α$\alpha$ β$\beta$ χ$\chi$ δ$\delta$ ε$\varepsilon$ Δ$\Delta$ η$\eta$
γ$\gamma$ Γ$\Gamma$ ι$\iota$ κ$\kappa$ λ$\lambda$ Λ$\Lambda$ μ$\mu$ ν$\nu$
ω$\omega$ Ω$\Omega$ π$\pi$ φ$\phi$ Φ$\Phi$ ψ$\psi$ Ψ$\Psi$ ρ$\rho$ σ$\sigma$
Σ$\Sigma$ τ$\tau$ θ$\theta$ Θ$\Theta$ ξ$\xi$ ζ$\zeta$

# Non-reduced valuation rings and descent for smooth blowup squares

Shane Kelly

###### Abstract.

We consider a class of non-reduced valuation rings, known in the literature as
_chain rings_. We observe that the Grothendieck topology generated by the
Zariski topology and smooth blowup squares is exactly the topology which has
chain rings for its local rings, and that sheaves for this topology are _not_
characterised by excision for smooth blowup squares.

This is collated from notes from early 2020. The motivation was to find a cdh
topology which can see nilpotents. The idea was the following. The cdh topos
is the classifying topos for hensel valuation rings. A finitary topology on
affine schemes of finite presentation can see nilpotents if and only if its
category of sheaves is the classifying topos for a class of rings which are
not all reduced, Obs.2.16. So one should look for a nice class of rings
related to valuation rings but which are not reduced. Chain rings are such a
class, but the associated topology was abandoned because its sheaves are not
characterised by excision for smooth blowup squares, Prop.1.3.

Upon a suggestion of Shuji Saito, this project turned into the procdh
topology.

##  1\. Summary

We summarise the main points here.

######  Definition 1.1.

A _chain ring_ is a ring whose poset of ideals is totally ordered. Let $S$ be
a qcqs scheme and $\mathrm{Sch}_{S}$ is the category of $S$-schemes of finite
presentation. The _chain topology_ on $\mathrm{Sch}_{S}$ has as coverings
those families $\\{Y_{i}\to X\\}_{i\in I}$ such that

$\amalg\hom(\mathrm{Spec}(R),Y_{i})\to\hom(\mathrm{Spec}(R),X)$

is surjective for every chain ring $R$.

Every valuation ring is a chain ring and every reduced chain ring is a
valuation ring, Lem.3.8. Every localisation and quotient of a chain ring is a
chain ring. I don’t know any examples of chain rings which are not quotients
of valuation rings. There are procdh local rings which are not chain
rings111E.g., $k[x,y]/\langle x^{2},y^{2}\rangle$. and chain rings which are
not procdh local rings.222E.g., $R/\langle xy\rangle$ where $R$ is any
valuation ring with three primes
$\mathfrak{m}\supseteq\mathfrak{p}\supseteq(0)$,
$x\in\mathfrak{m}\setminus\mathfrak{p}$ and $y\in\mathfrak{p}\setminus(0)$.

######  Proposition 1.2 (Proposition 4.3).

Let $\mathcal{U}=\\{Y_{i}\to X\\}_{i\in I}$ be a family of morphisms in
$\mathrm{Sch}_{\mathbb{Z}}$. The following are equivalent.

  1. (1)

$\mathcal{U}$ is a covering for the chain topology.

  2. (2)

$\mathcal{U}$ is refinable by a composition of pullbacks of families of the
form:

    1. (a)

Zariski coverings,

    2. (b)

the family

$\\{\\{0\\}\to\mathbb{A}^{2},Bl_{\mathbb{A}^{2}}\\{0\\}\to\mathbb{A}^{2}\\}.$

In other words, the chain topology is generated by the Zariski topology, and
blowup of the affine plane in the origin.

######  Proposition 1.3 (Corollary 4.6).

There exist chain sheaves $F$ such that

(1)
$\textstyle{F(\mathbb{A}^{2})\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{F(Bl_{\mathbb{A}^{2}}\\{0\\})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{F(\\{0\\})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{F(\mathbb{P}^{1})}$

is not a cartesian square.

In fact, if $τ\leq\mathrm{Zar}$ is a finitary topology on
$\mathrm{Sch}_{\mathbb{Z}}$ such that for all sheaves $F$ the square Eq.(1) is
cartesian, then $τ\leq rh$ and so $F(X)=F(X_{\mathrm{red}})$ for all sheaves
$F$ and $X\in\mathrm{Sch}_{\mathbb{Z}}$.

######  Observation 1.4 (Observation 2.15).

The rh topology on the category $\mathrm{Aff}_{\mathbb{Z}}$ of affine
$\mathbb{Z}$-schemes of finite presentation is generated by the two families

$\\{\mathrm{Spec}\ \mathbb{Z}[x]\to\mathrm{Spec}\
\mathbb{Z}[x,y]/(xy),\quad\mathrm{Spec}\ \mathbb{Z}[y]\to\mathrm{Spec}\
\mathbb{Z}[x,y]/(xy)\\}$

and

$\\{\mathrm{Spec}\ \mathbb{Z}[x,\tfrac{y}{x}]\to\mathrm{Spec}\
\mathbb{Z}[x,y],\quad\mathrm{Spec}\
\mathbb{Z}[\tfrac{x}{y},y]\to\mathrm{Spec}\ \mathbb{Z}[x,y]\\}$

in the sense that a family of morphisms $\\{U_{i}\to X\\}_{i\in I}$ in
$\mathrm{Aff}_{\mathbb{Z}}$ is covering for the rh topology if and only if it
is refinable by a composition of pullbacks of the above two families.

_Conventions and notation._ For a scheme $S$, we write $\mathrm{Sch}_{S}$ for
the category of $S$-schemes of finite presentation, and $\mathrm{Aff}_{S}$ for
the category of (absolutely) affine $S$-schemes of finite presentation.

We use the SGA conventions for pretopologies, topologies, and covering
families. That is, a _pretopology_ is a category equipped with a collection of
families $\\{U_{i}\to X\\}_{i\in I}$ satisfying certain axioms, while a
_topology_ is a category equipped with a collection of sieves
$R\subseteq\hom(-,X)$ satisfying certain axioms. It is an easy exercise to
show that a sieve $R\subseteq\hom(-,X)$ is a $τ$-covering sieve if and only if
its sheafification $a_{\tau}R\to a_{\tau}\hom(-,X)$ is an isomorphism c.f.,
[SGA72, Def.II.1.1(T2), Lemm.II.3.1(2), Thm.II.3.4].

If $τ$ is a topology, a family $\\{U_{i}\to X\\}_{i\in I}$ is called a _$τ$
-covering family_ if the sieve it generates $\cup_{i\in
I}image(\hom(-,U_{i})\to\hom(-,X))$ is a covering. If $τ_{\pi}$ is the
topology associated to a pretopology $\pi$, then one can show that the
$τ_{\pi}$-covering families are precisely those families which can be refined
by a $\pi$-covering family.

