# Generalized $n$-series and de Rham complexes

S. K. Devalapurkar and M. L. Misterka

###### Abstract.

The goal of this article is to study some basic algebraic and combinatorial
properties of “generalized $n$-series” over a commutative ring $R$, which are
functions $s:\mathbf{Z}_{\geq 0}\to R$ satisfying a mild condition. A special
example of generalized $n$-series is given by the $q$-integers
$\frac{q^{n}-1}{q-1}\in\mathbf{Z}[\\![q-1]\\!]$. Given a generalized
$n$-series $s$, one can define $s$-analogues of factorials (via
$n!_{s}=\prod_{i=1}^{n}s(n)$) and binomial coefficients. We prove that
Pascal’s identity, the binomial identity, Lucas’ theorem, and the Vandermonde
identity admit $s$-analogues; each of these specialize to their appropriate
$q$-analogue in the case of the $q$-integer generalized $n$-series. We also
study the growth rates of generalized $n$-series defined over the integers.
Finally, we define an $s$-analogue of the ($q$-)derivative, and prove
$s$-analogues of the Poincaré lemma and the Cartier isomorphism for the affine
line, as well as a pullback square due to Bhatt-Lurie.

Part of this work was done when the first author was supported by the PD Soros
Fellowship, Inpher, and NSF DGE-2140743

###### Contents

  1. 1 Introduction
    1. 1.1 Summary
    2. 1.2 Table of commonly-used notation
    3. 1.3 Acknowledgements
  2. 2 The $s$-Binomial Coefficients
    1. 2.1 A generalization of binomial coefficients
    2. 2.2 Number-theoretic properties of generalized $n$-series
    3. 2.3 The $s$-binomial theorem
    4. 2.4 The $s$-Lucas theorem
    5. 2.5 The $s$-Vandermonde identity
  3. 3 Generalized $n$-Series Over $\mathbf{Z}$
    1. 3.1 Lexicographically small nonnegative integer generalized $n$-series
    2. 3.2 An upper bound on lexicographically small nonnegative integer GNS
    3. 3.3 A lower bound on strictly increasing integer GNS
    4. 3.4 A more general $s$-Lucas theorem
  4. 4 Generalized $n$-Series and de Rham Complexes
    1. 4.1 Basic properties of the $s$-de Rham complex
    2. 4.2 $s$-analogues of the Poincaré lemma and Cartier isomorphism
    3. 4.3 Formal group law $n$-series and the $s$-derivative
    4. 4.4 A variant of the Weyl algebra
    5. 4.5 An analogue of the Bhatt-Lurie Cartesian square

##  1\. Introduction

###  1.1. Summary

Recent work of Bhatt, Drinfeld, Lurie, Morrow, Scholze, and others (see, e.g.,
[BMS18, BS19, Sch17, Dri21, Dri22, BL22]) has shown that $q$-deformations of
classical number-theoretic and algebro-geometric concepts play a central role
in arithmetic geometry. The basic premise behind the theory of
$q$-deformations is the idea that the $q$-integers
$[n]_{q}=\frac{q^{n}-1}{q-1}$ display many similarities to the ordinary
integers. This idea has a rich history111Any source recounting the history of
$q$-deformations would be most welcome to the authors!: the basic ideas date
back at least to Euler (e.g., [Eul53]) and Jacobi and the study of basic
hypergeometric series. A $q$-analogue of the derivative originates with
Jackson in 1909 (see [Jac09]). We refer the reader to the book [KC02] for an
exposition of $q$-deformed calculus.

The theory of formal group laws supplies a simultaneous generalization of both
ordinary integers and $q$-integers (see 4.3.1 for a quick summary of the
basics of formal group laws, and [Haz78, Rav86] for a detailed treatment).
Namely, every formal group law $F$ over a ring $R$ defines a sequence of power
series $\langle{n}\rangle$ over $R$ for every integer $n\in\mathbf{Z}$. In the
case of the the additive formal group law $x+y$, we have
$\langle{n}\rangle=n$; and in the case of the multiplicative formal group law
$x+y+xy$, one can identify $\langle{n}\rangle=[n]_{q}$. The goal of this
article is to explore whether certain aspects of $q$-deformed mathematics
(such as $q$-analogues of basic combinatorial formulae, and properties of the
$q$-de Rham complex of [Sch17]) admit generalizations to arbitrary formal
group laws. One of the primary motivations behind our investigation is the
unpublished observation of Arpon Raksit that homotopy-theoretic methods
naturally suggest studying “$F$-analogues” of the $q$-de Rham complexes
arising in the aforementioned work of Bhatt-Morrow-Scholze, as well as the
calculation of [Dev23a, Section 3.3].222Since the actual homotopy theory does
not play any role in this paper, we refer the interested reader to Remark
4.3.24 below for more.

Our primary observation is that one does not need the structure of a formal
group law to define and study these “$F$-analogues”. Instead, the following
significantly weaker structure suffices:

######  Definition (Definition 2.1.4).

Fix a ring $R$ (always assumed commutative with unit). A generalized
$n$-series (GNS) over $R$ is a function $s:\mathbf{Z}_{\geq 0}\to R$ such
that:

  1. (1)

$s(0)=0$,

  2. (2)

$s(n)$ is not a zero-divisor for any $n>0$,

  3. (3)

$s(n-k)\mid s(n)-s(k)$ for all $n>k>0$.

For instance, the map $s:\mathbf{Z}_{\geq 0}\to\mathbf{Z}[\\![q-1]\\!]$
sending $n\mapsto[n]_{q}=\frac{q^{n}-1}{q-1}$ defines a GNS.

In the body of this article, we show that this simple definition is sufficient
for proving several analogues of classical combinatorial identities, and is
also enough to study an “$s$-deformation” of the classical algebraic de Rham
complex. The results of this article do not rely on any sophisticated tools:
rather, the purpose is to demonstrate the efficiency of Definition 2.1.4. The
work done in this article seems closely related to Bhargava’s [Bha00], but we
have not attempted to make a comparison.

If $n\geq 0$, let $n!_{s}=\prod_{k=1}^{n}s(k)$, and let
$\binom{n}{j}_{s}=\frac{n!_{s}}{j!_{s}(n-j)!_{s}}$ denote the $s$-analogues of
the factorial and binomial coefficient, respectively. Our main combinatorial
results are the following; for the full statement of some of these results, we
refer the reader to the body of the text.

######  Theorem A.

Fix a GNS $s$ over $R$. The following hold:

  1. (1)

Pascal’s identity (Proposition 2.1.3):

$\binom{n}{k}_{s}=\binom{n-1}{k-1}_{s}+\frac{s(n)-s(k)}{s(n-k)}\binom{n-1}{k}_{s}.$

  2. (2)

An $s$-analogue of the binomial and $q$-binomial theorems; see Theorem 2.3.7.

  3. (3)

Lucas’ theorem (Proposition 2.4.8): suppose that $s(1)=1$ and

$s(a+b)\equiv s(a)+s(b)\pmod{s(a)s(b)}$

for all $a,b\in\mathbf{Z}_{>0}$. Then, for any prime $p$ and any nonnegative
integers $n_{1},n_{0},k_{1},k_{0}$ such that $n_{0},k_{0}<p$, we have

$\binom{n_{1}p+n_{0}}{k_{1}p+k_{0}}_{s}\equiv\binom{n_{1}}{k_{1}}\binom{n_{0}}{k_{0}}_{s}\pmod{s(p)}.$

  4. (4)

An analogue of the Vandermonde and $q$-Vandermonde identities; see Theorem
2.5.4.

In Section 3, we study the growth rate of generalized $n$-series over
$\mathbf{Z}$. For instance, we show in Theorem 3.3.1 that if $s(n)$ is a
strictly increasing generalized $n$-series over $\mathbf{Z}$ which is not a
scalar multiple of $n\mapsto[n]_{q}$ for any $q\in\mathbf{Z}_{>0}$, then
$s(n)=\Omega_{a}(a^{n})$ for all $a\geq 0$.

As one might expect given our motivation above, one important class of
examples of generalized $n$-series arises via formal group laws. Recall (see
4.3.1) that a formal group law over a commutative ring $R$ is a two-variable
power series $x+_{F}y\in R[\\![x,y]\\!]$ such that
$(x+_{F}y)+_{F}z=x+_{F}(y+_{F}z)$ and $x+_{F}y\equiv x+y\pmod{(x,y)^{2}}$. If
$n\geq 0$ is an integer, the $n$-series of $F$ is defined via the formula

$[n]_{F}(t)=\overbrace{t+_{F}t+_{F}\cdots+_{F}t}^{n}\in tR[\\![t]\\!].$

Let $\langle{n}\rangle_{F}=\frac{[n]_{F}(t)}{t}$. Suppose (for simplicity)
that $R$ is torsionfree. Then, the function $\mathbf{Z}_{\geq 0}\to
R[\\![t]\\!]$ sending $n\mapsto\langle{n}\rangle_{F}$ defines a GNS over
$R[\\![t]\\!]$ (Proposition 4.3.4).

One can define the $F$-de Rham complex of the affine line
$\mathbf{A}^{1}=\operatorname{Spec}R[x]$ as the cochain complex

$F\Omega_{\square,\mathbf{A}^{1}}=\left(R[\\![t]\\!][x]\xrightarrow{{\nabla_{F}}}R[\\![t]\\!][x]dx\right),\
x^{n}\mapsto\langle{n}\rangle_{F}x^{n-1}dx.$

This was first defined by Arpon Raksit in unpublished work. Many analytic
properties of the usual ($q$-)derivative continue to hold for the
$F$-derivative: for instance, we show (see Corollary 4.3.15) that there is an
explicit power series $F\mathrm{log}(x)$ which recovers the $q$-logarithm when
$F$ is the $q$-integer GNS, and which satisfies the property that
$\nabla_{F}(F\mathrm{log}(x))=1/x$.

Our main results regarding the $F$-de Rham complex can be summarized as
follows:

######  Theorem B (Theorem 4.3.20 and Theorem 4.5.10).

Let $R$ be a torsionfree (say) commutative ring, and let $F$ be a formal group
law over $R$.

  1. (1)

Let $R[\\![t]\\!]\langle{x}\rangle_{F}$ denote the ring
$R[\\![t]\\!][x,\frac{x^{n}}{[n]_{F}!}]_{n\geq 0}$. Then the Poincaré lemma
holds: the cohomology of the complex
$F\Omega_{\square,\mathbf{A}^{1}}\otimes_{R[\\![t]\\!][x]}R[\\![t]\\!]\langle{x}\rangle_{F}$
is concentrated in degree zero, where it is isomorphic to $R[\\![t]\\!]$.

  2. (2)

The Cartier isomorphism holds: after setting $\langle{p}\rangle_{F}=0$, the
$i$th cohomology of the complex $F\Omega_{\square,\mathbf{A}^{1}}$ is
isomorphic to the $i$th term of a Frobenius twist of
$F\Omega_{\square,\mathbf{A}^{1}}$.

  3. (3)

There is an analogue of the décalage isomorphism of [BO78, BS19] for
$F\Omega_{\square,\mathbf{A}^{1}}$.

  4. (4)

Using the aforementioned $F$-analogue $F\mathrm{log}(x)$ of the $q$-logarithm,
we prove a generalization of the Cartesian square of [BL22, Lemma 3.5.18].

Except for the final part, the above result in fact admits a generalization to
arbitary GNS (not just ones which arise from formal group laws), but the
statement is slightly more complicated; see Section 4.2. As with the
combinatorial results above, Theorem 4.3.20 is not technically involved;
however, it is supposed to serve as a blueprint for a more general program
that we outline at the end of Section 4.3. In particular, we state the (almost
certainly false) 4.3.22, stating that the assignment
$R[x_{1},\ldots,x_{n}]\mapsto(F\Omega_{\square,\mathbf{A}^{1}})^{\otimes_{R[\\![t]\\!]}n}$
should extend to a functor from the category of commutative $R$-algebras to
the $\infty$-category of ${\mathbf{E}_{\infty}}$-$R[\\![t]\\!]$-algebras.

