Lemma 85.29.3. 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

is bijective.

Lemma 85.29.3. 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 10.51.4) 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 29.36.7. 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 10.153.7 and 10.153.9). 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 85.29.2 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 $\mathop{\mathrm{Spec}}(R/\mathfrak m^ n)$. $\square$

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.

## Comments (0)

There are also: