## 86.9 Rig-surjective morphisms

For morphisms locally of finite type between locally Noetherian formal algebraic spaces a definition borrowed from [ArtinII] can be used. See Remark 86.9.10 for a discussion of what to do in more general cases.

Definition 86.9.1. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of formal algebraic spaces over $S$. Assume that $X$ and $Y$ are locally Noetherian and that $f$ is locally of finite type. We say $f$ is *rig-surjective* if for every solid diagram

\[ \xymatrix{ \text{Spf}(R') \ar@{..>}[r] \ar@{..>}[d] & X \ar[d]^ f \\ \text{Spf}(R) \ar[r]^-p & Y } \]

where $R$ is a complete discrete valuation ring and where $p$ is an adic morphism there exists an extension of complete discrete valuation rings $R \subset R'$ and a morphism $\text{Spf}(R') \to X$ making the displayed diagram commute.

We prove a few lemmas to explain what this means.

slogan
Lemma 86.9.2. Let $S$ be a scheme. Let $f : X \to Y$ and $g : Y \to Z$ be morphisms of formal algebraic spaces over $S$. Assume $X$, $Y$, $Z$ are locally Noetherian and $f$ and $g$ locally of finite type. Then if $f$ and $g$ are rig-surjective, so is $g \circ f$.

**Proof.**
Follows in a straightforward manner from the definitions (and Formal Spaces, Lemma 85.18.3).
$\square$

Lemma 86.9.3. Let $S$ be a scheme. Let $f : X \to Y$ and $Z \to Y$ be morphisms of formal algebraic spaces over $S$. Assume $X$, $Y$, $Z$ are locally Noetherian and $f$ and $g$ locally of finite type. If $f$ is rig-surjective, then the base change $Z \times _ Y X \to Z$ is too.

**Proof.**
Follows in a straightforward manner from the definitions (and Formal Spaces, Lemmas 85.18.9 and 85.18.4).
$\square$

Lemma 86.9.4. Let $S$ be a scheme. Let $f : X \to Y$ and $g : Y \to Z$ be morphisms of formal algebraic spaces over $S$. Assume $X$, $Y$, $Z$ locally Noetherian and $f$ and $g$ locally of finite type. If $g \circ f : X \to Z$ is rig-surjective, so is $g : Y \to Z$.

**Proof.**
Immediate from the definition.
$\square$

Lemma 86.9.5. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of formal algebraic spaces which is representable by algebraic spaces, étale, and surjective. Assume $X$ and $Y$ locally Noetherian. Then $f$ is rig-surjective.

**Proof.**
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, étale, and surjective. Hence $Z$ is nonempty. Pick a nonempty affine formal algebraic space $V$ and an étale 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 étale ring map, see Formal Spaces, Lemma 85.14.13. 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 15.43.4). Then $R' = A_\mathfrak m^\wedge $ is a complete discrete valuation ring (More on Algebra, Lemma 15.42.5). Applying Formal Spaces, Lemma 85.5.10. we find the desired morphism $\text{Spf}(R') \to V \to Z \to X$.
$\square$

Lemma 86.9.7. Let $S$ be a scheme. Let $f : X \to Y$ be a proper surjective morphism of locally Noetherian algebraic spaces over $S$. Let $T \subset |Y|$ be a closed subset and let $T' = |f|^{-1}(T) \subset |X|$. Then $X_{/T'} \to Y_{/T}$ is rig-surjective.

**Proof.**
The statement makes sense by Formal Spaces, Lemmas 85.15.6 and 85.18.10. Let $Y_ j \to Y$ be a jointly surjective family of étale morphism where $Y_ j$ is an affine scheme for each $j$. Denote $T_ j \subset Y_ j$ the inverse image of $T$. Then $\{ (Y_ j)_{/T_ j} \to Y_{/T}\} $ is a covering as in Formal Spaces, Definition 85.7.1. Moreover, setting $X_ j = Y_ j \times _ Y X$ and $T'_ j \subset |X_ j|$ the inverse image of $T$, we have

\[ (X_ j)_{/T'_ j} = (Y_ j)_{/T_ j} \times _{(Y_{/T})} X_{/T'} \]

By the discussion in Remark 86.9.6 we reduce to the case where $Y$ is an affine Noetherian scheme treated in the next paragraph.