###  1.2. Table of commonly-used notation

This article will introduce some notation which will be used heavily
throughout. For the reader’s convenience, we have summarized the commonly-used
ones in the table below.

Symbol | Definition | Location in text
---|---|---
$R[1/s]$ | $R[s(1)^{-1},s(2)^{-1},\cdots]$ | 1
$c_{s}(n,k)$ | $\frac{s(n)-s(k)}{s(n-k)}$ | 2
$(x+y)^{n}_{s}$ | Characterized by specific conditions | Definition 2.3.5
$C_{s}(n,k)$ | $\frac{s(n+k)-s(n)-s(k)}{s(n)s(k)}$ | 2.4.4
$\genfrac{(}{|}{0.0pt}{}{n}{j}\\!\\!\genfrac{|}{)}{0.0pt}{}{m}{k}_{s}$ | $\sum_{m<i_{1}<i_{2}<\cdots<i_{j}\leq m+n}\left(\prod_{\ell=1}^{j}c_{s}(i_{\ell},i_{\ell}-k+j-\ell)\right)$ | Definition 2.5.2
$\langle n\rangle_{F}(t)$ |  $\frac{[n]_{F}(t)}{t}$ for a FGL $F(x,y)$ | 4.3.1
$\ell_{F}(t)$, $\mathscr{E}_{F}(t)$ | Logarithm and exponential of a FGL | 4.3.1
$F\mathrm{log}(x)$ | $\frac{t}{\ell_{F}(t)}\mathrm{log}(x)$ | Corollary 4.3.15
$\mathbf{G}_{m}^{\sharp,F}$ | “$F$-divided power hull” of zero section of $\mathbf{G}_{m}$ | Definition 4.5.5

###  1.3. Acknowledgements

The first author is grateful to Ben Antieau, Dimitar Jetchev, and (especially)
Arpon Raksit for discussions on this topic, as well as Michael Kural for his
help in proving the main result of Section 4.5 in the case of the
multiplicative formal group law. The second author is grateful to Pavel
Etingof for suggesting that an $s$-analogue of the $q$-binomial theorem might
exist. Finally, both authors are grateful to the directors and organizers of
PRIMES-USA for the opportunity to collaborate, as well as for comments on this
article!

##  2\. The $s$-Binomial Coefficients

###  2.1. A generalization of binomial coefficients

Recall that the binomial coefficient $\binom{n}{k}$ is defined by

$\binom{n}{k}=\frac{n!}{k!(n-k)!}.$

The $q$-factorial and $q$-binomial coefficients are defined by

$[n]!_{q}=[1]_{q}\cdot[2]_{q}\cdot\cdots\cdot[n]_{q},\qquad\binom{n}{k}_{q}=\frac{[n]!_{q}}{[k]!_{q}[n-k]!_{q}},$

where $[n]_{q}=(q^{n}-1)/(q-1)\in\mathbf{Z}[\\![q-1]\\!]$. The similarity of
these two definitions hints that it might be interesting to study a
simultaneous generalization of the usual and $q$-binomial coefficients, where
the sequence of elements $n\in\mathbf{Z}$ and
$[n]_{q}\in\mathbf{Z}[\\![q-1]\\!]$ are replaced by a sequence of elements in
a commutative ring satisfying certain conditions.

######  Definition 2.1.1.

Let $R$ be a ring, and let $s:\mathbf{Z}_{\geq 0}\to R$ be a function such
that $s(0)=0$ and for all $n>0$, $s(n)$ is not a zero-divisor. For integers
$n\geq k\geq 0$, we define the $s$-factorial $n!_{s}$ by

$0!_{s}=1,\quad n!_{s}=\prod_{k=1}^{n}s(k),$

and the $s$-binomial coefficient $\binom{n}{k}_{s}$ by

$\binom{n}{k}_{s}=\frac{n!_{s}}{k!_{s}(n-k)!_{s}}.$

######  Remark 2.1.2.

In general, this quotient is undefined in $R$; however, it is always defined
in the localization

(1) $R[1/s]:=R[s(1)^{-1},s(2)^{-1},s(3)^{-1},\dots].$

We will soon restrict to the case where the $s$-binomial coefficients are
elements of $R$.

The $s$-binomial coefficient need not satisfy any nice properties, since there
are no restrictions placed on $s$. Our first observation is the following.

######  Proposition 2.1.3.

Let $R$ be a ring and let $s:\mathbf{Z}_{\geq 0}\to R$ be a function that
satisfies the following conditions:

  1. (1)

$s(0)=0$,

  2. (2)

$s(n)$ is not a zero-divisor for any $n>0$,

  3. (3)

$s(n-k)\mid s(n)-s(k)$ for all $n>k>0$.

Then, for all integers $n\geq k\geq 0$, the $s$-binomial coefficient
$\binom{n}{k}_{s}$ is an element of $R$, and the $s$-binomial coefficients
satisfy an “$s$-Pascal identity”: For all $n>k>0$,

$\binom{n}{k}_{s}=\binom{n-1}{k-1}_{s}+\frac{s(n)-s(k)}{s(n-k)}\binom{n-1}{k}_{s}.$

###### Proof.

Indeed, observe that in the localization $R[1/s]$, we have:

$\displaystyle\frac{s(n)-s(k)}{s(n-k)}\binom{n-1}{k}_{s}$
$\displaystyle=\frac{s(n)-s(k)}{s(n-k)}\frac{(n-1)!_{s}}{k!_{s}(n-k-1)!_{s}}$
$\displaystyle=\frac{s(n)\cdot(n-1)!_{s}}{k!_{s}(n-k)!_{s}}-\frac{s(k)\cdot(n-1)!_{s}}{k!_{s}(n-k)!_{s}}$
$\displaystyle=\frac{n!_{s}}{k!_{s}(n-k)!_{s}}-\frac{(n-1)!_{s}}{(k-1)!_{s}(n-k)!_{s}}$
$\displaystyle=\binom{n}{k}_{s}-\binom{n-1}{k-1}_{s}.$

It remains to show that $\binom{n}{k}_{s}\in R$ for all $n\geq k\geq 0$. We
will use induction on $n$.

The base case is clear, since $\binom{0}{0}_{s}=1\in R$. For the inductive
step, assume that for some fixed $n$ and for all $k$ with $n-1\geq k\geq 0$,
$\binom{n-1}{k}_{s}\in R$. Let $k$ be an integer such that $n\geq k\geq 0$. If
$k=0$, then $\binom{n}{k}_{s}=1\in R$. Otherwise, we can apply the $s$-Pascal
identity:

$\binom{n}{k}_{s}=\binom{n-1}{k-1}_{s}+\frac{s(n)-s(k)}{s(n-k)}\binom{n-1}{k}_{s}.$

By the inductive hypothesis, the two $s$-binomial coefficients on the right-
hand side are in $R$, and by condition (c) in the theorem statement,
$\frac{s(n)-s(k)}{s(n-k)}\in R$. Therefore, $\binom{n}{k}_{s}\in R$. This
completes the induction proof. ∎

Motivated by Proposition 2.1.3, we are led to the following:

######  Definition 2.1.4.

Let $R$ be a ring. A generalized $n$-series (GNS) over $R$ is a function
$s:\mathbf{Z}_{\geq 0}\to R$ such that the following conditions are true:

  1. (1)

$s(0)=0$,

  2. (2)

$s(n)$ is not a zero-divisor for any $n>0$,

  3. (3)

$s(n-k)\mid s(n)-s(k)$ for all $n>k>0$.

If $s$ is a generalized $n$-series, we will define

(2) $c_{s}(n,k):=\frac{s(n)-s(k)}{s(n-k)}.$

######  Example 2.1.5 (Integers).

The inclusion $s:\mathbf{Z}_{\geq 0}\to\mathbf{Z}$ is clearly a GNS over
$\mathbf{Z}$.

######  Example 2.1.6 ($q$-integers).

Consider the function $s:\mathbf{Z}_{\geq 0}\to\mathbf{Z}[\\![q-1]\\!]$ given
by $s(n)=[n]_{q}$. This defines a GNS: the first two conditions are satisfied,
since $[0]_{q}=0$, $[n]_{q}\neq 0$ for $n>0$, and $\mathbf{Z}[\\![q-1]\\!]$ is
an integral domain. For the third condition, note that

$\displaystyle[n]_{q}-[k]_{q}$
$\displaystyle=\frac{q^{n}-1}{q-1}-\frac{q^{k}-1}{q-1}=\frac{q^{n}-q^{k}}{q-1}$
$\displaystyle=q^{k}\left(\frac{q^{n-k}-1}{q-1}\right)=q^{k}[n-k]_{q}.$

Therefore, $s$ is a GNS over $\mathbf{Z}[\\![q-1]\\!]$, and we can apply
Proposition 2.1.3 to conclude that
$\binom{n}{k}_{q}\in\mathbf{Z}[\\![q-1]\\!]$ for all $n\geq k\geq 0$. The
$s$-Pascal identity reduces to the well-known $q$-Pascal identity:

$\binom{n}{k}_{q}=\binom{n-1}{k-1}_{q}+q^{k}\binom{n-1}{k}_{q}.$

######  Remark 2.1.7.

One can extend the definition of the $s$-binomial coefficients to allow
arbitrary integers $k$ by defining $\binom{n}{k}_{s}=0$ when $k<0$ or $k>n$.
Using this extended definition, the $s$-Pascal identity remains true when
$k=0$:

$\binom{n-1}{-1}_{s}+\frac{s(n)-s(0)}{s(n-0)}\binom{n-1}{0}_{s}=0+1\cdot
1=1=\binom{n}{0}_{s}.$

This relies on the condition $s(0)=0$. The fact that Pascal’s identity fails
for $k=0$ if $s(0)\neq 0$ is one motivation for including condition (1) in the
definition of GNS.

###  2.2. Number-theoretic properties of generalized $n$-series

In this section, we prove some number-theoretic properties of generalized
$n$-series, which will be useful later in this article. The main result of
this section is the following:

######  Theorem 2.2.1.

Let $s$ be a generalized $n$-series over a ring $R$. Then, for all
$a,b,n\in\mathbf{Z}_{\geq 0}$,

  1. (1)

$a\mid b\implies s(a)\mid s(b)$,

  2. (2)

$a\equiv b\pmod{n}\implies s(a)\equiv s(b)\pmod{s(n)}$,

  3. (3)

the ideals $(s(a),s(b))$ and $(s(\gcd(a,b)))$ are equal.

If $s(1)$ is a unit in $R$, then for all $a,n\in\mathbf{Z}_{\geq 0}$,

  1. (4)

$a\text{ is a unit in }\mathbf{Z}/n\implies s(a)\text{ is a unit in }R/s(n)$.

######  Remark 2.2.2.

In the case $R=\mathbf{Z}$, the equivalence of ideals in Theorem 2.2.1 is
equivalent to

$\gcd(s(a),s(b))=\pm s(\gcd(a,b)).$

We will prove Theorem 2.2.1 as a sequence of lemmas. Fix a generalized
$n$-series $s$ over a ring $R$.

######  Lemma 2.2.3.

Let $a,b\in\mathbf{Z}_{\geq 0}$. Then, $a\mid b$ implies $s(a)\mid s(b)$.

###### Proof.

We will use induction. Base case: $s(a)\mid s(0)$ because $s(0)=0$. Inductive
hypothesis: Let $n\in\mathbf{Z}_{>0}$, and assume that $s(a)\mid s(a(n-1))$.
Then, by the divisibility condition in the definition of generalized
$n$-series,

$s(a)\mid s(a(n-1))=s(an-a)\mid s(an)-s(a),$

so $s(a)\mid s(an)$. ∎

######  Lemma 2.2.4.

Let $a,b,n\in\mathbf{Z}_{\geq 0}$. If $a\equiv b\pmod{n}$ then

$s(a)\equiv s(b)\pmod{s(n)}.$

Another way to state this lemma is that $s$ induces a well-defined function
from $\mathbf{Z}/n$ to $R/s(n)$.

