hcju/stacksqa huggingface dataset:
Assume $\varphi$ is naively rig-flat. We first state some obvious consequences of the assumptions. Namely, let $f \in A$. Then $A, B, A_{\{f\}}, B_{\{f\}}$ are Noetherian adic topological rings. The maps $A \to A_{\{f\}} \to B_{\{f\}}$ and $A \to B \to B_{\{f\}}$ are adic and topologically of finite type. The ring maps $A \to A_{\{f\}}$ and $B \to B_{\{f\}}$ are flat as compositions of $A \to A_f$ and $B \to B_f$ and the completion maps which are flat by Algebra, Lemma \ref{algebra-lemma-completion-flat}. The quotients of each of the rings $A, B, A_{\{f\}}, B_{\{f\}}$ by $I$ is of finite type over $A/I$ and hence Jacobson too (Algebra, Proposition \ref{algebra-proposition-Jacobson-permanence}). \medskip\noindent Let $\mathfrak q' \subset B_{\{f\}}$ be rig-closed. It suffices to prove that $(B_{\{f\}})_{\mathfrak q'}$ is flat over $A_{\{f\}}$, see Lemma \ref{lemma-naively-rig-flat-continuous}. By Lemma \ref{lemma-rig-closed-jacobson} the primes $\mathfrak q \subset B$ and $\mathfrak p' \subset A_{\{f\}}$ and $\mathfrak p \subset A$ lying under $\mathfrak q'$ are rig-closed. We are going to apply Algebra, Lemma \ref{algebra-lemma-yet-another-variant-local-criterion-flatness} to the diagram $$ \xymatrix{ B_\mathfrak q \ar[r] & (B_{\{f\}})_{\mathfrak q'} \\ A_\mathfrak p \ar[u] \ar[r] & (A_{\{f\}})_{\mathfrak p'} \ar[u] } $$ with $M = B_\mathfrak q$. The only assumption that hasn't been checked yet is the fact that $\mathfrak p$ generates the maximal ideal of $(A_{\{f\}})_{\mathfrak p'}$. This follows from Lemma \ref{lemma-rig-closed-point-in-localization}; here we use that $\mathfrak p$ and $\mathfrak p'$ are rig-closed to see that $f$ maps to a unit of $A/\mathfrak p$ (this is the only step in the proof that fails without the Jacobson assumption). Namely, this tells us that $A/\mathfrak p \to A_{\{f\}}/\mathfrak p'$ is a finite inclusion of local rings (Lemma \ref{lemma-rig-closed-point-relative}) and $f$ maps to a unit in the second one.
Given $f \in A'$ the assumptions of the lemma remain true for the digram $$ \xymatrix{ B \ar[r] & (B')_{\{f\}} \\ A \ar[r] \ar[u]^\varphi & (A')_{\{f\}} \ar[u] } $$ Hence it suffices to prove that $\varphi'$ is naively rig-flat. \medskip\noindent Take a rig-closed prime ideal $\mathfrak q' \subset B'$. We have to show that $(B')_{\mathfrak q'}$ is flat over $A'$. We can choose an $f \in A$ which maps to a unit of $B'/\mathfrak q'$ such that the induced prime ideal $\mathfrak p''$ of $A_{\{f\}}$  is rig-closed, see Lemma \ref{lemma-rig-closed-point-after-localization}. To be precise, here $\mathfrak q'' = \mathfrak q' B'_{\{f\}}$ and $\mathfrak p'' = A_{\{f\}} \cap \mathfrak q''$. Consider the diagram $$ \xymatrix{ B'_{\mathfrak q'} \ar[r] & (B'_{\{f\}})_{\mathfrak q''} \\ A \ar[r] \ar[u] & A_{\{f\}} \ar[u] } $$ We want to show that the left vertical arrow is flat. The top horizontal arrow is faithfully flat as it is a local homomorphism of local rings and flat as $B'_{\{f\}}$ is the completion of a localization of the Noetherian ring $B'_f$. Similarly the bottom horizontal arrow is flat. Hence it suffices to prove that the right vertical arrow is flat. Finally, all the assumptions of the lemma remain true for the diagram $$ \xymatrix{ B_{\{f\}} \ar[r] & B'_{\{f\}} \\ A_{\{f\}} \ar[r] \ar[u] & A'_{\{f\}} \ar[u] } $$ This reduces us to the case discussed in the next paragraph. \medskip\noindent Take a rig-closed prime ideal $\mathfrak q' \subset B'$ and assume $\mathfrak p = A \cap \mathfrak q'$ is rig-closed as well. This implies also the primes $\mathfrak q = B \cap \mathfrak q'$ and $\mathfrak p' = A' \cap \mathfrak q'$ are rig-closed, see Lemma \ref{lemma-rig-closed-point-relative}. We are going to apply Algebra, Lemma \ref{algebra-lemma-yet-another-variant-local-criterion-flatness} to the diagram $$ \xymatrix{ B_\mathfrak q \ar[r] & B'_{\mathfrak q'} \\ A_\mathfrak p \ar[u] \ar[r] & A'_{\mathfrak p'} \ar[u] } $$ with $M = B_\mathfrak q$. The only assumption that hasn't been checked yet is the fact that $\mathfrak p$ generates the maximal ideal of $A'_{\mathfrak p'}$. This follows from Lemma \ref{lemma-rig-closed-point-etale}.