##  2\. Some first order logic

In this section, specifically Proposition 2.8, we recall a relationship
between topologies and ring theories, cf.[MM94, Chap.X] or [MR06, Chap.2]. The
reader familiar with the notions such as classifying topos, first order
theory, syntactical site, etc can skip this section.

######  Remark 2.1.

In broad strokes, the equivalence in Proposition 2.8 is the following. A ring
(= a set equipped with two binary relations $+$ and $\cdot$ and two elements
$0,1$ satisfying a list of axioms) of the form

(2) $\mathbb{Z}[x_{1},\dots,x_{n}]/\langle f_{1},\dots,f_{c}\rangle$

corresponds to a sentence (= a list of characters) of the form

(3) $\mathtt{f_{1}(x_{1},\dots,x_{n})=0\ \wedge\ \dots\ \wedge\
f_{c}(x_{1},\dots,x_{n})=0},$

cf.[MM94, §X.5]. Given a topology $τ$ on $\mathrm{Aff}_{\mathbb{Z}}$, we
associate the class $\mathcal{R}_{τ}$ of those rings $R$ such that for every
$τ$-covering family

(4) $\\{\mathrm{Spec}R_{i}\to\mathrm{Spec}R_{0}\\}_{i=1,\dots,m}$

the morphism of sets

(5) $\hom(R_{1},R)\sqcup\dots\sqcup\hom(R_{n},R)\to\hom(R_{0},R)$

is surjective. Conversely, given a class of rings $\mathcal{R}$, we make a
topology $τ_{\mathcal{R}}$ whose covering families are those families Eq.(4)
such that for every $R\in\mathcal{R}$, the morphism of sets Eq.(5) is
surjective. From the logic perspective, surjectivity of Eq.(5) is formalised
in the sentence

(6) $\mathtt{\forall a;\phi_{0}(a),\ \exists b_{1};\phi_{1}(a,b_{1})\ \vee\
\dots\ \vee\ \exists b_{n};\phi_{n}(a,b_{n})}.$

Here the sentences $\phi_{i}$ correspond to the $R_{0}$-algebras $R_{i}$, and
$\phi(a)=\phi(a_{1},\dots,a_{n})$ means substitution of elements $a_{i}\in R$
of the ring into the variables $x_{i}$, etc, in the obvious way. In this way
we obtain a correspondence:

Covering families Eq.(4) for $τ$ $\leftrightarrow$ Axioms Eq.(6) for
$\mathcal{R}$

Note that different collections of families can generate the same topology,
just as different sets of axioms can define the same class of rings. This
correspondence is a special case of the correspondence between coherent
theories and their classifying topoi, cf.[MR06, pg.49 and Thm.9.1.1].

Now let us be more precise. In order for the operations
$τ\mapsto\mathcal{R}_{τ}$ and $\mathcal{R}\mapsto τ_{\mathcal{R}}$ in Remark
2.1 to be mutually inverse, we need some restriction on which topologies and
which classes we consider. On the one hand, our topologies must have enough
points. This is assured for finitary topologies by Deligne’s completeness
theorem, Thm.2.10.333More generally, if $\mathrm{Sch}_{S}$ has countably many
objects (e.g., $S=\mathrm{Spec}(\mathbb{Z})$), and the topology is generated
by countably many coverings, each consisting of countably many morphisms, then
[MR06, Thm.6.2.4] assures that $\mathrm{Shv}(\mathrm{Sch}_{S})$ will have
enough points. In this case, we should make a more general version of
Definition 2.2 where countably many $\vee$ are allowed. Such a sentence could
be called _geometric_. On the other hand, our ring axioms are required to have
a very specific form. As such, it is convenient for us to make the following
nonstandard definition.

######  Definition 2.2.

Given a ring $R$, we will say that a sentence $\phi$ is _coherent_ if it is of
the form444The semicolons should be read as “such that”.

$\displaystyle\mathtt{\forall(a_{1},\dots,a_{n})\in R^{n};\quad f_{1}(a)=0\
\wedge\ }$ $\displaystyle\mathtt{\dots\ \wedge f_{I}(a)=0,}$
$\displaystyle\mathtt{\exists(b_{11},\dots,b_{1m_{1}})\in R^{m_{1}};\quad
g_{11}(a,b)=0\ \wedge\ }$ $\displaystyle\mathtt{\dots\ \wedge\
g_{1K_{1}}(a,b)=0}$
$\displaystyle\mathtt{\vee\quad\exists(b_{21},\dots,b_{1m_{2}})\in
R^{m_{2}};\quad g_{21}(a,b)=0\ \wedge\ }$ $\displaystyle\mathtt{\dots\ \wedge\
g_{1K_{2}}(a,b)=0}$ $\displaystyle\mathtt{\vdots}$
$\displaystyle\mathtt{\vee\quad\exists(b_{J1},\dots,b_{1m_{J}})\in
R^{m_{J}};\quad g_{J1}(a,b)=0\ \wedge\ }$ $\displaystyle\mathtt{\dots\ \wedge\
g_{JK_{J}}(a,b)=0}$

for some polynomials
$f_{i}\in\mathbb{Z}[x_{1},\dots,x_{n}],g_{jk}\in\mathbb{Z}[x_{1},\dots,x_{n},y_{1},\dots,y_{m_{j}}]$
with $i\in\\{1,\dots,I\\}$, $j\in\\{1,\dots,J\\}$, $k\in\\{1,\dots,K_{J}\\}$.

######  Remark 2.3.

The term _coherent_ is in the same sense as a coherent sheaf of
$\mathcal{O}_{X}$-modules and refers to the fact that the sentence $\phi$ is a
list of _finitely many_ symbols, cf.[MR06, pg.49]. The term _first order_ in
the title of this section refers to the fact that our variables will be
interpreted as elements of a ring as opposed to, say, subsets of a rings. So
for example we can’t formalise ascending or descending chain conditions on
ideals in this setup, or indeed, formalise the notion of ideals at all.

######  Definition 2.4.

Say that a class of rings $\mathcal{R}$ is _coherent_ if it can be defined by
a set of coherent sentences. That is, there exists a set
$\\{\phi_{γ}\\}_{γ\in\Gamma}$ of coherent sentences such that a ring belongs
to $\mathcal{R}$ if and only if every sentence $\phi_{γ}$ holds in $R$.

######  Example 2.5.

  1. (1)

The class of local rings is coherent: A ring $R$ is local if and only if for
every element $a\in R$, either $a$ or $1-a$ is a unit. That is,

$\mathtt{\forall a\in R,\qquad\exists b\in R;\ ab-1=0\quad\vee\quad\exists
c\in R;(1-a)c-1=0.}$

Here we have $g_{1}(x,y)=xy-1$, $g_{2}(x,z)=(1-x)z-1$ (and there are no
$f_{i}$, that is, $I=0$).

  2. (2)

The class of integral domains is coherent:

$\mathtt{\forall a,b\in R;ab=0,\qquad a=0,\quad\vee\quad b=0.}$

Here we have $f(x_{1},x_{2})=x_{1}x_{2}$, $g_{1}(x_{1},x_{2})=x_{1}$,
$g_{2}(x_{1},x_{2})=x_{2}$. There are no $y$, that is, $m_{1},m_{2}=0$.