###### Proof.

By the definition of congruence, $n\mid a-b$, so $s(n)\mid s(a-b)$ by Lemma
2.2.3. The definition of generalized $n$-series requires that

$s(a-b)\mid s(a)-s(b),$

so $s(n)\mid s(a)-s(b)$, which means that $s(a)\equiv s(b)\pmod{s(n)}$. ∎

######  Lemma 2.2.5.

Suppose that $s(1)$ is a unit in $R$. If $a,n\in\mathbf{Z}_{\geq 0}$ such that
$a$ is a unit in $\mathbf{Z}/n$, then $s(a)$ is a unit in $R/s(n)$.

###### Proof.

Let $b$ be the multiplicative inverse of $a$ modulo $n$. Then, $ab\equiv
1\pmod{n}$. By Lemmas 2.2.3 and 2.2.4,

$s(a)\mid s(ab)\equiv s(1)\pmod{s(n)}.$

So in the ring $R/s(n)$, $s(a)$ divides $s(1)$, which is a unit (because it is
a unit in $R$). Therefore, $s(a)$ is a unit in $R/s(n)$. ∎

######  Lemma 2.2.6.

Let $a,b\in\mathbf{Z}_{\geq 0}$. Then, we have the following equivalence of
ideals:

$\big{(}s(\gcd(a,b))\big{)}=\big{(}s(a),s(b)\big{)}.$

###### Proof.

Let $d=\gcd(a,b)$. By Lemma 2.2.3, $s(d)\mid s(a)$ and $s(d)\mid s(b)$, so
$s(a),s(b)\in\big{(}s(d)\big{)}$. This means that

$\big{(}s(d)\big{)}\supseteq\big{(}s(a),s(b)\big{)}.$

For the other direction, we can use Bézout’s identity to write $d=am+bn$ for
some $m,n\in\mathbf{Z}$. Taking this equation modulo $a$ gives $d\equiv
bn\pmod{a}$. By Lemma 2.2.4, $s(d)\equiv s(bn)\pmod{s(a)}$. Therefore, Lemma
2.2.3 implies that

$s(d)\in s(bn)+\big{(}s(a)\big{)}\subseteq\big{(}s(a),s(b)\big{)},$

and hence $\big{(}s(d)\big{)}\subseteq\big{(}s(a),s(b)\big{)}$. This shows
that the two ideals are equal. ∎

###  2.3. The $s$-binomial theorem

######  Recollection 2.3.1.

The binomial theorem and the $q$-binomial theorem are the following two
identities:

$\displaystyle(x+y)^{n}$
$\displaystyle=\sum_{k=0}^{n}\binom{n}{k}x^{n-k}y^{k},$
$\displaystyle(x+y)^{n}_{q}$
$\displaystyle=\sum_{k=0}^{n}\binom{n}{k}_{q}q^{k(k-1)/2}x^{n-k}y^{k},$

where $(x+y)^{n}_{q}$ is defined by

$(x+y)^{n}_{q}=\prod_{k=0}^{n-1}(x+q^{k}y)=(x+y)(x+qy)\cdots(x+q^{n-1}y).$

Note that the $x+y$ in the parentheses is part of the notation, and cannot be
treated as a sum; see [KC02] for this notation.

######  Remark 2.3.2.

For readers who are familiar with the $q$-Pochhammer symbol,
$(x+y)^{n}_{q}=x^{n}(-y/x;q)_{n}$, and $(a;q)_{n}=(1+(-a))^{n}_{q}$.

We will now state and prove an analogue of the binomial theorem for the
$s$-binomial coefficients. We begin by defining an analogue of the symbol
$(x+y)^{n}_{q}$. To motivate the definition, recall that the $q$-analogue
$(x+y)^{n}_{q}$ is the unique polynomial in $\mathbf{Z}[\\![q-1]\\!][x,y]$
such that the following properties hold:

  * •

The $q$-derivative with respect to $x$ of $(x+y)^{n}_{q}$ is
$[n]_{q}(x+y)^{n-1}_{q}$. This is analogous to the fact that the classical
derivative of $(x+y)^{n}$ with respect to $x$ is $n(x+y)^{n-1}$.

  * •

$(x+y)^{0}_{q}=1$.

  * •

If $y=-x$, then $(x+y)^{n}_{q}$ is $0$ for all $n>0$.

To define an $s$-analogue $(x+y)^{n}_{q}$ in a similar way, we need an
$s$-derivative; we will greatly expand on this notion in Section 4.

######  Definition 2.3.3.

Let $s:\mathbf{Z}_{\geq 0}\to R$ be a GNS. The $s$-derivative is the
$R$-linear map $\nabla_{s}:R[x]\to R[x]$ given on monomials by
$\nabla_{s}(x^{n})=s(n)x^{n-1}$.

######  Remark 2.3.4.

When $n=0$, we have $\nabla_{s}(x^{0})=s(0)x^{-1}$. This is not defined in
$R[x]$ unless $s(0)=0$, which is always true when $s$ is a GNS. Continuing
Remark 2.1.7, this observation is another reason for requiring $s(0)=0$ in the
definition of GNS.

We can now define $(x+y)^{n}_{s}$:

######  Definition 2.3.5.

Let $s$ be a GNS over $R$, so that $R[1/s]=R[s(1)^{-1},s(2)^{-1},\dots]$.
Define $(x+y)^{n}_{s}$ for $n\in\mathbf{Z}_{\geq 0}$ to be the unique
polynomial in $R[1/s][x,y]$ such that the following three conditions hold:

  1. (1)

$(x+y)^{0}_{s}=1$,

  2. (2)

$(x+(-x))^{n}_{s}=0$ for all $n>0$,

  3. (3)

$\nabla_{s,x}(x+y)^{n}_{s}=s(n)(x+y)^{n-1}_{s}$.

Here, $\nabla_{s,x}:R[1/s][x,y]\to R[1/s][x,y]$ is the operator given by the
“$s$-derivative with respect to $x$”: it is simply the $R[1/s][y]$-linear
extension of the $s$-derivative $\nabla_{s}:R[x]\to R[x]$ to $R[1/s][x,y]$.

######  Lemma 2.3.6.

The symbol $(x+y)^{n}_{s}$ in Definition 2.3.5 is well-defined: it exists and
is unique. Moreover, $(x+y)^{n}_{s}$ is a homogeneous polynomial of degree
$n$.

###### Proof.

We will use induction on $n$. For the base case $n=0$, observe that
$(x+y)^{0}_{s}=1$ by condition (1).

For the inductive step, fix $n>0$, and suppose that for all $k<n$,
$(x+y)^{k}_{s}$ is well-defined and homogeneous of degree $k$. We can
$s$-antidifferentiate $s(n)(x+y)^{n-1}_{s}$ using the $R[1/s][y]$-linear
operator $I_{s,x}:R[1/s][x,y]\to R[1/s][x,y]$ defined on monomials by
$I_{s,x}(x^{k})=s(k+1)^{-1}x^{k+1}$. By definition, this operator produces
polynomials with no term of $x$-degree $0$. Although the $s$-antiderivative

$f(x,y)=I_{s,x}(s(n)(x+y)^{n-1}_{s})$

is homogeneous of degree $n$ (since the operator $I_{s,x}$ increases
$x$-degree by $1$) and satisfies condition (3), it might not equal
$(x+y)^{n}_{s}$ because it does not have to satisfy condition (2). Since
$f(x,y)$ is homogeneous of degree $n$, $f(x,-x)$ is a scalar multiple of
$x^{n}$, say $ax^{n}$. Then, the polynomial

$g(x,y)=f(x,y)-a(-y)^{n}$

satisfies

$\nabla_{s,x}g(x,y)=\nabla_{s,x}f(x,y)=s(n)(x+y)^{n-1}_{s}$

and

$g(x,-x)=f(x,-x)-a(-(-x))^{n}=ax^{n}-ax^{n}=0.$

It follows that $(x+y)^{n}_{s}$ exists, and one possible value for it is
$g(x,y)$, which is homogeneous of degree $n$.

It remains to show that $(x+y)^{n}_{s}$ is unique. We know that any polynomial
$h(x,y)$ that satisfies the conditions of $(x+y)^{n}_{s}$ must match $g(x,y)$
in every term with positive $x$-degree, because their $s$-derivatives with
respect to $x$ are both $s(n)(x+y)^{n-1}_{s}$. Therefore, $h(x,y)-g(x,y)$ is a
scalar multiple of $y^{n}$, say $by^{n}$. Setting $y=-x$ gives
$b(-x)^{n}=h(x,-x)-g(x,-x)$, which is $0$ by condition (2), so $b=0$.
Therefore, $h(x,y)=g(x,y)$. This proves that $(x+y)^{n}_{s}$ is unique and is
equal to $g(x,y)$. ∎

Recall from Section 2 that we originally defined the $s$-binomial coefficients
as elements of the ring $R[1/s]=R[s(1)^{-1},s(2)^{-1},\dots]$, and later
proved (using the $s$-Pascal identity) that if $s$ is a GNS, then all the
$s$-binomial coefficients are elements of $R$. We will do something similar
for $(x+y)^{n}_{s}$ below, and we will use the $s$-binomial theorem as a lemma
in the proof that $(x+y)^{n}_{s}\in R[x,y]$. Here is the $s$-binomial theorem:

######  Theorem 2.3.7 ($s$-binomial theorem).

Let $s$ be a GNS over $R$. Then, as elements of $R[1/s][x,y]$, we have:

$(x+y)^{n}_{s}=\sum_{k=0}^{n}\binom{n}{k}_{s}x^{n-k}y^{k}(0+1)^{k}_{s}.$

###### Proof.

We will use induction on $n$, and the inductive step will mainly consist of
applying the $s$-antidifferentiation operator $I_{s,x}$ from the proof of
Lemma 2.3.6 to both sides. For the base case, observe that if $n=0$, both
sides are $1$.

For the inductive step, fix $n>0$, and assume that the $s$-binomial theorem is
true for $n-1$:

$(x+y)^{n-1}_{s}=\sum_{k=0}^{n-1}\binom{n-1}{k}_{s}x^{n-k-1}y^{k}(0+1)^{k}_{s}.$

Multiplying both sides by $s(n)$, applying $I_{s,x}$, and using
$R[1/s][y]$-linearity gives:

$\displaystyle I_{s,x}(s(n)(x+y)^{n-1}_{s})$
$\displaystyle=s(n)\sum_{k=0}^{n-1}\binom{n-1}{k}_{s}I_{s,x}(x^{n-k-1})y^{k}(0+1)^{k}_{s}$
$\displaystyle=\sum_{k=0}^{n-1}\frac{s(n)}{s(n-k)}\binom{n-1}{k}_{s}x^{n-k}y^{k}(0+1)^{k}_{s}.$

Notice that

$\frac{s(n)}{s(n-k)}\binom{n-1}{k}_{s}=\frac{s(n)(n-1)!_{s}}{s(n-k)k!_{s}(n-k-1)!_{s}}=\frac{n!_{s}}{k!_{s}(n-k)!_{s}}=\binom{n}{k}_{s}.$

This implies that

$I_{s,x}(s(n)(x+y)^{n-1}_{s})=\sum_{k=0}^{n-1}\binom{n}{k}_{s}x^{n-k}y^{k}(0+1)^{k}_{s}.$

The right-hand side almost looks like the right-hand side of the $s$-binomial
theorem that we are trying to prove, but the upper limit of the summation is
$n-1$ instead of $n$. To fix this, add $y^{n}(0+1)^{n}_{s}$ to both sides,
giving

$I_{s,x}(s(n)(x+y)^{n-1}_{s})+y^{n}(0+1)^{n}_{s}=\sum_{k=0}^{n}\binom{n}{k}_{s}x^{n-k}y^{k}(0+1)^{k}_{s}.$

We just have to show that the left-hand side is equal to $(x+y)^{n}_{s}$.