Assume $Y = \mathop{\mathrm{Spec}}(A)$ where $A$ is a Noetherian ring. This implies that $Y_{/T} = \text{Spf}(A^\wedge )$ where $A^\wedge $ is the $I$-adic completion of $A$ for some ideal $I \subset A$. Let $p : \text{Spf}(R) \to \text{Spf}(A^\wedge )$ be an adic morphism where $R$ is a complete discrete valuation ring. Let $K$ be the field of fractions of $R$. Consider the composition $A \to A^\wedge \to R$. Since $X \to Y$ is surjective, the fibre $X_ K = \mathop{\mathrm{Spec}}(K) \times _ Y X$ is nonempty. Thus we may choose an affine scheme $U$ and an étale morphism $U \to X$ such that $U_ K$ is nonempty. Let $u \in U_ K$ be a closed point (possible as $U_ K$ is affine). By Morphisms, Lemma 29.20.3 the residue field $L = \kappa (u)$ is a finite extension of $K$. Let $R' \subset L$ be the integral closure of $R$ in $L$. By More on Algebra, Remark 15.102.6 we see that $R'$ is a discrete valuation ring. Because $X \to Y$ is proper we see that the given morphism $\mathop{\mathrm{Spec}}(L) = u \to U_ K \to X_ K \to X$ extends to a morphism $\mathop{\mathrm{Spec}}(R') \to X$ over the given morphism $\mathop{\mathrm{Spec}}(R) \to Y$ (Morphisms of Spaces, Lemma 65.44.1). By commutativity of the diagram the induced morphisms $\mathop{\mathrm{Spec}}(R'/\mathfrak m_{R'}^ n) \to X$ are points of $X_{/T'}$ and we find

\[ \text{Spf}((R')^\wedge ) = \mathop{\mathrm{colim}}\nolimits \mathop{\mathrm{Spec}}(R'/\mathfrak m_{R'}^ n) \longrightarrow X_{/T'} \]

as desired (note that $(R')^\wedge $ is a complete discrete valuation ring by More on Algebra, Lemma 15.42.5; in fact in the current situation $R' = (R')^\wedge $ but we do not need this).
$\square$

Lemma 86.9.8. Let $A$ be a Noetherian ring complete with respect to an ideal $I$. Let $B$ be an $I$-adically complete $A$-algebra. If $A/I^ n \to B/I^ nB$ is of finite type and flat for all $n$ and faithfully flat for $n = 1$, then $\text{Spf}(B) \to \text{Spf}(A)$ is rig-surjective.

**Proof.**
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 85.5.10. 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 86.2.3 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 $\mathop{\mathrm{Spec}}(C_\mathfrak m) \setminus V(\mathfrak m_ R C_\mathfrak m)$ is a nonempty open. Hence We can pick a prime $\mathfrak q \subset \mathfrak m$ such that $\mathfrak q$ defines a closed point of $\mathop{\mathrm{Spec}}(C_\mathfrak m) \setminus \{ \mathfrak m\} $ and such that $\mathfrak q \not\in V(IC_\mathfrak m)$, see Properties, Lemma 28.6.4. 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 10.118.13). 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.
$\square$

Lemma 86.9.9. Let $A$ be a Noetherian ring complete with respect to an ideal $I$. Let $B$ be an $I$-adically complete $A$-algebra. Assume that

the $I$-torsion in $A$ is $0$,

$A/I^ n \to B/I^ nB$ is flat and of finite type for all $n$.

Then $\text{Spf}(B) \to \text{Spf}(A)$ is rig-surjective if and only if $A/I \to B/IB$ is faithfully flat.

**Proof.**
Faithful flatness implies rig-surjectivity by Lemma 86.9.8. 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 15.82.3). 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 85.5.10.

Assume $\text{Spf}(B) \to \text{Spf}(A)$ is rig-surjective. Choose a maximal ideal $I \subset \mathfrak m \subset A$. The open $U = \mathop{\mathrm{Spec}}(A_\mathfrak m) \setminus V(I_\mathfrak m)$ of $\mathop{\mathrm{Spec}}(A_\mathfrak m)$ is nonempty as the $I_\mathfrak m$-torsion of $A_\mathfrak m$ is zero (use Algebra, Lemma 10.61.4). 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 $\mathop{\mathrm{Spec}}(A_\mathfrak m) \setminus \{ \mathfrak m\} $, see Properties, Lemma 28.6.4. 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 10.118.13). 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 $\mathop{\mathrm{Spec}}(B/IB) \to \mathop{\mathrm{Spec}}(A/I)$ is surjective, whence $A/I \to B/IB$ is faithfully flat (Algebra, Lemma 10.38.16).
$\square$

Lemma 86.9.11. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of formal algebraic spaces. Assume $X$ and $Y$ are locally Noetherian, $f$ locally of finite type, and $f$ a monomorphism. Then $f$ is rig surjective if and only if every adic morphism $\text{Spf}(R) \to Y$ where $R$ is a complete discrete valuation ring factors through $X$.

**Proof.**
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 $\mathop{\mathrm{Spec}}(R'/\mathfrak m_ R^ n R') \to \mathop{\mathrm{Spec}}(R/\mathfrak m_ R^ n)$ is surjective and flat, hence the morphisms $\mathop{\mathrm{Spec}}(R/\mathfrak m_ R^ n) \to X$ factor through $X$ as $X$ satisfies the sheaf condition for fpqc coverings, see Formal Spaces, Lemma 85.25.1. In other words, $\text{Spf}(R) \to Y$ factors through $X$.
$\square$

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