  3. (3)

The class of valuation rings is coherent: A ring $R$ is a valuation ring if
and only if it is an integral domain, and for all $a,b\in R$ we have $a|b$ or
$b|a$. That is,

$\mathtt{\forall a,b\in R,\qquad\exists c\in R;ac=b\quad\vee\quad\exists d\in
R;a=bd.}$

  4. (4)

The class of henselian local rings is coherent, but writing a set of defining
sentences is not pleasant. The reader might take this as a challenge.

  5. (5)

For a fixed $n\in\mathbb{N}$, the class of rings of characteristic $n$ is
coherent, since they are characterised by: $\mathtt{\forall a\in R,na=0}$.
Here, $g(y)=ny$ and there are no $f$. On the other hand, the class of rings of
finite characteristic is not coherent, since we would need infinitely many
$\vee$’s (this class could be called “geometric”).

  6. (6)

It follows directly from the Definition 2.2 and Definition 2.4 that any class
of coherent rings is closed under filtered colimits. That is, if $\mathcal{R}$
is a coherent class of rings and $(R_{λ})_{λ\in\Lambda}$ is a filtered system
of rings in $\mathcal{R}$, then $\operatorname{colim}R_{λ}$ is also in
$\mathcal{R}$. Consequently, the class of Noetherian rings is not coherent.

  7. (7)

The class of $w$-local rings in the sense of [BS13] is not coherent. To see
this, note that the topology on $\mathrm{Sch}_{\mathbb{Z}}$ associated to the
class of $w$-local rings is the topology from Example 2.7(1), but the local
rings $R$ of this topology do not have to satisfy: the set of closed points of
$\mathrm{Spec}(R)$ is closed.

We make one more definition.

######  Definition 2.6.

A topology on $\mathrm{Sch}_{\mathbb{Z}}$ will be called _affine_ if it
induces an equivalence of categories
$\mathrm{Shv}(\mathrm{Aff}_{\mathbb{Z}})\cong\mathrm{Shv}(\mathrm{Sch}_{\mathbb{Z}})$
when we equip $\mathrm{Aff}_{\mathbb{Z}}$ with the induced topology.555A
family $\\{U_{i}\to X\\}_{i\in I}$ in $\mathrm{Aff}_{\mathbb{Z}}$ is a
covering family if and only if it is a covering family in
$\mathrm{Sch}_{\mathbb{Z}}$, [SGA72, Cor.III.3.3]. Equivalently, every
covering is refinable by one of the form $\\{\mathrm{Spec}(A_{i})\to
X\\}_{i\in I}$.

######  Example 2.7.

Any topology finer than the Zariski topology is affine. On the other hand,
there are affine topologies which are not finer than the Zariski topology.

  1. (1)

A silly example is the topology generated by singletons $\\{Y\to X\\}$ such
that $Y\in\mathrm{Sch}_{\mathbb{Z}}$ is a surjective disjoint union of open
affines of $X$. Sheaves for this topology do not necessarily satisfy
$F(X\sqcup X^{\prime})=F(X)\times F(X^{\prime})$.

  2. (2)

Another silly example is the topology generated by the families
$\\{\mathrm{Spec}(A)\to X\\}_{(\mathrm{Aff}_{\mathbb{Z}})_{/X}}$ consisting of
the set of _every_ morphism from an affine scheme towards $X$. Sheaves for
this topology are precisely those presheaves on $\mathrm{Sch}_{\mathbb{Z}}$
which are right Kan extended from $\mathrm{Aff}_{\mathbb{Z}}$.

  3. (3)

A less silly example is the topology generated by surjective proper morphisms,
and $\mathbb{A}^{n}$-bundles (for all $n\geq 0$). Indeed, by Nagata
compactification and Chow’s lemma, any $X$ admits a surjective proper morphism
$Y\to X$ with $Y$ quasi-projective. Then by Jouanolou’s trick, $Y$ admits a
surjective $\mathbb{A}^{n}$-bundle $E\to Y$ for some $n$ such that $E$ is
affine.

We introduce the above terminology to state the following proposition. Recall
that a topology is _finitary_ if every covering family contains a finite
subfamily which is also covering.

######  Proposition 2.8.

There is a bijection of posets between:

  1. (1)

affine finitary topologies $τ$ on $\mathrm{Sch}_{\mathbb{Z}}$, and

  2. (2)

coherent classes of rings.

The bijection sends a class $\mathcal{R}$ to the topology $τ_{\mathcal{R}}$
whose coverings are those families $\\{U_{i}\to X\\}_{i\in I}$ such that for
all $R$ in $\mathcal{R}$ the morphism of sets

(7) $\amalg_{i\in I}\hom(\mathrm{Spec}(R),U_{i})\to\hom(\mathrm{Spec}(R),X)$

is surjective. It sends a topology $τ$ to the class $\mathcal{R}_{τ}$ of rings
$R$ such that for all $τ$-covering families (7) is surjective.

######  Definition 2.9.

Given a ring $R$ and a family $\mathcal{U}=\\{U_{i}\to X\\}_{i\in I}$ let’s
name the condition:

  1. $(R\perp\mathcal{U})$

The morphism $\amalg_{i\in
I}\hom(\mathrm{Spec}(R),U_{i})\to\hom(\mathrm{Spec}(R),X)$ is surjective.

so we can refer to it more easily.

Proposition 2.8 is a formal consequence of Deligne’s completeness theorem.

######  Theorem 2.10 (Deligne, [Sta18, 00YQ]).

Suppose $C$ is a small category admitting finite limits equipped with a
finitary topology $τ$. Then the collection of fibre functors
$\phi:\mathrm{Shv}_{τ}(C)\to\mathrm{Set}$ is conservative.

That is, a morphism $f:F\to G$ of sheaves is an isomorphism if and only if
$\phi(f)$ is an isomorphism for every $\phi$ which preserves colimits and
finite limits.

######  Remark 2.11.

There is an equivalent statement of Deligne’s theorem: Suppose $C$ is a small
category admitting finite limits equipped with a finitary topology $τ$. Then a
family $\\{Y_{i}\to X\\}_{i\in I}$ is a covering family if and only if

$\amalg_{i\in I}\phi(a_{τ}\hom(-,Y_{i}))\to\phi(a_{τ}\hom(-,X))$

is a surjective morphism of sets for every $\phi$ which preserves colimits and
finite limits, [SGA72, Prop.IV.6.5].

We will also use the following well-known formal fact, explained in [GK15].

######  Lemma 2.12.

Suppose that $τ$ is an affine topology on $\mathrm{Sch}_{\mathbb{Z}}$. Then
there is a canonical bijection

$\mathcal{R}_{τ}\cong\left\\{\begin{array}[]{c}\textrm{ fibre functors }\\\
\textrm{ of }\mathrm{Shv}_{τ}(\mathrm{Sch}_{\mathbb{Z}})\end{array}\right\\}$

A ring $R\in\mathcal{R}_{τ}$ corresponds to the functor

$\phi_{R}:F\mapsto\underset{{\mathrm{Spec}(R)\to
X}}{\operatorname{colim}}F(X)$

where the colimit is over all $X\in\mathrm{Sch}_{\mathbb{Z}}$.