Let $g(x,y)$ be the left-hand side. The $s$-derivative with respect to $x$ of
$g(x,y)$ is $s(n)(x+y)^{n-1}_{s}$, because $I_{s,x}$ is an $s$-antiderivative
operator (a right inverse of $\nabla_{s,x}$) and $y^{n}(0+1)^{n}_{s}$ is
constant with respect to $x$. This is equal to the $s$-derivative of
$(x+y)^{n}_{s}$, so $g(x,y)$ matches $(x+y)^{n}_{s}$ in all terms with
positive $x$-degree. Both $g(x,y)$ and $(x+y)^{n}_{s}$ are homogeneous of
degree $n$, so the only terms with $x$-degree $0$ are the $y^{n}$ terms. The
coefficient of $y^{n}$ in $g(x,y)$ is $(0+1)^{n}_{s}$ because $I_{s,x}$ never
produces terms with $x$-degree $0$. The $y^{n}$ term of $(x+y)^{n}_{s}$ is
$(0+y)^{n}_{s}$, which is $y^{n}(0+1)^{n}_{s}$ by homogeneity, so the
coefficient of $y^{n}$ in $(x+y)^{n}_{s}$ is also $(0+1)^{n}_{s}$. Therefore,
$g(x,y)=(x+y)^{n}_{s}$, so

$(x+y)^{n}_{s}=\sum_{k=0}^{n}\binom{n}{k}_{s}x^{n-k}y^{k}(0+1)^{k}_{s}.$

This is what we needed to show for the inductive step. ∎

To complete this subsection, we will show that the coefficients of
$(x+y)^{n}_{s}$ are elements of $R$ for all GNS $s$ over $R$ and all
$n\in\mathbf{Z}_{\geq 0}$. This means that the $s$-binomial theorem is really
an equivalence of polynomials in $R[x,y]$, and can be stated without using the
larger ring $R[1/s][x,y]$.

######  Proposition 2.3.8.

Let $s$ be a GNS over a ring $R$. For all nonnegative integers $n$, we have
$(x+y)^{n}_{s}\in R[x,y]$.

###### Proof.

It suffices to show that $(0+1)^{n}_{s}\in R$ for all nonnegative integers
$n$, because using the $s$-binomial theorem, we could conclude that

$(x+y)^{n}_{s}=\sum_{k=0}^{n}\binom{n}{k}_{s}x^{n-k}y^{k}(0+1)^{k}_{s}\in
R[x,y].$

Setting $x=-1$ and $y=1$ in the $s$-binomial theorem, we get

$0=((-1)+1)^{n}_{s}=\sum_{k=0}^{n}\binom{n}{k}_{s}(-1)^{n-k}(0+1)^{k}_{s},$

so

(3) $(0+1)^{n}_{s}=\sum_{k=0}^{n-1}\binom{n}{k}_{s}(-1)^{n-k-1}(0+1)^{k}_{s}.$

This is a recurrence relation for $(0+1)^{n}_{s}$.

To prove that $(0+1)^{n}_{s}\in R$, we will use induction on $n$. For the base
case, note that $(0+1)^{0}_{s}=1$ which is an element of $R$. For the
inductive step, note that if $(0+1)^{k}_{s}\in R$ for all $k<n$, then by the
recurrence relation 3 and the fact that the $s$-binomial coefficients are in
$R$ (Proposition 2.1.3),

$(0+1)^{n}_{s}=\sum_{k=0}^{n-1}\binom{n}{k}_{s}(-1)^{n-k-1}(0+1)^{k}_{s}\in
R.$

This completes the induction. ∎

######  Remark 2.3.9.

Using the recurrence relation 3, it can be shown that

$\frac{(-1)^{n}(0+1)^{n}_{s}}{n!_{s}}=\sum_{\ell=1}^{n}\left(\sum_{k_{1}+\cdots+k_{\ell}=n}\left(\prod_{j=1}^{\ell}\frac{(-1)}{k_{\ell}!_{s}}\right)\right),$

where the inner sum is over all ordered $\ell$-tuples
$(k_{1},k_{2},\dots,k_{\ell})$ of positive integers that sum to $n$. Combining
the inner and outer sums, the right-hand side can be viewed as a sum over
compositions (ordered partitions) of $n$. Isolating $(0+1)^{n}_{s}$ gives

$(0+1)^{n}_{s}=\sum_{\pi\text{ composition of
}n}(-1)^{n-|\pi|}\binom{n}{\pi_{1},\pi_{2},\dots,\pi_{|\pi|}}_{s}.$

The summand is an $s$-multinomial coefficient, defined by

$\binom{n}{k_{1},k_{2},\dots,k_{n}}_{s}=\frac{n!_{s}}{k_{1}!_{s}\cdot
k_{2}!_{s}\cdot\cdots\cdot k_{n}!_{s}},$

and $|\pi|$ denotes the length of $\pi$.

###  2.4. The $s$-Lucas theorem

######  Recollection 2.4.1.

Lucas’s theorem says that for all primes $p$ and all integers
$n_{1},n_{0},k_{1},k_{0}\in\mathbf{Z}_{\geq 0}$ such that $n_{0},k_{0}<p$,

$\binom{n_{1}p+n_{0}}{k_{1}p+k_{0}}\equiv\binom{n_{1}}{k_{1}}\binom{n_{0}}{k_{0}}\pmod{p}.$

There is a $q$-analogue of this identity, known as the $q$-Lucas theorem:

$\binom{n_{1}p+n_{0}}{k_{1}p+k_{0}}_{q}\equiv\binom{n_{1}}{k_{1}}\binom{n_{0}}{k_{0}}_{q}\pmod{[p]_{q}}.$

This identity is also true if $p$ is composite, as long as we replace the
modulus $[p]_{q}$ with the cyclotomic polynomial $\Phi_{p}(q)$. Notice that
the first binomial coefficient on the right-hand side of the $q$-Lucas theorem
is not a $q$-binomial coefficient.

Here is an $s$-analogue of Lucas’s theorem:

######  Theorem 2.4.2 ($s$-Lucas theorem).

Let $s$ be a GNS over $R$ such that $s(1)=1$ and

(4) $s(a+b)\equiv s(a)+s(b)\pmod{s(a)s(b)}$

for all $a,b\in\mathbf{Z}_{>0}$. Then, for any prime $p$ and any nonnegative
integers $n_{1},n_{0},k_{1},k_{0}$ such that $n_{0},k_{0}<p$, we have

$\binom{n_{1}p+n_{0}}{k_{1}p+k_{0}}_{s}\equiv\binom{n_{1}}{k_{1}}\binom{n_{0}}{k_{0}}_{s}\pmod{s(p)}.$

######  Remark 2.4.3.

Define $\Phi_{n}(s)=\prod_{d\mid n}s(d)^{\mu(n/d)}$, where $\mu$ denotes the
Möbius function. Note that Möbius inversion implies the identity
$s(n)=\prod_{d\mid n}\Phi_{n}(s)$. One can prove a “composite version” of the
$s$-Lucas theorem where $s(p)$ is replaced by $\Phi_{n}(s)$, but the proof is
more complicated.

For the rest of this section, we fix the GNS $s$ over $R$. To simplify the
statement of the $s$-Lucas theorem, we will use the following:

######  Notation 2.4.4.

For integers $a,b\in\mathbf{Z}_{>0}$, we define

$C_{s}(a,b)=\frac{s(a+b)-s(a)-s(b)}{s(a)s(b)}\in R[1/s].$

The extra condition 4 in Theorem 2.4.2 is therefore equivalent to $C_{s}(a,b)$
being defined in $R$ for all $a,b\in\mathbf{Z}_{>0}$.

######  Lemma 2.4.5.

Suppose that $C_{s}(a,b)$ is defined in $R$ for all $a,b\in\mathbf{Z}_{>0}$.
Then, for all integers $m>0$ and $n>k\geq 0$,

$c_{s}(mn,mk)\equiv 1\pmod{s(m)}.$

###### Proof.

By definition,

$c_{s}(mn,mk)-1=\frac{s(mn)-s(mk)-s(mn-mk)}{s(mn-mk)}=s(mk)C_{s}(mk,mn-mk).$

By Lemma 2.2.3, $s(m)\mid s(mk)$, so the right-hand side is divisible by
$s(m)$. Therefore, $c_{s}(mn,mk)-1\equiv 0\pmod{s(m)}$. ∎

######  Remark 2.4.6.

A consequence of this lemma is the interesting fact that if we define the
“rescaled” GNS $s_{m}(n)=s(mn)$ for each positive integer $m$, then Pascal’s
identity for the $s_{m}$-binomial coefficients is the same as the usual
Pascal’s identity when we reduce modulo $s(m)$. This implies the following
congruence:

######  Lemma 2.4.7.

Suppose that $C_{s}(a,b)$ is defined for all $a,b\in\mathbf{Z}_{>0}$. For all
integers $m>0$ and $0\leq k\leq n$,

$\binom{n}{k}_{s_{m}}\equiv\binom{n}{k}\pmod{s(m)}.$

###### Proof.

We will use induction on $n$. For the base case, observe that if $n=0$ then
$k=0$, so both sides are $1$. For the inductive step, assume that the desired
claim is true when $n$ is replaced by $n-1$. Then:

$\displaystyle\binom{n}{k}_{s_{m}}$
$\displaystyle=\binom{n-1}{k-1}_{s_{m}}+c_{s_{m}}(n,k)\binom{n-1}{k}_{s_{m}}$
$\displaystyle\text{by the $s$-Pascal identity (\lx@cref{creftype~refnum}{thm:
s-Pascal identity})},$
$\displaystyle=\binom{n-1}{k-1}_{s_{m}}+c_{s}(mn,mk)\binom{n-1}{k}_{s_{m}}$
$\displaystyle\text{by definition of $c_{s_{m}}$},$
$\displaystyle\equiv\binom{n-1}{k-1}_{s_{m}}+\binom{n-1}{k}_{s_{m}}$
$\displaystyle\text{by Lemma \ref{lem: C defined implies c(mn, mk) is 1 mod
s(m)}},$ $\displaystyle\equiv\binom{n-1}{k-1}+\binom{n-1}{k}\qquad$
$\displaystyle\text{by inductive hypothesis},$
$\displaystyle=\binom{n}{k}\pmod{s(m)}$ $\displaystyle\text{by Pascal's
identity}.$

Therefore, by induction, this is true for all $m\in\mathbf{Z}_{>0}$. ∎

Theorem 2.4.2 is a consequence of a slight variant.

######  Proposition 2.4.8 (Another $s$-Lucas theorem).

Let $s$ be a GNS such that $s(1)=1$. Let $p$ be prime and let
$n_{1},n_{0},k_{1},k_{0}\in\mathbf{Z}_{\geq 0}$ such that $n_{0},k_{0}<m$.
Then,

$\binom{n_{1}p+n_{0}}{k_{1}p+k_{0}}_{s}\equiv\binom{n_{1}}{k_{1}}_{s_{p}}\binom{n_{0}}{k_{0}}_{s}\pmod{s(p)},$

where $s_{p}$ is the rescaled GNS from Remark 2.4.6.

###### Proof of Theorem 2.4.2.

This is an immediate consequence of Proposition 2.4.8 and Lemma 2.4.7. ∎

######  Remark 2.4.9.

Proposition 2.4.8 is just Theorem 2.4.2 but with $\binom{n_{1}}{k_{1}}$
replaced by $\binom{n_{1}}{k_{1}}_{s_{p}}$, and no requirement that
$C_{s}(a,b)$ be an element of $R$.