Let us recall what the statement signifies. First,  $\textit{WAdm}^{Noeth}$ is the category whose objects are adic Noetherian topological rings and whose morphisms are continuous ring homomorphisms. Consider a commutative diagram $$ \xymatrix{ B \ar[r] & (B')^\wedge \\ A \ar[r] \ar[u]^\varphi & (A')^\wedge \ar[u]_{\varphi'} } $$ satisfying the following conditions: $A$ and $B$ are adic Noetherian topological rings, $A \to A'$ and $B \to B'$ are \'etale ring maps, $(A')^\wedge = \lim A'/I^nA'$ for some ideal of definition $I \subset A$, $(B')^\wedge = \lim B'/J^nB'$ for some ideal of definition $J \subset B$, and $\varphi : A \to B$ and $\varphi' : (A')^\wedge \to (B')^\wedge$ are continuous. Note that $(A')^\wedge$ and $(B')^\wedge$ are adic Noetherian topological rings by Formal Spaces, Lemma \ref{formal-spaces-lemma-completion-in-sub}. We have to show \begin{enumerate} \item $\varphi$ is rig-flat $\Rightarrow \varphi'$ is rig-flat, \item if $B \to B'$ faithfully flat, then $\varphi'$ is rig-flat $\Rightarrow \varphi$ is rig-flat, and \item if $A \to B_i$ is rig-flat for $i = 1, \ldots, n$, then $A \to \prod_{i = 1, \ldots, n} B_i$ is rig-flat. \end{enumerate} Being adic and topologically of finite type satisfies conditions (1), (2), and (3), see Lemma \ref{lemma-finite-type}. Thus in verifying (1), (2), and (3) for the property ``rig-flat'' we may already assume our ring maps are all adic and topologically of finite type. Then (1) and (2) follow from Lemmas \ref{lemma-rig-flat-local-etale} and \ref{lemma-rig-flat-local-down}. We omit the trivial proof of (3).
The statement makes sense by Lemma \ref{lemma-rig-flat-axioms}. To see that it is true assume we have rig-flat morphisms $A \to B$ and $B \to C$ in $\textit{WAdm}^{Noeth}$. Then $A \to C$ is adic and topologically of finite type by Lemma \ref{lemma-composition-finite-type}. To finish the proof we have to show that for all $f \in A$ the map $A_{\{f\}} \to C_{\{f\}}$ is naively rig-flat. Since $A_{\{f\}} \to B_{\{f\}}$ and $B_{\{f\}} \to C_{\{f\}}$ are naively rig-flat, it suffices to show that compositions of naively rig-flat maps are naively rig-flat. This is a consequence of Algebra, Lemma \ref{algebra-lemma-composition-flat}.
The equivalence of (1) and (2) is Definition \ref{definition-rig-flat}. The equivalence of (2), (3), and (4) follows from the fact that being rig-flat is a local property of arrows of $\text{WAdm}^{Noeth}$ by Lemma \ref{lemma-rig-flat-axioms} and an application of the variant of Formal Spaces, Lemma \ref{formal-spaces-lemma-property-defines-property-morphisms} for morphisms between locally Noetherian algebraic spaces mentioned in Formal Spaces, Remark \ref{formal-spaces-remark-variant-Noetherian}.