######  Remark 2.13.

Note that since objects in $\mathrm{Sch}_{\mathbb{Z}}$ are finite
presentation, and we can write $\mathrm{Spec}\ R$ as a filtered limit of
affine $\mathbb{Z}$-schemes of finite presentation, for
$Y\in\mathrm{Sch}_{\mathbb{Z}}$ we have, [Gro66, Prop.8.13.1],

$\phi_{R}(\hom(-,Y))=\hom(\mathrm{Spec}(R),Y).$

###### Proof of Proposition 2.8..

First we show that $\mathcal{R}=\mathcal{R}_{τ_{\mathcal{R}}}$ for every
coherent class $\mathcal{R}$. Let $\mathcal{R}$ be a coherent class and choose
a set of defining sentences $\\{\phi_{γ}\\}_{γ\in\Gamma}$. To a coherent
sentence $\phi$ as in Definition 2.2, we associate the family

$\mathcal{U}_{\phi}=\biggl{\\{}Y_{\phi k}:=\underset{}{\mathrm{Spec}\
\tfrac{\mathbb{Z}[x_{1},\dots,x_{n},y_{1},\dots,y_{m_{j}}]}{\langle
f_{i},g_{jk}\rangle}}\to X_{\phi}:=\mathrm{Spec}\
\tfrac{\mathbb{Z}[x_{1},\dots,x_{n}]}{\langle f_{i}\rangle}\biggr{\\}}_{j\in
J}.$

Then we have

$R\in\mathcal{R}\iff\forall γ\in\Gamma;(R\perp\mathcal{U}_{\phi_{γ}}).$

Now note that one can concretely describe the coverings of $τ_{\mathcal{R}}$.
They are those families which are refinable by a composition of pullbacks of a
families of the form $\mathcal{U}_{\phi_{γ}}$; $γ\in\Gamma$. The condition
$(R\perp\mathcal{U})$ is preserved by pullback, composition, and corefinement
in $\mathcal{U}$, in the sense that if $(R\perp\\{U_{i}{\to}X\\})$ holds then
$(R\perp\\{Y\times_{X}U_{i}\to Y\\})$ holds, etc. So it follows that (for a
fixed ring $R$) we have

$\forall γ\in\Gamma;(R\perp\mathcal{U}_{\phi_{γ}})\iff\forall\mathcal{U}\in
τ_{\mathcal{R}};(R\perp\mathcal{U}),$

where $\mathcal{U}\in τ$ means $\mathcal{U}$ is a covering family for $τ$.
Combining the two display equations gives
$\mathcal{R}=\mathcal{R}_{τ_{\mathcal{R}}}$, since by definition,

$R\in\mathcal{R}_{τ_{\mathcal{R}}}\iff\forall\mathcal{U}\in
τ_{\mathcal{R}};(R\perp\mathcal{U}).$

Now we show $τ=τ_{\mathcal{R}_{τ}}$. As above, it follows immediately from the
definitions that $τ_{\mathcal{R}_{τ}}\leq τ$. Indeed, $\mathcal{U}\in
τ\implies\forall R\in\mathcal{R}_{τ};(R\perp\mathcal{U})\iff\mathcal{U}\in
τ_{\mathcal{R}_{τ}}$.

For the converse, suppose $\mathcal{U}=\\{U_{i}\to X\\}_{i\in I}$ is a
$τ_{\mathcal{R}_{τ}}$-covering family. By definition, this means that
$(R\perp\mathcal{U})$ holds for every $R$ in $\mathcal{R}_{τ}$. That is, for
every $R$ in $\mathcal{R}_{τ}$ the morphism

$\amalg\phi_{R}(\hom(-,U_{i}))\to\phi_{R}(\hom(-,X))$

is surjective, cf.Remark 2.13. By Lemma 2.12, this means precisely that
$\amalg\phi a_{τ}\hom(-,U_{i})\to\phi a_{τ}\hom(-,X)$ is surjective for every
fibre functor of $\mathrm{Shv}_{τ}(\mathrm{Sch}_{\mathbb{Z}})$. By Deligne’s
theorem in the form of Remark 2.11, it follows that $\mathcal{U}$ is a
$τ$-covering family. ∎

######  Remark 2.14.

Looking at the proof of Proposition 2.8 one notes that
$\mathcal{R}=\mathcal{R}_{τ_{\mathcal{R}}}$ also holds for _geometric_
classes, where by geometric we mean that we allow infinitely many $\vee$ in
the sentences in Definition 2.2 (i.e., indexed by a small set). On the other
hand, the $\tau=\tau_{\mathcal{R}_{\tau}}$ direction only uses the existence
of a conservative family of fibre functors. So the same proof gives the
following more general statement.

There is a bijection of posets between:

  1. (1)

affine topologies $τ$ on $\mathrm{Sch}_{\mathbb{Z}}$ admitting a conservative
family of fibre functors, and

  2. (2)

geometric classes of rings.

If one starts with an arbitrary affine topology $τ$ on
$\mathrm{Sch}_{\mathbb{Z}}$ then $τ_{\mathcal{R}_{τ}}$ is the coarsest
topology whose family of fibre functors is conservative such that $τ\leq
τ_{\mathcal{R}_{τ}}$. Conversely, if we start with an arbitrary class of rings
$\mathcal{R}$ then $\mathcal{R}_{τ_{\mathcal{R}}}\supseteq\mathcal{R}$ is the
class of those rings which satisfy every geometric sentence satisfied by all
rings in $\mathcal{R}$.

We can apply Proposition 2.8 to make the following cute observations.

######  Observation 2.15.

  1. (1)

The Zariski topology on $\mathrm{Aff}_{\mathbb{Z}}$ is generated by the family

$\\{\mathbb{A}^{1}\setminus\\{0\\}\to\mathbb{A}^{1},\mathbb{A}^{1}\setminus\\{1\\}\to\mathbb{A}^{1}\\}$

in the sense that every Zariski covering is refinable by a composition of
pullbacks of this family. One can also show this directly without too much
difficult (exercise).

Note, that this covering does not generate the Zariski topology on
$\mathrm{Sch}_{\mathbb{Z}}$, since there are no surjective morphisms from
proper schemes towards $\mathbb{A}^{1}$.666Incidentally, the Zariski topology
on $\mathrm{Sch}_{\mathbb{Z}}$ is generated by
$\\{\mathbb{P}^{1}\setminus\\{0\\}\to\mathbb{P}^{1},\mathbb{P}^{1}\setminus\\{\infty\\}\to\mathbb{P}^{1}\\}$
(exercise).