Before we prove Proposition 2.4.8 in full generality, we will prove the
special case where $n_{0}=k_{0}=0$. Then, we will use this case as a lemma in
the proof of the general result.

###### Proof of Theorem 2.4.8 in the case $n_{0}=k_{0}=0$.

Writing out the definition of the $s$-binomial coefficient
$\binom{pn}{pk}_{s}$, we get

$\binom{pn}{pk}_{s}=\frac{s(pn)s(pn-1)\cdots s(pn-pk+1)}{s(pk)s(pk-1)\cdots
s(1)}.$

By Lemma 2.2.5, $s(pn-j)$ is a unit modulo $s(p)$ for all $j$ not divisible by
$p$. Together with Lemma 2.2.4, this implies that we can cancel $s(pn-j)$ with
$s(pk-j)$ (since they are congruent units modulo $s(p)$). After all of this
cancellation, we are left with

$\binom{pn}{pk}_{s}=\frac{s(pn)s(p(n-1))\cdots
s(p(n-k+1))}{s(pk)s(p(k-1))\cdots s(p)},$

which is just $\binom{n}{k}_{s_{p}}$. This means that

$\binom{pn}{pk}_{s}\equiv\binom{n}{k}_{s_{p}}\pmod{s(p)}.$

Combining this with the congruence of $\binom{n}{k}_{s_{p}}$ and
$\binom{n}{k}$, we get

$\binom{pn}{pk}_{s}\equiv\binom{n}{k}\pmod{s(p)},$

which is the special case of Theorem 2.4.8 where $n_{0}=k_{0}=0$. ∎

We will now extend this to any value of $n_{0}$ in the valid range $0\leq
n_{0}<p$.

###### Proof of Theorem 2.4.8 in the case $k_{0}=0$.

We proved Proposition 2.4.8 above for $n_{0}=0$, so let $0<n_{0}<p$. Notice
that the definition of $s$-binomial coefficients implies that

$s(n-k)\binom{n}{k}_{s}=\frac{n!_{s}}{k!_{s}(n-k-1)!_{s}}=s(n)\binom{n-1}{k}_{s}.$

Therefore,

$s(p(n_{1}-k)+n_{0})\binom{pn_{1}+n_{0}}{pk}_{s}=s(pn_{1}+n_{0})\binom{pn_{1}+n_{0}-1}{pk}_{s}.$

Since $n_{0}$ is a unit modulo $p$, we see that $s(p(n_{1}-k)+n_{0})$ is a
unit modulo $s(p)$. But $s(p(n_{1}-k)+n_{0})$ is congruent to
$s(pn_{1}+n_{0})$ by Lemma 2.2.4. Since both are units, this implies that

$\binom{pn_{1}+n_{0}}{pk}_{s}\equiv\binom{pn_{1}+n_{0}-1}{pk}_{s}\pmod{s(p)}.$

A simple induction proof gives

$\binom{pn_{1}+n_{0}}{pk}_{s}\equiv\binom{pn_{1}}{pk}_{s}\pmod{s(p)}.$

It follows that

$\binom{pn_{1}+n_{0}}{pk}_{s}\equiv\binom{pn_{1}}{pk}_{s}\equiv\binom{n_{1}}{k}=\binom{n_{1}}{k}\binom{n_{0}}{0}_{s}\pmod{s(p)},$

which is exactly the $s$-Lucas theorem where $k_{0}=0$. ∎

Finally, we will extend this by induction to any value of $k_{0}$ between $0$
and $p-1$.

###### Proof of Theorem 2.4.8 in full generality.

We will use induction on $k_{0}$. We already proved the base case $k_{0}=0$
above. Suppose we have shown that the induction hypothesis

$\binom{pn_{1}+n_{0}}{pk_{1}+k_{0}-1}_{s}\equiv\binom{n_{1}}{k_{1}}\binom{n_{0}}{k_{0}-1}_{s}\pmod{s(p)}$

is true for some $k_{0}-1$ between $0$ and $p-2$ (or $0<k_{0}<p$). We will
show that it is true with $k_{0}-1$ replaced by $k_{0}$. Notice that for all
integers $1\leq k\leq n$,

$s(k)\binom{n}{k}_{s}=\frac{n!_{s}}{(k-1)!_{s}(n-k)!_{s}}=s(n-k+1)\binom{n}{k-1}_{s},$

so

$s(pk_{1}+k_{0})\binom{pn_{1}+n_{0}}{pk_{1}+k_{0}}_{s}=s(p(n_{1}-k_{1})+n_{0}-k_{0}+1)\binom{pn_{1}+n_{0}}{pk_{1}+k_{0}-1}_{s}.$

Reducing modulo $s(p)$ gives

$s(k_{0})\binom{pn_{1}+n_{0}}{pk_{1}+k_{0}}_{s}\equiv
s(n_{0}-k_{0}+1)\binom{pn_{1}+n_{0}}{pk_{1}+k_{0}-1}\pmod{s(p)}.$

So

$\displaystyle s(k_{0})\binom{pn_{1}+n_{0}}{pk_{1}+k_{0}}_{s}$
$\displaystyle\equiv s(n_{0}-k_{0}+1)\binom{pn_{1}+n_{0}}{pk_{1}+k_{0}-1}$
$\displaystyle\equiv\binom{n_{1}}{k_{1}}s(n_{0}-k_{0}+1)\binom{n_{0}}{k_{0}-1}_{s}$
$\displaystyle=\binom{n_{1}}{k_{1}}s(k_{0})\binom{n_{0}}{k_{0}}_{s}\pmod{s(p)}.$

Since $0<k_{0}<p$, $s(k_{0})$ is a unit modulo $s(p)$, so we can cancel it
from both sides, and we get the congruence we wanted. Therefore, the induction
proof is complete. ∎

When the ring $R$ is $\mathbf{Z}$, there is a version of the $s$-Lucas theorem
which allows $p$ to be any natural number, not just a prime; see Theorem
3.4.4.

###  2.5. The $s$-Vandermonde identity

######  Recollection 2.5.1.

The classical Vandermonde identity states that

$\binom{m+n}{k}=\sum_{j}\binom{m}{k-j}\binom{n}{j};$

this admits a $q$-analogue, known as the the $q$-Vandermonde identity:

$\binom{m+n}{k}_{q}=\sum_{j}\binom{m}{k-j}_{q}\binom{n}{j}_{q}q^{j(m-k+j)}.$

To state an $s$-analogue of the Vandermonde identity, we need a definition.

######  Definition 2.5.2.

For $m,n,k,j\in\mathbf{Z}$ with $0\leq j\leq n$ and $0\leq k-j\leq m$, define

$\genfrac{(}{|}{0.0pt}{}{n}{j}\\!\\!\genfrac{|}{)}{0.0pt}{}{m}{k}_{s}=\sum_{\begin{subarray}{c}I\subseteq(m,m+n]\cap\mathbf{Z}\\\
|I|=j\end{subarray}}\left(\prod_{\ell=1}^{j}c_{s}(i_{\ell},n-(k-j+\ell))\right),$

where $I=\\{i_{1}<\cdots<i_{j}\\}$. We also set
$\genfrac{(}{|}{0.0pt}{}{n}{j}\\!\\!\genfrac{|}{)}{0.0pt}{}{m}{k}_{s}=0$ for
$j<0$ or $j>n$.

######  Remark 2.5.3.

For $s(n)=n$, the product inside the sum in Definition 2.5.2 is $1$, so the
sum is just the number of subsets of $(m,m+n]\cap\mathbf{Z}$ of size $j$,
which is $\binom{n}{j}$. In that sense,
$\genfrac{(}{|}{0.0pt}{}{n}{j}\\!\\!\genfrac{|}{)}{0.0pt}{}{m}{k}_{s}$ is an
$s$-analogue of the subset-counting definition of $\binom{n}{j}$ which depends
on two extra parameters $m$ and $k$. The other type of $s$-binomial
coefficient $\binom{n}{j}_{s}$ is an $s$-analogue of the algebraic definition
of $\binom{n}{j}$.

######  Theorem 2.5.4 ($s$-Vandermonde identity).

For all $m,n,k\in\mathbf{Z}_{\geq 0}$ with $k\leq m+n$,

$\binom{m+n}{k}_{s}=\sum_{j}\binom{m}{k-j}_{s}\genfrac{(}{|}{0.0pt}{}{n}{j}\\!\\!\genfrac{|}{)}{0.0pt}{}{m}{k}_{s},$

where the sum is taken over all $j\in\mathbf{Z}$ such that $0\leq j\leq n$ and
$0\leq k-j\leq m$.

######  Question 2.5.5.

As explained in [Sas18], the $q$-Vandermonde identity for $\binom{m+n}{k}_{q}$
can be understood as arising via a motivic cellular decomposition of the
Grassmannian $\mathrm{Gr}_{k}(\mathbf{C}^{m+n})$. Is there a motivic
interpretation of Theorem 2.5.4? (A similar question can also be asked for the
$s$-analogues of the other combinatorial identities proved elsewhere in this
article.)

The proof of Theorem 2.5.4 requires a preliminary lemma, which can be viewed
as an analogue of Pascal’s identity for
$\genfrac{(}{|}{0.0pt}{}{n}{j}\\!\\!\genfrac{|}{)}{0.0pt}{}{m}{k}_{s}$.

######  Lemma 2.5.6.

For all $m,n,k,j\in\mathbf{Z}$ such that $0\leq k-j\leq m$,

$\genfrac{(}{|}{0.0pt}{}{n}{j}\\!\\!\genfrac{|}{)}{0.0pt}{}{m}{k}_{s}=\genfrac{(}{|}{0.0pt}{}{n-1}{j}\\!\\!\genfrac{|}{)}{0.0pt}{}{m}{k}_{s}+c_{s}(m+n,m+n-k)\genfrac{(}{|}{0.0pt}{}{n-1}{j-1}\\!\\!\genfrac{|}{)}{0.0pt}{}{m}{k-1}_{s}.$

Notice that this includes the cases $j<0$ and $j>n$.

###### Proof.

We will split up the sum in the definition of
$\genfrac{(}{|}{0.0pt}{}{n}{j}\\!\\!\genfrac{|}{)}{0.0pt}{}{m}{k}_{s}$ into
two sums, based on whether $I$ contains $m+n$ or not. If $m+n\notin I$, then
$I$ ranges over all $j$-element subsets of $(m,m+n-1]\cap\mathbf{Z}$. If
$m+n\in I$, then we remove it, and we get a $(j-1)$-element subset of
$(m,m+n-1]\cap\mathbf{Z}$. Then, we have to pull out the $\ell=j$ term of the
product, since $i_{j}=m+n$. So we have

$\displaystyle\genfrac{(}{|}{0.0pt}{}{n}{j}\\!\\!\genfrac{|}{)}{0.0pt}{}{m}{k}_{s}$
$\displaystyle=\sum_{\begin{subarray}{c}I\subseteq(m,m+n]\cap\mathbf{Z}\\\
|I|=j\end{subarray}}\left(\prod_{\ell=1}^{j}c_{s}(i_{\ell},i_{\ell}-(k-j+\ell))\right)$
$\displaystyle=\sum_{\begin{subarray}{c}I\subseteq(m,m+n]\cap\mathbf{Z}\\\
|I|=j\\\ m+n\in
I\end{subarray}}\left(\prod_{\ell=1}^{j}c_{s}(i_{\ell},i_{\ell}-(k-j+\ell))\right)+\sum_{\begin{subarray}{c}I\subseteq(m,m+n-1]\cap\mathbf{Z}\\\
|I|=j\end{subarray}}\left(\prod_{\ell=1}^{j}c_{s}(i_{\ell},i_{\ell}-(k-j+\ell))\right)$
$\displaystyle=\sum_{\begin{subarray}{c}I\subseteq(m,m+n-1]\cap\mathbf{Z}\\\
|I|=j-1\end{subarray}}c_{s}(m+n,m+n-(k-j+j))\left(\prod_{\ell=1}^{j-1}c_{s}(i_{\ell},i_{\ell}-(k-j+\ell))\right)+\genfrac{(}{|}{0.0pt}{}{n-1}{j}\\!\\!\genfrac{|}{)}{0.0pt}{}{m}{k}_{s}$
$\displaystyle=c_{s}(m+n,m+n-k)\genfrac{(}{|}{0.0pt}{}{n-1}{j-1}\\!\\!\genfrac{|}{)}{0.0pt}{}{m}{k-1}_{s}+\genfrac{(}{|}{0.0pt}{}{n-1}{j}\\!\\!\genfrac{|}{)}{0.0pt}{}{m}{k}_{s}.$