Let $I$ and $I'$ be ideals of definitions of $A$. Then there exists an integer $c \geq 0$ such that $I^c \subset I'$ and $(I')^c \subset I$. Hence $B$ is rig-smooth over $(A, I)$ if and only if $B$ is rig-smooth over $(A, I')$. This follows from Definition \ref{definition-rig-smooth-homomorphism}, the inclusions $I^c \subset I'$ and $(I')^c \subset I$, and the fact that the naive cotangent complex $\NL_{B/A}^\wedge$ is independent of the choice of ideal of definition of $A$ by Remark \ref{remark-NL-well-defined-topological}.
The statement makes sense by Lemma \ref{lemma-rig-smooth-continuous}. To see that it is true assume we have morphisms $B \to A$ and $B \to C$ in $\textit{WAdm}^{Noeth}$ and that $B \to A$ is rig-smooth and $B \to C$ is adic (Formal Spaces, Definition \ref{formal-spaces-definition-adic-homomorphism}). Then we can choose an ideal of definition $I \subset B$ such that the topology on $A$ and $C$ is the $I$-adic topology. In this situation it follows immediately that $A \widehat{\otimes}_B C$ is rig-smooth over $(C, IC)$ by Lemma \ref{lemma-base-change-rig-smooth-homomorphism}.
We strongly urge the reader to find their own proof and not read the proof that follows. The statement makes sense by Lemma \ref{lemma-rig-smooth-continuous}. To see that it is true assume we have rig-smooth morphisms $A \to B$ and $B \to C$ in $\textit{WAdm}^{Noeth}$. Then we can choose an ideal of definition $I \subset A$ such that the topology on $C$ and $B$ is the $I$-adic topology. By Lemma \ref{lemma-exact-sequence-NL} we obtain an exact sequence $$ \xymatrix{ C \otimes_B H^0(\NL_{B/A}^\wedge) \ar[r] & H^0(\NL_{C/A}^\wedge) \ar[r] & H^0(\NL_{C/B}^\wedge) \ar[r] & 0 \\ H^{-1}(\NL_{B/A}^\wedge \otimes_B C) \ar[r] & H^{-1}(\NL_{C/A}^\wedge) \ar[r] & H^{-1}(\NL_{C/B}^\wedge) \ar[llu] } $$ Observe that $H^{-1}(\NL_{B/A}^\wedge \otimes_B C)$ and $H^{-1}(\NL_{C/B}^\wedge)$ are annihilated by a power of $I$; this follows from Lemma \ref{lemma-equivalent-with-artin-smooth} part (2) combined with More on Algebra, Lemmas \ref{more-algebra-lemma-two-term-base-change} and \ref{more-algebra-lemma-base-change-property-ext-1-annihilated} (to deal with the base change by $B \to C$). Hence $H^{-1}(\NL_{C/A}^\wedge)$ is annihilated by a power of $I$. Next, by the characterization of rig-smooth algebras in Lemma \ref{lemma-equivalent-with-artin-smooth} part (2) which in turn refers to More on Algebra, Lemma \ref{more-algebra-lemma-ext-1-annihilated} part (5) we can choose $f_1, \ldots, f_s \in IB$ and $g_1, \ldots, g_t \in IC$ such that $V(f_1, \ldots, f_s) = V(IB)$ and $V(g_1, \ldots, g_t) = V(IC)$ and such that $H^0(\NL_{B/A}^\wedge)_{f_i}$ is a finite projective $B_{f_i}$-module and $H^0(\NL_{C/B}^\wedge)_{g_j}$ is a finite projective $C_{g_j}$-module. Since the cohomologies in degree $-1$ vanish upon localization at $f_ig_j$ we get a short exact sequence $$ 0 \to (C \otimes_B H^0(\NL_{B/A}^\wedge))_{f_ig_j} \to H^0(\NL_{C/A}^\wedge)_{f_ig_j} \to H^0(\NL_{C/B}^\wedge)_{f_ig_j} \to 0 $$ and we conclude that $H^0(\NL_{C/A}^\wedge)_{f_ig_j}$ is a finite projective $C_{f_ig_j}$-module as an extension of same. Thus by the criterion in Lemma \ref{lemma-equivalent-with-artin-smooth} part (2) and via that the criterion in More on Algebra, Lemma \ref{more-algebra-lemma-ext-1-annihilated} part (4) we conclude that $C$ is rig-smooth over $(A, I)$.
