## 87.21 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 87.21.2 for a discussion of what to do in more general cases.

Definition 87.21.1. 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 will see in the lemmas below that this notion behaves reasonably well in the context of locally Noetherian formal algebraic spaces and morphisms which are locally of finite type. In the next remark we discuss options for modifying this definition to a wider class of morphisms of formal algebraic spaces.

slogan
Lemma 87.21.3. 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 86.24.3).
$\square$

Lemma 87.21.4. 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 86.24.9 and 86.24.4).
$\square$

Lemma 87.21.5. Let $S$ be a scheme. Let $f : X \to Y$ and $g : Y \to Z$ be morphisms locally of finite type of locally Noetherian formal algebraic spaces over $S$. If $g \circ f$ is rig-surjective and $g$ is a monomorphism, then $f$ is rig-surjective.

**Proof.**
Use Lemma 87.21.4 and that $f$ is the base change of $g \circ f$ by $g$.
$\square$

Lemma 87.21.6. 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 87.21.7. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of locally Noetherian formal algebraic spaces which is representable by algebraic spaces, étale, and surjective. 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 86.19.13. 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 15.44.4). Then $R' = A_\mathfrak m^\wedge $ is a complete discrete valuation ring (More on Algebra, Lemma 15.43.5). Applying Formal Spaces, Lemma 86.9.10. we find the desired morphism $\text{Spf}(R') \to V \to Z \to X$.
$\square$

The upshot of the lemmas above is that we may check whether $f : X \to Y$ is rig-surjective, étale locally on $Y$.

Lemma 87.21.8. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of locally Noetherian formal algebraic spaces which is locally of finite type. Let $\{ g_ i : Y_ i \to Y\} $ be a family of morphisms of formal algebraic spaces which are representable by algebraic spaces and étale such that $\coprod g_ i$ is surjective. Then $f$ is rig-surjective if and only if each $f_ i : X \times _ Y Y_ i \to Y_ i$ is rig-surjective.

**Proof.**
Namely, if $f$ is rig-surjective, so is any base change (Lemma 87.21.4). 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 87.21.7 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 87.21.3). Since this morphism factors through $X \to Y$ we see that $X \to Y$ is rig-surjective by Lemma 87.21.6.
$\square$

Lemma 87.21.9. 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 86.9.10. 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 87.2.3 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 10.119.13). 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 87.21.10. 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 87.21.9. 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 15.88.3). 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 86.9.10.

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 10.62.4). 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 10.119.13). 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 10.39.16).
$\square$

Lemma 87.21.11. 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 86.32.1. In other words, $\text{Spf}(R) \to Y$ factors through $X$.
$\square$

Lemma 87.21.12. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of formal algebraic spaces. Assume that $X$ and $Y$ are locally Noetherian and $f$ is a closed immersion. The following are equivalent

$f$ is rig-surjective, and

for every affine formal algebraic space $V$ and every morphism $V \to Y$ which is representable by algebraic spaces and étale the morphism $X \times _ Y V \to V$ corresponds to a surjective morphism $B \to A$ in $\textit{WAdm}^{Noeth}$ whose kernel $J$ has the following property: $IJ^ n = 0$ for some ideal of definition $I$ of $B$ and some $n \geq 1$.

**Proof.**
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 86.29.5.

Assume (1). By Lemma 87.21.4 we see that $\text{Spf}(A) \to \text{Spf}(B)$ is rig-surjective. Let $I \subset B$ be an ideal of definition. Since $B$ is adic, $I^ m \subset B$ is an ideal of definition for all $m \geq 1$. If $I^ m J^ n \not= 0$ for all $n, m \geq 1$, then $IJ$ is not nilpotent, hence $V(IJ) \not= \mathop{\mathrm{Spec}}(B)$. Thus we can find a prime ideal $\mathfrak p \subset B$ with $\mathfrak p \not\in V(I) \cup V(J)$. Observe that $I(B/\mathfrak p) \not= B/\mathfrak p$ hence we can find a maximal ideal $\mathfrak p + I \subset \mathfrak m \subset B$. By Algebra, Lemma 10.119.13 we can find a discrete valuation ring $R$ and an injective local ring homomorphism $(B/\mathfrak p)_\mathfrak m \to R$. Clearly, the ring map $B \to R$ cannot factor through $A = B/J$. According to Lemma 87.21.11 this contradicts the fact that $\text{Spf}(A) \to \text{Spf}(B)$ is rig-surjective. Hence for some $n, m$ we do have $I^ n J^ m = 0$ which shows that (2) holds.

Assume (2). By Lemma 87.21.8 it suffices to show that $\text{Spf}(A) \to \text{Spf}(B)$ is rig-surjective. Pick an ideal of definition $I \subset B$ and an integer $n$ such that $I J^ n = 0$. Consider a ring map $B \to R$ where $R$ is a discrete valuation ring and the image of $I$ is nonzero. Since $R$ is a domain, we conclude the image of $J$ in $R$ is zero. Hence $B \to R$ factors through the surjection $B \to A$ and we are done by definition of rig-surjective morphisms.
$\square$

Lemma 87.21.13. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of formal algebraic spaces. Assume that $X$ and $Y$ are locally Noetherian and $f$ is a closed immersion. The following are equivalent

$f$ is rig-smooth and rig-surjective,

$f$ is rig-étale and rig-surjective, and

for every affine formal algebraic space $V$ and every morphism $V \to Y$ which is representable by algebraic spaces and étale the morphism $X \times _ Y V \to V$ corresponds to a surjective morphism $B \to A$ in $\textit{WAdm}^{Noeth}$ whose kernel $J$ has the following property: $IJ = 0$ for some ideal of definition $I$ of $B$.