Rearranging this completes the proof. ∎

###### Proof of Theorem 2.5.4.

We will prove this by induction on $n$. For the base case, observe that if
$n=0$, then all $j$ in the range of summation satisfy $0\leq j\leq 0$, so
either the sum is empty (if $k-0>m$) or its only term is $j=0$ (if $k-0\leq
m$). The conditions of the theorem force $k\leq m+n=m$, so the sum has exactly
one term:

$\sum_{j}\binom{m}{k-j}_{s}\genfrac{(}{|}{0.0pt}{}{n}{j}\\!\\!\genfrac{|}{)}{0.0pt}{}{m}{k}_{s}=\binom{m}{k}_{s}\genfrac{(}{|}{0.0pt}{}{0}{0}\\!\\!\genfrac{|}{)}{0.0pt}{}{m}{k}_{s}.$

We want to show that this is $\binom{m}{k}_{s}$. The definition of
$\genfrac{(}{|}{0.0pt}{}{0}{0}\\!\\!\genfrac{|}{)}{0.0pt}{}{m}{k}_{s}$ gives
the rather degenerate identity

$\genfrac{(}{|}{0.0pt}{}{0}{0}\\!\\!\genfrac{|}{)}{0.0pt}{}{m}{k}_{s}=\sum_{\begin{subarray}{c}I\subseteq(m,m]\cap\mathbf{Z}\\\
|I|=0\end{subarray}}\left(\prod_{\ell=1}^{0}c_{s}(i_{\ell},i_{\ell}-(k+\ell))\right)=\sum_{I\subseteq\emptyset}1=1,$

which completes the base case.

For the inductive step, suppose that the theorem is true with $n$ replaced by
$n-1$. We will first apply a version of the $s$-Pascal identity to
$\binom{m+n}{k}_{s}$ that has been “flipped” using the identity
$\binom{n}{k}_{s}=\binom{n}{n-k}_{s}$:

$\displaystyle\binom{m+n}{k}_{s}$ $\displaystyle=\binom{m+n}{m+n-k}_{s}$
$\displaystyle=\binom{m+n-1}{m+n-k-1}_{s}+c_{s}(m+n,m+n-k)\binom{m+n-1}{m+n-k}_{s}$
$\displaystyle=\binom{m+n-1}{k}_{s}+c_{s}(m+n,m+n-k)\binom{m+n-1}{k-1}_{s}.$

The inductive hypothesis gives:

$\displaystyle\binom{m+n-1}{k}_{s}+c_{s}(m+n,m+n-k)\binom{m+n-1}{k-1}_{s}$
$\displaystyle=\sum_{j}\binom{m}{k-j}_{s}\genfrac{(}{|}{0.0pt}{}{n-1}{j}\\!\\!\genfrac{|}{)}{0.0pt}{}{m}{k}_{s}+c_{s}(m+n,m+n-k)\sum_{j}\binom{m}{k-1-j}_{s}\genfrac{(}{|}{0.0pt}{}{n-1}{j}\\!\\!\genfrac{|}{)}{0.0pt}{}{m}{k-1}_{s}$
$\displaystyle=\sum_{j}\binom{m}{k-j}_{s}\genfrac{(}{|}{0.0pt}{}{n-1}{j}\\!\\!\genfrac{|}{)}{0.0pt}{}{m}{k}_{s}+c_{s}(m+n,m+n-k)\sum_{j}\binom{m}{k-j}_{s}\genfrac{(}{|}{0.0pt}{}{n-1}{j-1}\\!\\!\genfrac{|}{)}{0.0pt}{}{m}{k-1}_{s}$
$\displaystyle=\sum_{j}\binom{m}{k-j}_{s}\left(\genfrac{(}{|}{0.0pt}{}{n-1}{j}\\!\\!\genfrac{|}{)}{0.0pt}{}{m}{k}_{s}+c_{s}(m+n,m+n-k)\genfrac{(}{|}{0.0pt}{}{n-1}{j-1}\\!\\!\genfrac{|}{)}{0.0pt}{}{m}{k-1}_{s}\right).$

By Lemma 2.5.6, this is equal to

$\sum_{j}\binom{m}{k-j}_{s}\genfrac{(}{|}{0.0pt}{}{n}{j}\\!\\!\genfrac{|}{)}{0.0pt}{}{m}{k}_{s},$

which completes the induction. ∎

##  3\. Generalized $n$-Series Over $\mathbf{Z}$

###  3.1. Lexicographically small nonnegative integer generalized $n$-series

When doing computations with generalized $n$-series, it is useful to have some
examples over $\mathbf{Z}$ that are lexicographically small (close to $0$ for
small values of $n$). We will restrict to integer GNS that are always
nonnegative. We have already seen some examples of relatively small integer
generalized $n$-series: $s(n)=n$ and $s(n)=[n]_{q}$. Before looking at an
algorithm for constructing lexicographically small GNS, we need a definition:

######  Definition 3.1.1.

Let $R$ be a ring, and let $N\in\mathbf{Z}_{\geq 0}$. A partial generalized
$n$-series is a function $s:\\{0,1,\dots,N\\}\to R$ such that $s(0)=0$, $s(n)$
is not a zero-divisor for $0<n\leq N$, and for all $0\leq k\leq n\leq N$,
$s(n-k)\mid s(n)-s(k)$. An extension of a partial GNS $s$ is a GNS which
agrees with $s$ on the domain of $s$.

Given a partial GNS

$s:\\{0,1,\dots,N\\}\to\mathbf{Z}_{\geq 0}$

we can use a greedy algorithm to construct the lexicographically smallest
nonnegative GNS $\widetilde{s}$ which is an extension of $s$.

######  Definition 3.1.2.

Suppose we are given a partial GNS $s:\\{0,1,\dots,N\\}\to\mathbf{Z}_{\geq
0}$. Define a function $\widetilde{s}:\mathbf{Z}_{\geq 0}\to\mathbf{Z}_{\geq
0}$ in the following way: $\widetilde{s}(n)=s(n)$ for $0\leq n\leq N$, and for
each $n>N$ we define $\widetilde{s}(n)$ in terms of the previous values of
$\widetilde{s}(k)$ to be the smallest positive integer that makes the
restriction $\widetilde{s}|_{\\{0,1,\dots,n\\}}$ a partial GNS.

The fact that $\widetilde{s}$ is well-defined is nontrivial. We have to show
that at each step of the algorithm, $\widetilde{s}(n)$ exists. We will
actually prove a stronger theorem:

######  Theorem 3.1.3.

Consider the following recursive construction of an arbitrary nonnegative GNS
over $\mathbf{Z}$: Start with $s(0)=0$, and for each $N$ starting with $1$ in
increasing order, choose $s(N)$ to be an _arbitrary_ integer such that
$s|_{\\{0,1,\dots,N\\}}$ is a partial GNS. No matter what choices are made, it
is always possible to continue (there are never any contradictions).

###### Proof.

Suppose we have a partial GNS $s$ with domain $\\{0,1,\dots,N-1\\}$, and we
want to choose $s(N)$. We have to show that there exists $s(N)$ such that for
all $0<k<N$, $s(N-k)\mid s(N)-s(k)$. This is equivalent to $s(N)\equiv
s(k)\pmod{s(N-k)}$, so we really have a system of linear congruences. This
system has a solution by the generalized Chinese Remainder Theorem as long as
no two congruences contradict each other. We want to show that if we reduce
two of the congruences $s(N)\equiv s(N-k)\pmod{s(k)}$ and $s(N)\equiv
s(N-j)\pmod{s(j)}$ modulo $\gcd(s(k),s(j))$, they become the same congruence.
That is, we want to show that

$s(N-k)\equiv s(N-j)\pmod{\gcd(s(k),s(j))}.$

By Lemma 2.2.6, $\gcd(s(k),s(j))=\pm s(\gcd(k,j))$. And since

$N-k\equiv N\equiv N-j\pmod{\gcd(k,j)},$

the desired congruence must be true by Lemma 2.2.4. ∎

We have seen that many number-theoretic properties of the positive integers
remain true for arbitrary GNS. However, there are some important differences
between $s(n)=n$ and other GNS. For example, if $n=p^{j}m$ with $p\nmid m$,
then $n/p^{j}$ is a unit modulo $p$. This is generally false for other
generalized $n$-series; using Theorem 3.1.3, we can construct a
counterexample.

######  Example 3.1.4.

Let $s$ be an extension of the partial GNS with domain $\\{0,1,\dots,6\\}$
whose values are $0,1,2,3,10,11,12$. One can check manually that this forms a
partial GNS. For this GNS,

$\frac{s(6)}{s(2)}=\frac{12}{2}=6\equiv 0\pmod{s(2)}.$

###  3.2. An upper bound on lexicographically small nonnegative integer GNS

In this section, we will write partial generalized $n$-series as lists of
numbers, where the first item in the list is $s(0)$. For example:

######  Example 3.2.1.

Define a partial generalized $n$-series $0,1,3$ via the function
$s:\\{0,1,2\\}\to\mathbf{Z}_{\geq 0}$ given by $0\mapsto 0$, $1\mapsto 1$,
$2\mapsto 3$. If we apply the algorithm of Definition 3.1.2 to the partial
generalized $n$-series $0,1,k$ where $k\in\mathbf{Z}_{>0}$, we get a
generalized $n$-series $0,1,k,1,k,\dots$, where the $n$-th term is $0$ if
$n=0$, $1$ if $n$ is odd, and $k$ if $n$ is even and nonzero. If we start with
a longer partial generalized $n$-series $s$, say $0,1,3,4$, then the
generalized $n$-series $\widetilde{s}$ starts with

$0,1,3,4,9,19,552,22081,$

and the next few terms are

$219440979,2669857856653708,6558922971496604200448626056129.$

This seems to grow very fast when $n$ increases, almost doubling the number of
digits each term. As we will see, $\widetilde{s}(n)$ is bounded by

$\displaystyle\widetilde{s}(n)$ $\displaystyle<12^{2^{n-4}},\qquad n\geq
4,\text{ and}$ $\displaystyle\widetilde{s}(n)$
$\displaystyle=\Omega_{a}(a^{n}),\qquad\forall a\geq 0.$

The upper bound (a special case of Theorem 3.2.5) is proved below, and the
lower bound (a special case of Theorem 3.3.1) is proved in the next
subsection.

Let us first show that $\widetilde{s}$ is strictly increasing.

######  Lemma 3.2.2.

Let $N\geq 3$, and let $s:\\{0,1,\dots,N\\}\to\mathbf{Z}_{\geq 0}$ be a
strictly increasing partial generalized $n$-series. Then, $\widetilde{s}$ is
also strictly increasing.

######  Remark 3.2.3.

The condition $N\geq 3$ is important, because we already saw that if $s=0,1,3$
(so $N=2$) then $\widetilde{s}=0,1,3,1,3,\dots$, which is not strictly
increasing.