Let $\overline{\mathfrak q} \subset C$ be a prime ideal which does not contain an ideal of definition. Since the modules in question are finite it suffices to show that $$ H^{-1}(\NL_{B/A}^\wedge \otimes_B C)_{\overline{\mathfrak q}} \to H^{-1}(\NL_{C/A}^\wedge)_{\overline{\mathfrak q}} $$ is injective. As in the proof of Lemma \ref{lemma-exact-sequence-NL} choose presentations $B = A\{x_1, \ldots, x_r\}/J$, $C = B\{y_1, \ldots, y_s\}/J'$, and $C = A\{x_1, \ldots, x_r, y_1, \ldots, y_s\}/K$. Looking at the diagram in the proof of Lemma \ref{lemma-exact-sequence-NL} we see that it suffices to show that $J/J^2 \otimes_B C \to K/K^2$ is injective after localization at the prime ideal $\mathfrak q \subset A\{x_1, \ldots, x_r, y_1, \ldots, y_s\}$ corresponding to $\overline{\mathfrak q}$. Please compare with More on Algebra, Lemma \ref{more-algebra-lemma-transitive-lci-at-end} and its proof. This is the same as  asking $J/KJ \to K/K^2$ to be injective after localization at $\mathfrak q$. Equivalently, we have to show that $J_\mathfrak q \cap K^2_\mathfrak q = (KJ)_\mathfrak q$. By Lemma \ref{lemma-rig-smooth-rig-flat} we know that $(K/J)_\mathfrak q = J'_\mathfrak q$ is generated by a regular sequence. Hence the desired intersection property follows from More on Algebra, Lemma \ref{more-algebra-lemma-conormal-sequence-H1-regular-ideal} (and the fact that an ideal generated by a regular sequence is $H_1$-regular, see More on Algebra, Section \ref{more-algebra-section-ideals}).
The equivalence of (1) and (2) is Definition \ref{definition-rig-smooth}. The equivalence of (2), (3), and (4) follows from the fact that being rig-smooth is a local property of arrows of $\text{WAdm}^{Noeth}$ by Lemma \ref{lemma-rig-smooth-axioms} and an application of the variant of Formal Spaces, Lemma \ref{formal-spaces-lemma-property-defines-property-morphisms} for morphisms between locally Noetherian algebraic spaces mentioned in Formal Spaces, Remark \ref{formal-spaces-remark-variant-Noetherian}.
Let $I$ and $I'$ be ideals of definitions of $A$. Then there exists an integer $c \geq 0$ such that $I^c \subset I'$ and $(I')^c \subset I$. Hence $B$ is rig-\'etale over $(A, I)$ if and only if $B$ is rig-\'etale over $(A, I')$. This follows from Definition \ref{definition-rig-etale-homomorphism}, the inclusions $I^c \subset I'$ and $(I')^c \subset I$, and the fact that the naive cotangent complex $\NL_{B/A}^\wedge$ is independent of the choice of ideal of definition of $A$ by Remark \ref{remark-NL-well-defined-topological}.
The statement makes sense by Lemma \ref{lemma-rig-etale-continuous}. To see that it is true assume we have morphisms $B \to A$ and $B \to C$ in $\textit{WAdm}^{Noeth}$ and that $B \to A$ is rig-\'etale and $B \to C$ is adic (Formal Spaces, Definition \ref{formal-spaces-definition-adic-homomorphism}). Then we can choose an ideal of definition $I \subset B$ such that the topology on $A$ and $C$ is the $I$-adic topology. In this situation it follows immediately that $A \widehat{\otimes}_B C$ is rig-\'etale over $(C, IC)$ by Lemma \ref{lemma-base-change-rig-etale-homomorphism}.
