Lemma 84.26.3. 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

is bijective.

Lemma 84.26.3. 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 64.4.1. In particular $T$ is locally Noetherian, see Morphisms, Lemma 29.14.6. 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 29.34.7. 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 10.149.7 and 10.149.9). Denote $x'_ n$ the restriction of $x_ n$ to $\mathop{\mathrm{Spec}}(R'/\mathfrak m^ nR')$. By More on Morphisms of Spaces, Lemma 73.16.8 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 84.26.2 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 $\mathop{\mathrm{Spec}}(R/\mathfrak m^ n)$. $\square$

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