  2. (2)

The rh-topology on $\mathrm{Aff}_{\mathbb{Z}}$ is generated by the two
families

$\\{\mathrm{Spec}\ \mathbb{Z}[x]\to\mathrm{Spec}\
\mathbb{Z}[x,y]/(xy),\quad\mathrm{Spec}\ \mathbb{Z}[y]\to\mathrm{Spec}\
\mathbb{Z}[x,y]/(xy)\\}$ $\\{\mathrm{Spec}\
\mathbb{Z}[x,\tfrac{y}{x}]\to\mathrm{Spec}\
\mathbb{Z}[x,y],\quad\mathrm{Spec}\
\mathbb{Z}[\tfrac{x}{y},y]\to\mathrm{Spec}\ \mathbb{Z}[x,y]\\}$

Now a more serious observation.

######  Observation 2.16.

A finitary affine topology $τ$ on $\mathrm{Sch}_{\mathbb{Z}}$ has
$F(X)=F(X_{\mathrm{red}})$ for every sheaf $F$ and
$X\in\mathrm{Sch}_{\mathbb{Z}}$ if and only if every $R\in R_{τ}$ is reduced.

Hence, if we want a “cdh-topology” which can see nilpotents, the obvious
choice is $τ_{\mathcal{R}}$ for some class of nonreduced rings $\mathcal{R}$
containing henselian valuation rings.

##  3\. Non-reduced valuation rings; definition

Recall the following.

######  Lemma 3.1.

Suppose that $R$ is an integral domain. The following conditions are
equivalent.

  1. (1)

The poset of ideals of $R$ is totally ordered.

  2. (2)

Every finitely generated ideal of $R$ is principal, and $R$ is local.

  3. (3)

For every $x,y\in R$ either $x$ divides $y$ or $y$ divides $x$.

###### Proof.

$(1)\Rightarrow(2)$. For locality, notice that since the ideals are totally
ordered, the union of the non-proper ideals is again a non-proper ideal,
necessarily maximal. For principality, suppose that $I$ is an ideal with $n$
generators $f_{1},\dots,f_{n}$. Since the poset of ideals is totally ordered,
either $(f_{n})\subseteq(f_{n-1})$ or $(f_{n-1})\subseteq(f_{n})$. Hence, $I$
can be generated by $n-1$ elements. By induction on $n$, the ideal $I$ is
principal.

$(2)\Rightarrow(3)$. Suppose $I$, $J$ are two ideals, principal by assumption,
so $I=(x)$ and $J=(y)$. Again by assumption $(x,y)$ is principal, so there
exists $z$ with $(x,y)=(z)$. That is, there exist $a,b,c,d$ with $z=ax+by$ and
$x=cz$, $y=dz$. Substituting the former into the latter gives $(1-ca)x=cby$.
Since $R$ is local, either $ca$ is a unit, or $1-ca$ is a unit. If $ca$ is a
unit then $c$ is a unit so $(y)\subseteq(x,y)=(z)=(cz)=(x)$. That is, $x$
divides $y$. If $1-ca$ is a unit then $(x)=((1-ca)x)=(cby)\subseteq(y)$. That
is, $y$ divides $x$.

$(3)\Rightarrow(1)$. If there exist ideals $I$, $J$ with $I\not\subseteq J$
and $J\not\subseteq I$, then there exist $x,y$ with $x\in I$, $y\in J$,
$x\notin J$, $y\notin I$. But then we cannot have $x$ divides $y$ or $y$
divides $x$. ∎

######  Definition 3.2.

An integral domain satisfying the equivalent conditions of Lemma 3.1 is known
as a _valuation ring_. Noetherian valuation rings are called _discrete
valuation rings_.

######  Remark 3.3.

The term “valuation” comes from the fact that the abelian group
$\Gamma:=(\mathrm{Frac}R)^{*}/R^{*}$ has a canonical total ordering, induced
by the divisibility relation. The canonical map $v:R\setminus\\{0\\}\to\Gamma$
is called the _valuation_ of $R$. If $R$ is Noetherian we necessarily have
$\Gamma\cong(\mathbb{Z},+)$ via a choice of generator for the maximal ideal
$\mathfrak{m}$.

Now observe that in the proof of Lemma 3.1, the assumption that $R$ was an
integral domain was never used.

######  Definition 3.4.

A ring (not necessarily an integral domain) satisfying the equivalent
conditions of Lemma 3.1 is known as a _chain ring_.

######  Example 3.5.

  1. (1)

Every valuation ring is a chain ring.

  2. (2)

Every localisation of a chain ring is a chain ring.

  3. (3)

Every quotient of a chain ring is a chain ring.

  4. (4)

Conversely, one can show that every Noetherian chain ring is a field, or a
quotient of a discrete valuation ring, Prop.5.1.

  5. (5)

It is also possible to show that if $R$ is a chain ring over $\mathbb{F}_{p}$
and the Frobenius is surjective, then $R$ is a quotient of a (perfect)
valuation ring, Prop.5.4.

######  Remark 3.6.

We do not know examples of any chain rings which are not quotients of
valuation rings. One may conjecture that every chain ring is a quotient of a
valuation ring.

######  Question 3.7.

Is every chain ring the quotient of a valuation ring?

Clearly every chain ring which is an integral domain is a valuation ring (by
definition). But in fact it suffices to be reduced. This follows from the more
general fact that the minimal prime is the nilradical.

######  Lemma 3.8.

Suppose that $R$ is a chain ring. Then $R$ has a unique minimal prime, and it
is the nilradical of $R$. Consequently, if $R$ is a reduced chain ring, then
$R$ is a valuation ring.

######  Remark 3.9.

In particular, this shows that chain rings $R$ have the curious property that
the nilradical $Nil(R)$ is the unique minimal prime ideal $\mathfrak{n}$, and
the set of zero divisors $ZD(R)$ is the unique maximal ideal $\mathfrak{m}$.

$Nil(R)=\mathfrak{n},\qquad ZD(R)=\mathfrak{m}.$

###### Proof.

First recall that the nilradical $\mathfrak{n}$ is the intersection of all
prime ideals. Since the ideals in a chain ring are totally ordered, we are
reduced to showing that this intersection is again a prime ideal. Suppose that
$ab\in\mathfrak{n}$. For all primes $\mathfrak{p}$, either $a\in\mathfrak{p}$
or $b\in\mathfrak{p}$. If both $a,b$ are in all primes, then
$a,b\in\mathfrak{n}$. If there is a prime which does not contain, say $a$,
then all smaller primes do not contain $a$, so all smaller primes must contain
$b$. So $b=\cap\mathfrak{p}=\mathfrak{n}$.

For the “consequently” part, if $R$ is a reduced chain ring,
$\mathfrak{n}=(0)$, i.e., $(0)$ is prime, so $R$ is an integral domain. ∎

The following lemma may or may not be useful.

######  Lemma 3.10.

Let $A\neq 0$ be a zero dimensional chain ring. Then $A$ is a colimit (even
union) of local Artin rings (but not necessarily Artin chain rings).