The statement makes sense by Lemma \ref{lemma-rig-etale-continuous}. To see that it is true assume we have rig-\'etale morphisms $A \to B$ and $B \to C$ in $\textit{WAdm}^{Noeth}$. Then we can choose an ideal of definition $I \subset A$ such that the topology on $C$ and $B$ is the $I$-adic topology. By Lemma \ref{lemma-exact-sequence-NL} we obtain an exact sequence $$ \xymatrix{ C \otimes_B H^0(\NL_{B/A}^\wedge) \ar[r] & H^0(\NL_{C/A}^\wedge) \ar[r] & H^0(\NL_{C/B}^\wedge) \ar[r] & 0 \\ H^{-1}(\NL_{B/A}^\wedge \otimes_B C) \ar[r] & H^{-1}(\NL_{C/A}^\wedge) \ar[r] & H^{-1}(\NL_{C/B}^\wedge) \ar[llu] } $$ There exists a $c \geq 0$ such that for all $a \in I$  multiplication by $a^c$ is zero on $\NL_{B/A}^\wedge$ in $D(B)$ and $\NL_{C/B}^\wedge$ in $D(C)$. Then of course multiplication by $a^c$ is zero on $\NL_{B/A}^\wedge \otimes_B C$ in $D(C)$ too. Hence $H^0(\NL_{B/A}^\wedge) \otimes_A C$,  $H^0(\NL_{C/B}^\wedge)$,  $H^{-1}(\NL_{B/A}^\wedge \otimes_B C)$, and $H^{-1}(\NL_{C/B}^\wedge)$ are annihilated by $a^c$. From the exact sequence we obtain that multiplication by $a^{2c}$ is zero on $H^0(\NL_{C/A}^\wedge)$ and $H^{-1}(\NL_{C/A}^\wedge)$. It follows from Lemma \ref{lemma-equivalent-with-artin} that $C$ is rig-\'etale over $(A, I)$ as desired.
The statement makes sense by Lemma \ref{lemma-rig-etale-continuous}. To see that it is true assume we have maps $A \to B$ and $B \to C$ in $\textit{WAdm}^{Noeth}$ with $A \to C$ and $A \to B$ rig-\'etale. We have to show that $B \to C$ is rig-\'etale. Then we can choose an ideal of definition $I \subset A$ such that the topology on $C$ and $B$ is the $I$-adic topology. By Lemma \ref{lemma-exact-sequence-NL} we obtain an exact sequence $$ \xymatrix{ C \otimes_B H^0(\NL_{B/A}^\wedge) \ar[r] & H^0(\NL_{C/A}^\wedge) \ar[r] & H^0(\NL_{C/B}^\wedge) \ar[r] & 0 \\ H^{-1}(\NL_{B/A}^\wedge \otimes_B C) \ar[r] & H^{-1}(\NL_{C/A}^\wedge) \ar[r] & H^{-1}(\NL_{C/B}^\wedge) \ar[llu] } $$ There exists a $c \geq 0$ such that for all $a \in I$  multiplication by $a^c$ is zero on $\NL_{B/A}^\wedge$ in $D(B)$ and $\NL_{C/A}^\wedge$ in $D(C)$. Hence $H^0(\NL_{B/A}^\wedge) \otimes_A C$, $H^0(\NL_{C/A}^\wedge)$, and $H^{-1}(\NL_{C/A}^\wedge)$ are annihilated by $a^c$. From the exact sequence we obtain that multiplication by $a^{2c}$ is zero on $H^0(\NL_{C/B}^\wedge)$ and $H^{-1}(\NL_{C/B}^\wedge)$. It follows from Lemma \ref{lemma-equivalent-with-artin} that $C$ is rig-\'etale over $(B, IB)$ as desired.
The equivalence of (1) and (2) is Definition \ref{definition-rig-etale}. The equivalence of (2), (3), and (4) follows from the fact that being rig-\'etale is a local property of arrows of $\text{WAdm}^{Noeth}$ by Lemma \ref{lemma-rig-etale-axioms} and an application of the variant of Formal Spaces, Lemma \ref{formal-spaces-lemma-property-defines-property-morphisms} for morphisms between locally Noetherian algebraic spaces mentioned in Formal Spaces, Remark \ref{formal-spaces-remark-variant-Noetherian}.
