The Stacks project

Proof. If $X$ is affine, say $X = \mathop{\mathrm{Spec}}(B)$, then we see from Lemma 86.9.10 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 $\mathop{\mathrm{Spec}}(A) \to \mathop{\mathrm{Spec}}(B)$. $\square$

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 86.33.1 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. $\square$

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

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

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 66.4.1. 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 $k'/k$ and a $k'$-point $u_0 : \mathop{\mathrm{Spec}}(k') \to U$ such that

commutes. Let $R \subset R'$ be the finite étale extension of Noetherian complete local rings which induces $k'/k$ 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 75.16.8 we can find an element $(u'_ n) \in \mathop{\mathrm{lim}}\nolimits \mathop{\mathrm{Mor}}\nolimits _ S(\mathop{\mathrm{Spec}}(R'/\mathfrak m^ nR'), U)$ mapping to $(x'_ n)$. By Lemma 86.33.2 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

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 $\mathop{\mathrm{Spec}}(R/\mathfrak m^ n)$. $\square$

Proof. The statement makes sense because $X_{/T}$ is adic* by Lemma 86.20.8 (and hence we're allowed to use the terminology adic for morphisms, see Definition 86.23.2). Let $p$ be given. By Lemma 86.33.3 we get a unique morphism $g : \mathop{\mathrm{Spec}}(R) \to X$ corresponding to the composition $\text{Spf}(R) \to X_{/T} \to X$. Let $Z \subset X$ be the reduced induced closed subspace structure on $T$. The incusion morphism $Z \to X$ corresponds to a morphism $Z \to X_{/T}$. Since $p$ is adic it is representable by algebraic spaces and we find

is an algebraic space endowed with a closed immersion to $\text{Spf}(R)$. (Equality holds because $X_{/T} \to X$ is a monomorphism.) Thus this fibre product is equal to $\mathop{\mathrm{Spec}}(R/J)$ for some ideal $J \subset R$ wich contains $\mathfrak m_ R^{n_0}$ for some $n_0 \geq 1$. This implies that $\mathop{\mathrm{Spec}}(R) \times _ X Z$ is a closed subscheme of $\mathop{\mathrm{Spec}}(R)$, say $\mathop{\mathrm{Spec}}(R) \times _ X Z = \mathop{\mathrm{Spec}}(R/I)$, whose intersection with $\mathop{\mathrm{Spec}}(R/\mathfrak m_ R^ n)$ for $n \geq n_0$ is equal to $\mathop{\mathrm{Spec}}(R/J)$. In algebraic terms this says $I + \mathfrak m_ R^ n = J + \mathfrak m_ R^ n = J$ for all $n \geq n_0$. By Krull's intersection theorem this implies $I = J$ and we conclude. $\square$