppf covering we conclude that $x'$ descends to a morphism $x : \mathop{\mathrm{Spec}}(R) \to X$. We omit the proof that $x_ n$ is the restriction of $x$ to $\mathop{\mathrm{Spec}}(R/\mathfrak m^ n)$. $\square$

Comments (2)

Comment #1058 by Matthieu Romagny on

In the statement of Lemma 65.23.3 (0AQH), it would be better to write : Let be an -scheme, with a complete local Noetherian ring.

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 0AQE. Beware of the difference between the letter 'O' and the digit '0'.