###### Proof.

Notice that $s(2)>s(1)\geq 1$. Let $n>N$. We will show that
$\widetilde{s}(n)>\widetilde{s}(n-1)$. By Lemma 2.2.4, $\widetilde{s}(n)\equiv
s(1)\pmod{s(n-1)}$, so either $\widetilde{s}(n)=s(1)$ or
$\widetilde{s}(n)>s(n-1)$. The first case cannot happen because
$\widetilde{s}(n)\equiv s(2)\pmod{\widetilde{s}(n-2)}$, and
$\widetilde{s}(n-2)\geq s(2)>s(1)$. Therefore,
$\widetilde{s}(n)>\widetilde{s}(n-1)$. ∎

Given the value of $\widetilde{s}(k)$ for all $1\leq k<n$, there is an upper
bound on $\widetilde{s}(n)$, as long as $\widetilde{s}$ is strictly
increasing:

######  Lemma 3.2.4.

Let $s:\\{0,1,\dots,N\\}\to\mathbf{Z}$ be a strictly increasing partial
generalized $n$-series with $N\geq 3$. Then, for all $n>N$,

$\widetilde{s}(n)<\mathrm{lcm}\\{\widetilde{s}(k)\mid 0<k<n\\}.$

###### Proof.

Recall that the proof of the existence of $\widetilde{s}$ (Theorem 3.1.3)
constructs $\widetilde{s}(n)$ from $\widetilde{s}(k)$ for all $0<k<n$ using
the generalized Chinese Remainder Theorem. The congruences are
$x\equiv\widetilde{s}(n-k)\pmod{\widetilde{s}(k)}$ for each $0<k<n$, so the
generalized CRT proves the existence of a unique solution $x$ modulo
$L=\mathrm{lcm}\\{\widetilde{s}(k)\mid 0<k<n\\}$. Therefore, there exists a
solution to the system of congruences with $0<x\leq L$. The smallest positive
solution is $\widetilde{s}(n)$ by Definition 3.1.2, so $\widetilde{s}(n)\leq
L$. We just have to show that $\widetilde{s}(n)\neq L$.

Suppose for contradiction that $\widetilde{s}(n)=L$. Then, by Lemma 2.2.4,
$L\equiv\widetilde{s}(1)\pmod{\widetilde{s}(n-1)}$. But by the definition of
$L$, $L$ is divisible by $\widetilde{s}(n-1)$, so $\widetilde{s}(1)$ is also
divisible by $\widetilde{s}(n-1)$. Since $n-1>N-1>1$,
$\widetilde{s}(1)<\widetilde{s}(n-1)$ by Lemma 3.2.2, which is a
contradiction. ∎

Using the fact that the LCM of a set is at most its product, we can prove the
following upper bound:

######  Theorem 3.2.5.

Let $s:\\{0,1,\dots,N\\}\to\mathbf{Z}$ be a strictly increasing partial
generalized $n$-series with $N\geq 3$. Then, for all $n>N$,

$\widetilde{s}(n)<\Pi^{2^{n-(N+1)}},$

where $\Pi=\prod_{k=1}^{N}s(k)$.

###### Proof.

We will use strong induction. Suppose that $n>N$ and for all $k$ strictly
between $N$ and $n$,

$\widetilde{s}(k)<\Pi^{2^{k-(N+1)}}.$

By Lemma 3.2.4 and the inductive hypothesis,

$\displaystyle\widetilde{s}(n)$
$\displaystyle<\mathrm{lcm}\\{\widetilde{s}(k)\mid
0<k<n\\}\leq\prod_{k=1}^{n-1}\widetilde{s}(k)$
$\displaystyle\leq\left(\prod_{k=1}^{N}s(k)\right)\left(\prod_{k=N+1}^{n-1}\Pi^{2^{k-(N+1)}}\right)$
$\displaystyle=\Pi\cdot\prod_{k=0}^{n-(N+2)}\Pi^{2^{k}}=\Pi\cdot\Pi^{\left(\sum_{k=0}^{n-(N+2)}2^{k}\right)}$
$\displaystyle=\Pi\cdot\Pi^{2^{n-(N+1)}-1}=\Pi^{2^{n-(N+1)}}.$

This completes the induction. ∎

######  Remark 3.2.6.

Another way to think about this theorem is that the recursively defined
sequence

$a_{n}=\begin{cases}s(n)&\text{if }0\leq n\leq N,\\\
\prod_{k=1}^{n-1}a_{k}&\text{if }n>N\end{cases}$

is exactly equal to $\Pi^{2^{n-(N+1)}}$ for all $n>N$, and we know that
$\widetilde{s}(n)$ satisfies this recurrence but with the equality replaced by
$<$ whenever $n>N$, so it should be true that $\widetilde{s}(n)<a_{n}$ for all
$n>N$.

###  3.3. A lower bound on strictly increasing integer GNS

We proved an upper bound on the lexicographically smallest extension of the
partial GNS $0,1,3,4$ in the previous section, but we also stated a lower
bound. The purpose of this subsection is to prove this lower bound by proving
a more general result.

######  Theorem 3.3.1.

Let $s(n)$ be a strictly increasing generalized $n$-series over $\mathbf{Z}$
that is not a scalar multiple of $n\mapsto[n]_{q}$ for any
$q\in\mathbf{Z}_{>0}$. Then, $s(n)=\Omega_{a}(a^{n})$ for all $a\geq 0$.

In this theorem statement, we use the convention that $[n]_{1}=n$. To prove
Theorem 3.3.1, we will use the following lemma:

######  Lemma 3.3.2.

Let $s$ satisfy the conditions of Theorem 3.3.1, and fix a nonnegative integer
$a$. Then, for all sufficiently large $n$,

$s(n+1)\neq as(n)+s(1).$

###### Proof.

Let $k\in\mathbf{Z}_{\geq 0}$ such that $s(k+1)\neq as(k)+s(1)$. Such a $k$
must exist, because otherwise $s(n+1)=as(n)+s(1)$ for all
$n\in\mathbf{Z}_{\geq 0}$. This would imply by induction that
$s(n)=s(1)[n]_{q}$ with $q=a$ for all $n$, which contradicts our assumption
about $s$. Next, choose an integer $N$ large enough so that

$s(N-k)>\max\\{s(k+1),as(k)+s(1)\\}.$

This is always possible, because $s:\mathbf{Z}_{\geq 0}\to\mathbf{Z}$ is
strictly increasing and therefore unbounded.

To prove the lemma, we will show that for all $n\geq N$, we have $s(n+1)\neq
as(n)+s(1)$. Suppose for contradiction that there exists $n\geq N$ with

$s(n+1)=as(n)+s(1).$

Taking this equation modulo $s(n-k)$ and using Lemma 2.2.4 gives

$s(k+1)\equiv as(k)+s(1)\pmod{s(n-k)}.$

But $n\geq N$, so

$s(n-k)\geq s(N-k)>\max\\{s(k+1),as(k)+s(1)\\},$

so the modulus is greater than both sides of the congruence, which means that
it is an equality. This contradicts the fact that $s(k+1)\neq as(k)+s(1)$. ∎

###### Proof of Theorem 3.3.1.

We want to show that $s(n)=\Omega_{a}(a^{n})$ for all $a\geq 0$. It suffices
to prove this for $a\in\mathbf{Z}_{\geq 0}$. For each integer $b$ with
$0<b<a$, apply Lemma 3.3.2 with $a$ replaced by $b$. This gives an integer
$N_{b}$ such that for all $n\geq N_{b}$,

$s(n+1)\neq bs(n)+s(1).$

Let

$N=\max\\{N_{b}\mid 0<b<a\\}.$

We claim that for all $n\geq N$, we have $s(n+1)\geq as(n)+s(1)$.

To see this, let $n\geq N$. By Lemma 2.2.4,

$s(n+1)\equiv s(1)\pmod{s(n)},$

so $s(n+1)=bs(n)+s(1)$ for some $b\in\mathbf{Z}$. Since $s$ is strictly
increasing, $b$ must be positive. We want to show that $b\geq a$. If $b<a$,
then we defined $N_{b}$ above, and $n\geq N\geq N_{b}$. So $s(n+1)\neq
bs(n)+s(1)$, which is a contradiction. Therefore, $b\geq a$, so

$s(n+1)\geq as(n)+s(1).$

To complete the proof, notice that the claim implies (by induction) that for
all $n\geq N$, $s(n)\geq a^{n-N}s(N)$, which means that
$s(n)=\Omega_{a}(a^{n})$ as $n\to\infty$. ∎

###  3.4. A more general $s$-Lucas theorem

The $s$-Lucas theorem (Proposition 2.4.8) is a congruence modulo $s(p)$, where
$s$ is a GNS and $p$ is prime. The $q$-Lucas theorem has a more general form
which allows $p$ to be composite:

$\binom{n_{1}p+n_{0}}{k_{1}p+k_{0}}_{q}\equiv\binom{n_{1}}{k_{1}}\binom{n_{0}}{k_{0}}_{q}\pmod{\Phi_{p}(q)}.$

In this subsection, we will state and prove an $s$-analogue of the more
general $q$-Lucas theorem in the case $n_{0}=k_{0}=0$.

Recall the $s$-analogue $\Phi_{n}(s)$ of the cyclotomic polynomial from Remark
2.4.3.

######  Theorem 3.4.1.

If $s$ is a GNS over $\mathbf{Z}$, then $\Phi_{n}(s)\in\mathbf{Z}$ for all
$n>0$.

We will prove that $\Phi_{n}(s)\in\mathbf{Z}$ by showing that

$\Phi_{n}(s)=\frac{s(n)}{\mathrm{lcm}\\{s(n/p)\mid p\text{ prime factor of
}n\\}}.$

Here is a lemma:

######  Lemma 3.4.2.

Let $S$ be an arbitrary multiset of positive integers. Then,

$\mathrm{lcm}(S)=\prod_{\begin{subarray}{c}\text{\emph{multiset }}A\subseteq
S\\\ A\neq\emptyset\end{subarray}}\gcd(A)^{(-1)^{|A|-1}}.$

This lemma can be proved using the tools of elementary number theory. Notice
that when $s$ is a two-element multiset $\\{a,b\\}$, this formula reduces to
$\mathrm{lcm}(a,b)=ab/\gcd(a,b)$.

###### Proof of Lemma 3.4.2.

We can decompose the right-hand side into its prime-power factors. Let $p$ be
a prime, and let $T$ be the multiset $\\{v_{p}(a)\mid a\in S\\}$, where
$v_{p}(a)$ is the exponent of $p$ in the prime factorization of $a$. The
exponent of $p$ in the right-hand side of the equality we are trying to prove
is

$\sum_{\begin{subarray}{c}\text{multiset }B\subseteq T\\\
B\neq\emptyset\end{subarray}}(-1)^{|B|-1}\mathrm{min}(B),$

because a GCD of powers of $p$ is $p$ to the power of the minimum exponent. We
want to show that this is equal to $v_{p}(\mathrm{lcm}(S))$, which is
$\max(T)$. Write $T=\\{a_{1},a_{2},\dots,a_{k}\\}$, where $k=|T|$ and
$a_{1}\leq a_{2}\leq\cdots\leq a_{n}$. We will count the number of times each
$a_{j}$ is counted in the sum. Since $\max(T)=a_{n}$, we have to show that
$a_{n}$ is counted once and $a_{j}$ is counted $0$ times for all $1\leq j<n$.