###### Proof.

Consider any finite set of elements $a_{1},\dots,a_{n}\subseteq A$ and the
induced morphism $\mathbb{Z}[t_{1},\dots,t_{n}]\to A$. Let
$\mathfrak{p}\subseteq\mathbb{Z}[t_{1},\dots,t_{n}]$ be the preimage of the
unique prime of $A$. Let
$\mathcal{O}=\mathbb{Z}[t_{1},\dots,t_{n}]_{\mathfrak{p}}$ be the localisation
at $\mathfrak{p}$ so we now have a local morphism of local rings
$\mathcal{O}\to A$. Let $\mathfrak{m}_{\mathcal{O}}$ be the maximal ideal of
$\mathcal{O}$, and $f_{1},\dots,f_{m}$ a finite set of generators. They are
sent inside the maximal ideal $\mathfrak{m}_{A}$ of $A$, as $\mathcal{O}\to A$
is a local homomorphism. But $\mathfrak{m}_{A}$ consists of nilpotents. So
there is some integer $N$ such that $\mathcal{O}\to A$ factors as
$\mathcal{O}/\mathfrak{m}_{\mathcal{O}}^{N}\to A$. So now we have a morphism
from an Artin ring whose image contains all the original $a_{1},\dots,a_{n}$.
Since $R$ is non-zero, the kernel of
$\mathcal{O}/\mathfrak{m}_{\mathcal{O}}^{N}\to A$ is contained in
$\mathfrak{m}_{\mathcal{O}}$, and quotienting by this, we obtain an injective
morphism $\mathcal{O}/ker\subseteq A$ from an Artin ring
$A_{0}:=\mathcal{O}/ker$. ∎

##  4\. Chain rings as local rings

######  Definition 4.1.

Let $τ_{\operatorname{chain}}$ be the topology such that
$\mathcal{R}_{τ_{\operatorname{chain}}}$ is the class of chain rings. Let’s
call this the _chain topology_.

In the spirit of Observation 2.15 we have:

######  Observation 4.2.

The chain ring topology on $\mathrm{Aff}_{\mathbb{Z}}$ is generated by the
family

$\\{\mathrm{Spec}\ \mathbb{Z}[x,\tfrac{y}{x}]\to\mathrm{Spec}\
\mathbb{Z}[x,y],\quad\mathrm{Spec}\
\mathbb{Z}[\tfrac{x}{y},y]\to\mathrm{Spec}\ \mathbb{Z}[x,y]\\}.$

In fact, the chain ring topology can also be described in terms of smooth
blowup squares.

######  Proposition 4.3.

Let $\mathcal{U}=\\{Y_{i}\to X\\}_{i\in I}$ be a family of morphisms in
$\mathrm{Sch}_{\mathbb{Z}}$. The following are equivalent.

  1. (1)

For every chain ring $R$ the morphism

$\amalg\hom(\mathrm{Spec}(R),Y_{i})\to\hom(\mathrm{Spec}(R),X)$

is surjective. That is, $\mathcal{U}$ is a covering for the chain topology.

  2. (2)

$\mathcal{U}$ is refinable by a composition of pullbacks of families of the
form:

    1. (a)

Zariski coverings,

    2. (b)

smooth blowups, i.e, families of the form

$\\{Z\to X,Bl_{X}Z\to X\\}.$

where $Z$ and $X$ are smooth.

  3. (3)

$\mathcal{U}$ is refinable by a composition of pullbacks of families of the
form:

    1. (a)

Zariski coverings,

    2. (b)

the families

$\\{\\{0\\}\to\mathbb{A}^{n},Bl_{\mathbb{A}^{n}}\\{0\\}\to\mathbb{A}^{n}\\},$

for $n\geq 2$.

  4. (4)

$\mathcal{U}$ is refinable by a composition of pullbacks of families of the
form:

    1. (a)

Zariski coverings,

    2. (b)

the family

$\\{\\{0\\}\to\mathbb{A}^{2},Bl_{\mathbb{A}^{2}}\\{0\\}\to\mathbb{A}^{2}\\}.$

###### Proof.

Clearly $(1)\iff(4)\implies(3)\implies(2)$, so it suffices to show that the
families $\\{Z\to X,Bl_{X}Z\to X\\}$ in (2) satisfy (1). Suppose that $R$ is a
chain ring and $\mathrm{Spec}(R)\to X$ is any morphism. We will show that
$\mathrm{Spec}(R)\to X$ factors through $Bl_{X}Z$. Let $x\in X$ be the image
of the closed point of $\mathrm{Spec}(R)$. Since $X$ and $Z$ are smooth, there
exists an open affine neighbourhood $U$ of $x$, and an étale morphism
$f:U\to\mathbb{A}^{d}$ such that $Z\cap U=f^{-1}\mathbb{A}^{d-c}$ where $d$ is
the dimension of $X$ and $c$ the local codimension of $Z$. Replacing $X$ with
$U$, we can assume the morphism is $f:X\to\mathbb{A}^{d}$. Since it is étale,
it is flat, so $Bl_{X}Z$ is the pullback of
$Bl_{\mathbb{A}^{d}}\mathbb{A}^{d-c}$. So in fact we can assume
$X=\mathbb{A}^{d}$ and $Z=\mathbb{A}^{d-c}$. But this is the pullback of
$Bl_{\mathbb{A}^{c}}\\{0\\}$ so we assume $d=c$.

Let $a_{1},\dots,a_{d}$ be the images in $R$ of
$x_{1},\dots,x_{d}\in\mathbb{Z}[x_{1},\dots,x_{d}]$. Since $R$ is a chain
ring, there is one $a_{i}$ which divides the others. So
$\mathrm{Spec}(R)\to\mathrm{Spec}\ \mathbb{Z}[x_{1},\dots,x_{d}]$ factors
through $\mathrm{Spec}\
\mathbb{Z}[\tfrac{x_{1}}{x_{i}},\dots,x_{i},\dots,\tfrac{x_{d}}{x_{i}}]$. But
this is one of the standard opens of the blowup $Bl_{X}Z$. So
$\mathrm{Spec}(R)\to X$ factors through $Bl_{X}Z$. ∎

An obvious question is if the chain ring topology fits into Voevodsky’s theory
of (bounded, complete, regular) cd-structures. It does not.

######  Proposition 4.4.

Suppose that we are given a collection $P$ of cartesian squares

$\textstyle{B\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{Y\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{A\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{i}$$\textstyle{X}$

in $\mathrm{Sch}_{\mathbb{Z}}$ such that $i$ is a categorical monomorphism.
Let $τ$ be the topology generated by $P$,777So $τ$-covering families are those
which are refinable by composition of pullbacks of families of the form
$\\{A\to X,Y\to X\\}$ for some square in $P$., suppose $τ$ is affine, and let
$\mathcal{R}$ be the class of $τ$-local rings. Then the following are
equivalent.

  1. (1)

A presheaf $F$ is a sheaf if and only if

$\textstyle{F(B)}$$\textstyle{\ignorespaces\ignorespaces\ignorespaces\ignorespaces
F(Y)}$$\textstyle{F(A)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\ignorespaces\ignorespaces\ignorespaces\ignorespaces
F(X)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$

is cartesian for all pullbacks of squares in $P$.

  2. (2)

The image of every square of $P$ in
$\mathrm{Shv}_{τ}(\mathrm{Sch}_{\mathbb{Z}})$ is cocartesian.