Let us observe that given $V$ and $V \to Y$ as in (2) without any further assumption on $f$ we see that the morphism $X \times_Y V \to V$ corresponds to a surjective morphism $B \to A$ in $\textit{WAdm}^{Noeth}$ by Formal Spaces, Lemma \ref{formal-spaces-lemma-closed-immersion-into-countably-indexed}. \medskip\noindent We have (2) $\Rightarrow$ (1) by Lemma \ref{lemma-rig-etale-rig-smooth-morphism}. \medskip\noindent Proof of (3) $\Rightarrow$ (2). Assume (3). By Lemma \ref{lemma-rig-etale-morphisms} it suffices to show that the ring maps $B \to A$ occuring in (3) are rig-\'etale in the sense of Definition \ref{definition-rig-etale-continuous-homomorphism}. Let $I$ be as in (3). The naive cotangent complex $\NL_{A/B}^\wedge$ of $A$ over $(B, I)$ is the complex of $A$-modules given by putting $J/J^2$ in degree $-1$. Hence $A$ is rig-\'etale over $(B, I)$ by Definition \ref{definition-rig-etale-homomorphism}. \medskip\noindent Assume (1) and let $V$ and $B \to A$ be as in (3). By Definition \ref{definition-rig-smooth} we see that $B \to A$ is rig-smooth. Choose any ideal of definition $I \subset B$. Then $A$ is rig-smooth over $(B, I)$. As above the complex $\NL_{A/B}^\wedge$ is  given by putting $J/J^2$ in degree $-1$. Hence by Lemma \ref{lemma-equivalent-with-artin-smooth} we see that $J/J^2$ is annihilated by a power $I^n$ for some $n \geq 1$. Since $B$ is adic, we see that $I^n$ is an ideal of definition of $B$ and the proof is complete.
Let $p : \text{Spf}(R) \to Y$ be an adic morphism where $R$ is a complete discrete valuation ring. Let $Z = \text{Spf}(R) \times_Y X$. Then $Z \to \text{Spf}(R)$ is representable by algebraic spaces, \'etale, and surjective. Hence $Z$ is nonempty. Pick a nonempty affine formal algebraic space $V$ and an \'etale morphism $V \to Z$ (possible by our definitions). Then $V \to \text{Spf}(R)$ corresponds to $R \to A^\wedge$ where $R \to A$ is an \'etale ring map, see Formal Spaces, Lemma \ref{formal-spaces-lemma-etale}. Since $A^\wedge \not = 0$ (as $V \not = \emptyset$) we can find a maximal ideal $\mathfrak m$ of $A$ lying over $\mathfrak m_R$. Then $A_\mathfrak m$ is a discrete valuation ring (More on Algebra, Lemma \ref{more-algebra-lemma-Dedekind-etale-extension}). Then $R' = A_\mathfrak m^\wedge$ is a complete discrete valuation ring (More on Algebra, Lemma \ref{more-algebra-lemma-completion-dvr}). Applying Formal Spaces, Lemma \ref{formal-spaces-lemma-morphism-between-formal-spectra}. we find the desired morphism $\text{Spf}(R') \to V \to Z \to X$.
Namely, if $f$ is rig-surjective, so is any base change (Lemma \ref{lemma-base-change-rig-surjective}). Conversely, if all $f_i$ are rig-surjective, so is $\coprod f_i : \coprod X \times_Y Y_i \to \coprod Y_i$. By Lemma \ref{lemma-etale-covering-rig-surjective} the morphism $\coprod g_i : \coprod Y_i \to Y$ is rig-surjective. Hence $\coprod X \times_Y Y_i \to Y$ is rig-surjective (Lemma \ref{lemma-composition-rig-surjective}). Since this morphism factors through $X \to Y$ we see that $X \to Y$ is rig-surjective by Lemma \ref{lemma-permanence-rig-surjective}.