**Proof.**
Let $I$ and $J$ be ideals of a ring $B$ such that $IJ^ n = 0$ and $I(J/J^2) = 0$. Then $I^ nJ = 0$ (proof omitted). Hence this lemma follows from a trivial combination of Lemmas 87.20.9 and 87.21.12.
$\square$

Lemma 87.21.14. Let $S$ be a scheme. Let $f : X \to Y$ and $g : Y \to Z$ be morphisms of locally Noetherian formal algebraic spaces over $S$. Assume

$g$ is locally of finite type,

$f$ is rig-smooth (resp. rig-étale) and rig-surjective,

$g \circ f$ is rig-smooth (resp. rig-étale)

then $g$ is rig-smooth (resp. rig-étale).

**Proof.**
We will prove this in the rig-smooth case and indicate the necessary changes to prove the rig-étale case at the end of the proof. Consider a commutative diagram

\[ \xymatrix{ X \times _ Y V \ar[r] \ar[d] & V \ar[d] \ar[r] & W \ar[d] \\ X \ar[r] & Y \ar[r] & Z } \]

with $V$ and $W$ affine formal algebraic spaces, $V \to Y$ and $W \to Z$ representable by algebraic spaces and étale. We have to show that $V \to W$ corresponds to a rig-smooth map of adic Noetherian topological rings, see Definition 87.18.1. We may write $V = \text{Spf}(B)$ and $W = \text{Spf}(C)$ and that $V \to W$ corresponds to an adic ring map $C \to B$ which is topologically of finite type, see Lemma 87.11.5.

We will use below without further mention that $X \times _ Y V \to V$ is rig-smooth and rig-surjective, see Lemmas 87.18.3 and 87.21.4. Also, the composition $X \times _ Y V \to V \to W$ is rig-smooth since $g \circ f$ is rig-smooth.

Let $I \subset C$ be an ideal of definition. The module Assume $C \to B$ is not rig-smooth to get a contradiction. This means that there exists a prime ideal $\mathfrak q \subset B$ not containing $IB$ such that either $H^{-1}(\mathop{N\! L}\nolimits _{B/C}^\wedge )_\mathfrak p$ is nonzero or $H^0(\mathop{N\! L}\nolimits _{B/C}^\wedge )_\mathfrak p$ is not a finite free $B_\mathfrak q$-module. See Lemma 87.4.2; some details omitted. We may choose a maximal ideal $IB + \mathfrak q \subset \mathfrak m$. By Algebra, Lemma 10.119.13 we can find a complete discrete valuation ring $R$ and an injective local ring homomorphism $(B/\mathfrak q)_\mathfrak m \to R$.

After replacing $R$ by an extension, we may assume given a lift $\text{Spf}(R) \to X \times _ Y V$ of the adic morphism $\text{Spf}(R) \to V = \text{Spf}(B)$. Choose an étale covering $\{ \text{Spf}(A_ i) \to X \times _ Y V\} $ as in Formal Spaces, Definition 86.11.1. By Lemma 87.21.7 we may assume $\text{Spf}(R) \to X \times _ Y V$ lifts to a morphism $\text{Spf}(R) \to \text{Spf}(A_ i)$ for some $i$ (this might require replacing $R$ by another extension). Set $A = A_ i$. Consider the ring maps

\[ C \to B \to A \to R \]

Let $\mathfrak p \subset A$ be the kernel of the map $A \to R$ and note that $\mathfrak p$ lies over $\mathfrak q$. We know that $C \to A$ and $B \to A$ are rig-smooth. In particular the ring map $B_\mathfrak q \to A_\mathfrak p$ is flat by Lemma 87.17.6. Consider the associated exact sequence

\[ \xymatrix{ & H^0(\mathop{N\! L}\nolimits _{B/C}^\wedge ) \otimes _ B A_\mathfrak p \ar[r] & H^0(\mathop{N\! L}\nolimits _{A/C}^\wedge )_\mathfrak p \ar[r] & H^0(\mathop{N\! L}\nolimits _{A/B}^\wedge )_\mathfrak p \ar[r] & 0 \\ 0 \ar[r] & H^{-1}(\mathop{N\! L}\nolimits _{B/C}^\wedge \otimes _ B A)_\mathfrak p \ar[r] & H^{-1}(\mathop{N\! L}\nolimits _{A/C}^\wedge )_\mathfrak p \ar[r] & H^{-1}(\mathop{N\! L}\nolimits _{A/B}^\wedge )_\mathfrak p \ar[llu] } \]

of Lemmas 87.3.5 and 87.17.7. Given the rig-smoothness of $C \to A$ and $B \to A$ we conclude that $H^{-1}(\mathop{N\! L}\nolimits _{B/C}^\wedge \otimes _ B A)_\mathfrak p = 0$ and that $H^0(\mathop{N\! L}\nolimits _{B/C}^\wedge ) \otimes _ B A_\mathfrak p$ is finite free as a kernel of a surjection of finite free $A_\mathfrak p$-modules. Since $B_\mathfrak q \to A_\mathfrak p$ is flat and hence faithfully flat, this implies that $H^{-1}(\mathop{N\! L}\nolimits _{B/C}^\wedge )_\mathfrak q = 0$ and that $H^0(\mathop{N\! L}\nolimits _{B/C}^\wedge )_\mathfrak q$ is finite free which is the contradiction we were looking for.

In the rig-étale case one argues in exactly the same manner but the conclusion obtained is that both $H^{-1}(\mathop{N\! L}\nolimits _{B/C}^\wedge )_\mathfrak q$ and $H^0(\mathop{N\! L}\nolimits _{B/C}^\wedge )_\mathfrak q$ are zero.
$\square$

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