  3. (3)

For every $R\in\mathcal{R}$ the square

$\textstyle{\phi_{R}(B)\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\phi_{R}(Y)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\phi_{R}(A)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\phi_{R}(X)}$

is a cocartesian square of sets.

  4. (4)

For every $R\in\mathcal{R}$ the morphism

$\phi_{R}(Y)\setminus\phi_{R}(B)\to\phi_{R}(X)\setminus\phi_{R}(A)$

is injective.

######  Remark 4.5.

Pay attention that the morphism in part (4) is not the morphism
$\phi_{R}(Y\setminus B)\to\phi_{R}(X\setminus A)$.

###### Proof.

(1) $\iff$ (2) is essentially just Yoneda, colimits being universal in topoi,
and the factorisation $B\to(A\sqcup Y)\times_{X}(A\sqcup
Y)\rightrightarrows(A\sqcup Y)$.

(2) $\iff$ (3) is the fact discussed above that the $\phi_{R}$ form a
conservative family of fibre functors (consider the morphism $A\sqcup_{B}Y\to
X$ in $\mathrm{Sch}_{τ}(\mathrm{Sch}_{\mathbb{Z}})$).

(3) $\iff$ (4) is the fact that a cartesian square of sets with monic
horizontal morphisms is cocartesian if and only if the “complement” morphism
is a monomorphism. ∎

######  Corollary 4.6.

Suppose $τ\leq\mathrm{Zar}$ is a finitary topology on
$\mathrm{Sch}_{\mathbb{Z}}$ such that the image of

$\textstyle{\mathbb{P}^{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{Bl_{\mathbb{A}^{2}}\\{0\\}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\\{0\\}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathbb{A}^{2}}$

in $\mathrm{Shv}_{τ}(\mathrm{Sch}_{\mathbb{Z}})$ is cocartesian. Then all
local rings are valuation rings. In particular, $τ\leq rh$ and all
$X_{\mathrm{red}}\to X$ are sent to isomorphisms in
$\mathrm{Shv}_{τ}(\mathrm{Sch}_{\mathbb{Z}})$.

###### Proof.

First consider the cd-structure which is the union of the Zariski cd-
structure, and the square above. By Proposition 4.3 the local rings of $τ$ are
all chain rings. On the other hand, by Proposition 4.4 the morphism888This is
the composition of one of the standard opens $U\subseteq
Bl_{\mathbb{A}^{2}}\\{0\\}$ and the canonical morphism
$Bl_{\mathbb{A}^{2}}\\{0\\}\to\mathbb{A}^{2}$.

$R\times R\to R\times R;\qquad(x,y)\mapsto(x,xy)$

is injective when restricted to the preimage of $R\times
R\setminus\\{(0,0)\\}$. In particular, if there exist $x,y\in R$ with $x\neq
0$ and $xy=0$, then we must have $y=0$. That is, there are no nonzero zero
divisors. A chain ring without nonzero zerodivisors is a valuation ring by
Lemma 3.8. Hence $τ\leq rh$ by Proposition 2.8. ∎

##  5\. Chain rings as quotients of valuation rings

######  Proposition 5.1.

Suppose that $A$ is a Noetherian chain ring. Then either:

  1. (1)

$A$ is a field,

  2. (2)

$A$ is a dvr,

  3. (3)

$A=k[[t]]/(t^{n})$ for some field $k$ and some integer $n\geq 1$, or

  4. (4)

$A=\Lambda/(p^{n})$ for some complete discrete valuation ring $\Lambda$ whose
uniformiser is a prime $p\in\mathbb{Z}$ and $n\geq 1$.

###### Proof.

Let $A$ be a Noetherian chain ring. Let $t$ be a nonzero divisor, and $e$ a
zero divisor. We claim that either $t$ is a unit or $e=0$. This implies that
either:

  1. (1)

all nonzero divisors are units, or

  2. (2)

$A$ is an integral domain.

Proof that either $t\in A^{*}$ or $e=0$. Let $a\in A$ be such that $at=e$ or
$t=ae$. We cannot have $t=ae$ because $t$ is not a zero divisor. So $at=e$.
The same is true for $t^{n}$ in place of $t$ for all $n\geq 1$. So we find a
sequence of elements $a_{1},a_{2},a_{3},\dots$, such that $a_{n}t^{n}=e$.
Notice that $a_{n}t^{n}=a_{n+1}t^{n+1}$ so $(a_{n}-a_{n+1}t)t^{n}=0$. Since
$t$ is not a zero divisor, we find that $a_{n}=a_{n+1}t$, or in other words,
$(a_{n})\subseteq(a_{n+1})$. Since $A$ is Noetherian, this sequence of ideals
stabilises. So there is some $n$ such that $a_{n}$ and $a_{n+1}$ differ by a
unit. That is, $a_{n}=ua_{n}t$ for some unit $u$. Then $a_{n}(1-ut)=0$. If $t$
is not a unit, then $(1-ut)$ is a unit, so $a_{n}=0$, and therefore
$e=a_{n}t^{n}=0$.

If $A$ is a Noetherian integral domain, then it is a field or a dvr, so assume
that $A$ is not an integral domain. By the above, this implies all nonzero
divisors in $A$ are units. Since $A$ is local, this implies that the maximal
ideal—the set of non-units—is exactly the set of zero divisors. Using a
similar argument to the above we claim that we can show that all zero divisors
are nilpotent.

Proof that all zero divisors are nilpotent. Suppose that $xy=0$, with $x,y\neq
0$, and that for some $n\in\mathbb{N}$ we have found

$y_{-1}=0,\quad y_{0}=y,\quad y_{1},\dots,y_{n}$

such that $y_{n}=y_{n+1}x$ for all $n=-1,\dots,n{-}1$. Note that by induction
we have $y_{i}x^{i}=y$ (it may help to think of $y_{i}$ as $y/x^{i}$). Now we
must have $x|y_{n}$ or $y_{n}|x$. In the case $n=0$, without loss of
generality we can assume $y=y_{1}x$ for some $y_{1}$. In the case $n>0$, if
$x=zy_{n}$ then $xx^{n}x=zy_{n}x^{n}x=zyx=0$ and $y^{n+2}=(y_{1}x)^{n+2}=0$,
so $x$ and $y$ are nilpotent. If not, then $y_{n}=y_{n+1}x$ and we continue
the sequence. So either $x$ and $y$ are nilpotent, or we can find $y_{n}$ for
all $n\in\mathbb{N}$ with $y_{n}=y_{n+1}x$. Since $A$ is Noetherian, the
sequence of ideals $\dots\subseteq(y_{n})\subseteq(y_{n+1})\subseteq$
stabilises, and we have $y_{n+1}=zy_{n}$ for some $n$. Since we also have
$y_{n}=y_{n+1}x$ we find that $y_{n+1}(1-zx)=0$. Since $x$ is a zero divisor,
$zx$ cannot be a unit, so $1-zx$ is a unit, and we find that $y_{n+1}=0$,
implying that $y_{n+1}x^{n+1}=y=0$, a contradiction.