We will use without further mention that morphisms between formal spectra are given by continuous maps between the corresponding topological rings, see Formal Spaces, Lemma \ref{formal-spaces-lemma-morphism-between-formal-spectra}. Let $\varphi : A \to R$ be a continuous map into a complete discrete valuation ring $A$. This implies that $\varphi(I) \subset \mathfrak m_R$. On the other hand, since we only need to produce the lift $\varphi' : B' \to R'$ in the case that $\varphi$ corresponds to an adic morphism, we may assume that $\varphi(I) \not = 0$. Thus we may consider the base change $C = B \widehat{\otimes}_A R$, see Remark \ref{remark-base-change} for example. Then $C$ is an $\mathfrak m_R$-adically complete $R$-algebra such that $C/\mathfrak m_R^n C$ is of finite type and flat over $R/\mathfrak m_R^n$ and such that $C/\mathfrak m_R C$ is nonzero. Pick any maximal ideal $\mathfrak m \subset C$ lying over $\mathfrak m_R$. By flatness (which implies going down) we see that $\Spec(C_\mathfrak m) \setminus V(\mathfrak m_R C_\mathfrak m)$ is a nonempty open. Hence We can pick a prime $\mathfrak q \subset \mathfrak m$ such that $\mathfrak q$ defines a closed point of $\Spec(C_\mathfrak m) \setminus \{\mathfrak m\}$ and such that $\mathfrak q \not \in V(IC_\mathfrak m)$, see Properties, Lemma \ref{properties-lemma-complement-closed-point-Jacobson}. Then $C/\mathfrak q$ is a dimension $1$-local domain and we can find $C/\mathfrak q \subset R'$ with $R'$ a discrete valuation ring (Algebra, Lemma \ref{algebra-lemma-exists-dvr}). By construction $\mathfrak m_R R' \subset \mathfrak m_{R'}$ and we see that $C \to R'$ extends to a continuous map $C \to (R')^\wedge$ (in fact we can pick $R'$ such that $R' = (R')^\wedge$ in our current situation but we do not need this). Since the completion of a discrete valuation ring is a discrete valuation ring, we see that the assumption gives a commutative diagram of rings $$ \xymatrix{ (R')^\wedge & C \ar[l] & B \ar[l] \\ R \ar[u] & R \ar[l] \ar[u] & A \ar[l] \ar[u] } $$ which gives the desired lift.
Faithful flatness implies rig-surjectivity by Lemma \ref{lemma-faithfully-flat-rig-surjective}. To prove the converse we will use without further mention that the vanishing of $I$-torsion is equivalent to the vanishing of $I$-power torsion (More on Algebra, Lemma \ref{more-algebra-lemma-torsion-free}). We will also use without further mention that morphisms between formal spectra are given by continuous maps between the corresponding topological rings, see Formal Spaces, Lemma \ref{formal-spaces-lemma-morphism-between-formal-spectra}. \medskip\noindent Assume $\text{Spf}(B) \to \text{Spf}(A)$ is rig-surjective. Choose a maximal ideal $I \subset \mathfrak m \subset A$. The open $U = \Spec(A_\mathfrak m) \setminus V(I_\mathfrak m)$ of $\Spec(A_\mathfrak m)$ is nonempty as the $I_\mathfrak m$-torsion of $A_\mathfrak m$ is zero (use Algebra, Lemma \ref{algebra-lemma-Noetherian-power-ideal-kills-module}). Thus we can find a prime $\mathfrak q \subset A_\mathfrak m$ which defines a point of $U$ (i.e., $IA_\mathfrak m \not \subset \mathfrak q$) and which corresponds to a closed point of $\Spec(A_\mathfrak m) \setminus \{\mathfrak m\}$, see Properties, Lemma \ref{properties-lemma-complement-closed-point-Jacobson}. Then $A_\mathfrak m/\mathfrak q$ is a dimension $1$ local domain. Thus we can find an injective local homomorphism of local rings $A_\mathfrak m/\mathfrak q \subset R$ where $R$ is a discrete valuation ring (Algebra, Lemma \ref{algebra-lemma-exists-dvr}). By construction $IR \subset \mathfrak m_R$ and we see that $A \to R$ extends to a continuous map $A \to R^\wedge$. Since the completion of a discrete valuation ring is a discrete valuation ring, we see that the assumption gives a commutative diagram of rings $$ \xymatrix{ R' & B \ar[l] \\ R^\wedge \ar[u] & A \ar[l] \ar[u] } $$ Thus we find a prime ideal of $B$ lying over $\mathfrak m$. It follows that $\Spec(B/IB) \to \Spec(A/I)$ is surjective, whence $A/I \to B/IB$ is faithfully flat (Algebra, Lemma \ref{algebra-lemma-ff-rings}).
One direction is trivial. For the other, suppose that $\text{Spf}(R) \to Y$ is an adic morphism such that there exists an extension of complete discrete valuation rings $R \subset R'$ with $\text{Spf}(R') \to \text{Spf}(R) \to X$ factoring through $Y$. Then $\Spec(R'/\mathfrak m_R^n R') \to \Spec(R/\mathfrak m_R^n)$ is surjective and flat, hence the morphisms $\Spec(R/\mathfrak m_R^n) \to X$ factor through $X$ as $X$ satisfies the sheaf condition for fpqc coverings, see Formal Spaces, Lemma \ref{formal-spaces-lemma-sheaf-fpqc}. In other words, $\text{Spf}(R) \to Y$ factors through $X$.