So all zero divisors are nilpotent. We have seen previously that the set of
zero divisors is the maximal ideal, so we deduce that $A/\mathfrak{m}=A_{red}$
is a field. Or in other words, $A$ is a dimension zero local ring, or in other
words, a local Artin ring. In particular, it is complete. The result then
follows from Cohen’s structure theorem, [Stacks Project, 0323]. ∎

######  Lemma 5.2.

Suppose $A$ is a chain ring in positive characteristic. Then the colimit
perfection $A_{\mathrm{perf}}=\operatorname{colim}(A\stackrel{{\scriptstyle
Frob}}{{\to}}A\stackrel{{\scriptstyle Frob}}{{\to}}\dots)$ is also a chain
ring.

###### Proof.

Any two elements $x,y\in A_{\mathrm{perf}}$ are in the image of some $A$,
where one divides the other. ∎

######  Remark 5.3.

The passage $A\mapsto A_{\mathrm{perf}}$ loses all information about
nilpotents. Indeed, if $a\in A$ satisfies $a^{n}=0$ then there is some $m$ for
which $a^{p^{m}}=0$. So $A_{\mathrm{perf}}$ is reduced. As we observed above,
this implies it is a valuation ring. In fact, it is the colimit perfection of
the valuation ring $A_{red}$.

######  Proposition 5.4.

Suppose that $A$ is a chain ring in positive characteristic $p$. The inverse
limit perfection

$ A^{\mathrm{perf}}=\varprojlim(\dots\stackrel{{\scriptstyle
Frob}}{{\to}}A\stackrel{{\scriptstyle Frob}}{{\to}}A)$

is a valuation ring. In particular, if the Frobenius is surjective, then $A$
is the quotient of a (perfect) valuation ring.

We isolate the following lemma for readibility.

######  Lemma 5.5.

Suppose that $A$ is a chain ring of characteristic $p$, and $xy=0$. If
$y^{p}\neq 0$, then $y|x$ and $x^{p}=0$.

###### Proof.

If $y|x$, then $x=yz$ so $x^{p}=x^{p-1}yz=x^{p-2}xyz=0$. If $x|y$, by symmetry
$y^{p}=0$, a contradiction. ∎

###### Proof.

Suppose $a=(\dots,a_{2},a_{1}),b\in
A^{\mathrm{perf}}\subseteq\prod_{\mathbb{N}}A$ are nonzero elements. That is,
we have $a_{i},b_{i}\in A$ and

(8) $a_{i+1}^{p}=a_{i},b_{i+1}^{p}=b_{i}$

and

(9) $a_{n},b_{n}\neq 0$

for some $n$. Now if $ab=0$, then we have $0=a_{n+1}b_{n+1}$ so by Lemma 5.5,
since $0\neq a_{n}=a_{n+1}^{p}$ we must have $b_{n}=b_{n+1}^{p}=0$,
contradicting the assumption that $b_{n}\neq 0$. So $A^{\mathrm{perf}}$ is an
integral domain.

Next, we show that $A^{\mathrm{perf}}$ is a chain ring. As above, let $a,b$ be
two nonzero elements, such that $a_{n},b_{n}$ are nonzero. Since $A$ is a
chain ring, we have either $b_{n}|a_{n}$ or $a_{n}|b_{n}$. Suppose
$b_{n}|a_{n}$. We will show that $b|a$ by constructing a $c$ with $a=bc$.

First, note that we have $b_{i}|a_{i}$ for all $i\leq n$ because
$b_{i}=b_{i+1}^{p}$ and $a_{i}=a_{i+1}^{p}$. In the other direction we must
also have $b_{n+1}|a_{n+1}$. Indeed, if $b_{n+1}=ca_{n+1}$ for some non-unit
$c$, then $b_{n}=c^{p}a_{n}=c^{p}b_{n}c_{n}$ for some $c_{n}$ so
$(1-c^{p}c_{n})b_{n}=0$. By assumption $c$ is not a unit, so $c^{p}c_{n}$ is
not a unit so because $A$ is local, $(1-c^{p}c_{n})$ is a unit, so $b_{n}=0$,
a contradiction.

So by induction $b_{i}|a_{i}$ for all $i\in\mathbb{N}$. For each $i$, choose
$c_{i}\in A_{i}$ such that

(10) $a_{i}=b_{i}c_{i}.$

We do not necessarily have $c_{i+1}^{p}=c_{i}$, but we claim that upon
replacing $c_{i}$ with $c_{i}^{\prime}:=c_{i+2}^{p^{2}}$ for all $i\geq n$, we
have both

$a_{i}=b_{i}c_{i}^{\prime}\quad\textrm{ and
}\quad(c_{i+1}^{\prime})^{p}=c_{i}^{\prime}$

for all $i\geq n$, and therefore $a=bc$, or in other words, $b|a$ (of course
for $i<n$ we set $c_{i}^{\prime}=(c_{n}^{\prime})^{p^{i}}$ ).

To begin with, since
$b_{i}c_{i+1}^{p}\stackrel{{\scriptstyle\eqref{equa:aabb}}}{{=}}(b_{i+1}c_{i+1})^{p}\stackrel{{\scriptstyle\eqref{equa:abc}}}{{=}}a_{i+1}^{p}\stackrel{{\scriptstyle\eqref{equa:aabb}}}{{=}}a_{i}\stackrel{{\scriptstyle\eqref{equa:abc}}}{{=}}b_{i}c_{i}$,
we see that $b_{i}(c_{i+1}^{p}-c_{i})=0$ for all $i$. On the other hand, if
$b_{i}\neq 0$, then $b_{i+1}^{p}\neq 0$ and by Lemma 5.5
$(c_{i+2}^{p}-c_{i+1})^{p}=0$. In other words, $b_{i}\neq 0$ implies

(11) $c_{i+2}^{p^{2}}=c_{i+1}^{p}.$

At the beginning we assumed $b_{n}\neq 0$, so we have (11) for all $i\geq n$.

Set $c_{i}^{\prime}:=c_{i+2}^{p^{2}}$ for $i\geq n$. Then for $i\geq n$ we
have

$a_{i}\stackrel{{\scriptstyle\eqref{equa:aabb}}}{{=}}a_{i+2}^{p^{2}}\stackrel{{\scriptstyle\eqref{equa:abc}}}{{=}}b_{i+2}^{p^{2}}c_{i+2}^{p^{2}}\stackrel{{\scriptstyle
def}}{{=}}b_{i+2}^{p^{2}}c_{i}^{\prime}\stackrel{{\scriptstyle\eqref{equa:aabb}}}{{=}}b_{i}c_{i}^{\prime}$

and

$(c_{i+1}^{\prime})^{p}\stackrel{{\scriptstyle
def}}{{=}}(c_{i+3}^{p^{2}})^{p}\stackrel{{\scriptstyle\eqref{equa:cp2}}}{{=}}(c_{i+2}^{p})^{p}\stackrel{{\scriptstyle
def}}{{=}}c_{i}^{\prime}$

as claimed. ∎

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