Let $1\leq j\leq n$ and let $1\leq k\leq n$. We want to count how many subsets
$B\subseteq T$ have minimum $j$ and cardinality $n$. Such a subset must
include $a_{j}$, but can have any combination of $k-1$ elements $a_{\ell}$
with $\ell>j$. There are $\binom{n-j}{k-1}$ choices for these remaining
elements. Therefore, the number of times $a_{j}$ is counted in the sum is

$\sum_{k=1}^{n}(-1)^{k-1}\binom{n-j}{k-1}=\sum_{k=0}^{n-1}(-1)^{k}\binom{n-j}{k}.$

Since $\binom{n-j}{k}$ is zero for $k>n-j$, this is just the alternating sum
of row $n-j$ of Pascal’s triangle, which is $0$ if $n-j>0$ and $1$ if $n-j=0$.
Therefore, $a_{j}$ is counted $0$ times for $j<n$ and $1$ time for $j=n$ in
the sum

$\sum_{\begin{subarray}{c}B\subseteq P\\\
B\neq\emptyset\end{subarray}}(-1)^{|B|-1}\mathrm{min}(B),$

so the sum is equal to $a_{n}$. This means that the exponents of $p$ in the
left-hand and right-hand sides of the equality in the theorem statement are
both $a_{n}$. Combining this fact for each prime $p$ proves the lemma. ∎

###### Proof of Theorem 3.4.1.

Let $P$ be the set of prime factors of $n$, and let

$r(n)=\frac{s(n)}{\mathrm{lcm}\\{s(n/p)\mid p\in P\\}}.$

Notice that $r(n)\in\mathbf{Z}$ because Lemma 2.2.3 implies that $s(n/p)\mid
s(n)$ for all prime factors $p$ of $n$. We want to show that
$\Phi_{n}(s)=r(n)$.

By the definition of $r(n)$,

$\frac{s(n)}{r(n)}=\mathrm{lcm}\left\\{s\left(\frac{n}{p}\right)\;\middle|\;p\in
P\right\\}.$

By Lemma 3.4.2 with $S=\\{s(n/p)\mid p\in P\\}$, this is equal to

$\prod_{\begin{subarray}{c}A\subseteq P\\\
A\neq\emptyset\end{subarray}}\gcd\left\\{s\left(\frac{n}{p}\right)\;\middle|\;p\in
A\right\\}^{(-1)^{|A|-1}}.$

Lemma 2.2.6 and the fact that $s$ is nonnegative imply that $s$ preserves
GCDs, so

$\gcd\left\\{s\left(\frac{n}{p}\right)\;\middle|\;p\in
A\right\\}=s\left(\gcd\left\\{\frac{n}{p}\;\middle|\;p\in
A\right\\}\right)=s\left(\frac{n}{\mathrm{lcm}(A)}\right).$

This implies that

$\frac{s(n)}{r(n)}=\prod_{\begin{subarray}{c}A\subseteq P\\\
A\neq\emptyset\end{subarray}}s\left(\frac{n}{\mathrm{lcm}(A)}\right)^{(-1)^{|A|-1}}.$

Dividing both sides by $s(n)$ and taking the reciprocal of both sides gives

$r(n)=\prod_{A\subseteq
P}s\left(\frac{n}{\mathrm{lcm}(A)}\right)^{(-1)^{|A|}}.$

(On the right-hand side, we removed the condition $A\neq\emptyset$ and
multiplied the exponent by $-1$.)

We have to count how many ways each divisor $d$ of $n$ can be written as
$\mathrm{lcm}(A)$ for $A\subseteq P$. Since the elements of $A$ are distinct
primes, $\mathrm{lcm}(A)$ is the product of the elements of $A$. Therefore,
$d=\mathrm{lcm}(A)$ is square-free, and given $d$ there is a unique choice of
$A$ (the set of prime factors of $d$). So we get

$r(d)=\prod_{\begin{subarray}{c}d\mid n\\\ d\text{ square-
free}\end{subarray}}s\left(\frac{n}{d}\right)^{\mu(d)},$

using the fact that $\mu(d)$ is $(-1)^{|A|}$ if $A$ is the set of prime
factors of a square-free number $d$. Since $\mu(d)=0$ for $d$ not square-free,
we can remove the restriction that $d$ is square-free:

$r(n)=\prod_{d\mid n}s\left(\frac{n}{d}\right)^{\mu(d)}=\prod_{d\mid
n}s(d)^{\mu(n/d)}=\Phi_{n}(s).$

This is what we wanted to show. ∎

Before proving the $s$-Lucas theorem for composite $p$, we will prove a lemma
about $\Phi_{n}(s)$.

######  Lemma 3.4.3.

Let $s$ be a nonnegative integer GNS, let $m\in\mathbf{Z}_{>0}$, and let
$a,b\in\mathbf{Z}_{\geq 0}$ such that

$a\equiv b\not\equiv 0\pmod{m}.$

Define

$d=\gcd(a,m)=\gcd(b,m).$

Then,

$\frac{s(a)}{s(d)}\equiv\frac{s(b)}{s(d)}\pmod{\Phi_{m}(s)}.$

Additionally, both sides of this congruence are units modulo $\Phi_{m}(s)$.
Equivalently, the function $\mathbf{Z}_{\geq 0}\to\mathbf{Z}_{\geq 0}$ defined
by $a\mapsto s(a)/s(\gcd(a,m))$ induces a well-defined function

$(\mathbf{Z}/m)\setminus\\{0\\}\to(\mathbf{Z}/\Phi_{m}(s))^{\times}.$

###### Proof.

By Lemma 2.2.4,

$s(a)\equiv s(b)\pmod{s(m)}.$

Since $\gcd(s(m),s(d))=s(d)$, this implies that

$\frac{s(a)}{s(d)}\equiv\frac{s(b)}{s(d)}\quad\left(\\!\\!\\!\\!\\!\\!\mod{\frac{s(m)}{s(d)}}\right).$

Notice that

$\displaystyle\gcd\left\\{\frac{s(m)}{s(d)}\;\middle|\;d\text{ proper divisor
of }m\right\\}$ $\displaystyle=\frac{s(m)}{\mathrm{lcm}\\{s(d)\mid d\text{
proper divisor of }m\\}}$ $\displaystyle=\frac{s(m)}{\mathrm{lcm}\\{s(m/p)\mid
p\text{ prime factor of }m\\}}$ $\displaystyle=\Phi_{m}(s).$

In the second step of this chain of equalities, we used the fact that for
every proper divisor $d$ of $m$ there exists a prime factor $p$ of $m$ such
that $d\mid m/p$. We also used Lemma 2.2.3. We can conclude that
$\Phi_{m}(s)\mid s(m)/s(d)$ for all proper divisors $d$ of $m$. In particular,
for $d=\gcd(a,m)=\gcd(b,m)$, we have

$\frac{s(a)}{s(d)}\equiv\frac{s(b)}{s(d)}\pmod{\Phi_{m}(s)}.$

To complete the proof, we just have to show that $s(a)/s(d)$ is a unit modulo
$\Phi_{m}(s)$. This is true because $\Phi_{m}(s)\mid s(m)/s(d)$ implies that

$\gcd\left(\Phi_{m}(s),\frac{s(a)}{s(d)}\right)\mid\gcd\left(\frac{s(m)}{s(d)},\frac{s(a)}{s(d)}\right)=\frac{\gcd(s(m),s(a))}{s(d)}=\frac{s(d)}{s(d)}=1.\qed$

######  Theorem 3.4.4 (“Composite version” of the $s$-Lucas theorem).

Let $m\in\mathbf{Z}_{>0}$ and let $n,k\in\mathbf{Z}_{\geq 0}$. Then,

$\binom{nm}{km}_{s}\equiv\binom{n}{k}_{s_{m}}\pmod{\Phi_{m}(s)}.$

###### Proof.

By definition,

$\binom{nm}{km}_{s}=\frac{s(nm)s(nm-1)\cdots s(nm-km+1)}{s(km)s(km-1)\cdots
s(1)}=\frac{\prod_{j=1}^{km}s(nm-km+j)}{\prod_{j=1}^{km}s(j)}.$

Our goal will be to cancel each $s(nm-km+j)$ with the corresponding $s(j)$
modulo $\Phi_{m}(s)$. The problem is that in general, these are not units
modulo $\Phi_{m}(s)$. To fix this, we will divide $s(nm-km+j)$ and $s(j)$ by
$s(\gcd(j,m))$:

$\binom{nm}{km}_{s}=\frac{\prod_{j=1}^{km}s(nm-
km+j)/s(\gcd(j,m))}{\prod_{j=1}^{km}s(j)/s(\gcd(j,m))}.$

Since $j\equiv nm-km+j\pmod{m}$, we can apply Lemma 3.4.3 for all $j$ not
divisible by $m$, and we get that

$\frac{s(nm-km+j)}{s(\gcd(j,m))}\quad\text{and}\quad\frac{s(j)}{s(\gcd(j,m))}$

are congruent units modulo $\Phi_{m}(s)$.

Therefore, we can cancel $s(nm-km+j)/s(\gcd(j,m))$ with $s(j)/s(\gcd(j,m))$
modulo $\Phi_{m}(s)$ for every $j$ not divisible by $m$. We get

$\displaystyle\binom{nm}{km}_{s}$ $\displaystyle=\frac{\prod_{j=1}^{km}s(nm-
km+j)/s(\gcd(j,m))}{\prod_{j=1}^{km}s(j)/s(\gcd(j,m))}$
$\displaystyle\equiv\frac{\prod_{\ell=1}^{k}s((n-k+\ell)m)/s(m)}{\prod_{\ell=1}^{k}s(\ell
m)/s(m)}$
$\displaystyle=\frac{\prod_{\ell=1}^{k}s((n-k+\ell)m)}{\prod_{\ell=1}^{k}s(\ell
m)}$ $\displaystyle=\binom{n}{k}_{s_{m}}\pmod{\Phi_{m}(s)}.$

We changed product indices from $j$ to $\ell=j/m$, because the terms with
$m\nmid j$ were canceled in the second step. ∎

######  Remark 3.4.5.

When $m$ is prime, Theorem 3.4.4 reduces to the case $n_{0}=k_{0}=0$ of the
$s$-Lucas theorem (Proposition 2.4.8).

##  4\. Generalized $n$-Series and de Rham Complexes

###  4.1. Basic properties of the $s$-de Rham complex

In this subsection, we will define the “$s$-de Rham complex” using the
$s$-derivative of Definition 2.3.3. Recall that this is the $R$-linear map
$\nabla_{s}:R[x]\to R[x]$ given on monomials by
$\nabla_{s}(x^{n})=s(n)x^{n-1}$. Write
$\mathbf{A}^{1}=\operatorname{Spec}R[x]$ to denote the affine line over $R$.

######  Definition 4.1.1.

The $s$-de Rham complex for $R[x]$ is the $2$-term complex

$s\Omega_{\square,\mathbf{A}^{1}}:=(R[x]\xrightarrow{{\nabla_{s}}}R[x]dx).$

Here, the square indicates the dependence of
$s\Omega_{\square,\mathbf{A}^{1}}$ on the choice of coordinate $x$.

######  Remark 4.1.2.

It is easy to generalize the $s$-de Rham complex to several variables (e.g.,
by defining $s\Omega_{\mathbf{A}^{n}}$ to be
$s\Omega_{\square,\mathbf{A}^{1}}^{\otimes_{R}n}$). Since proving
multivariable analogues of the results below is straightforward, we will only
study the case of a single variable.

######  Example 4.1.3.

Let $s:\mathbf{Z}_{\geq 0}\to\mathbf{Z}[\\![q-1]\\!]$ denote the $q$-integer
GNS from Example 2.1.6. Then Definition 4.1.1 is precisely the $q$-de Rham
complex of [Sch17].

The $s$-Pascal identity of Proposition 2.1.3 can be restated in terms of the
$s$-derivative:

######  Lemma 4.1.4.

There is an equality of operators:

$\left[\frac{\nabla_{s}^{k}}{k!_{s}},x\right]=\frac{\nabla_{s}^{k-1}}{(k-1)!_{s}}.$

###### Proof.

Let $n>k\geq 0$ be integers. Since $\nabla_{s}(x^{n})=s(n)x^{n-1}$, we have

$\nabla_{s}^{k}(x^{n})=s(n)s(n-1)\cdots
s(n-k+1)x^{n-k}=\frac{n!_{s}}{(n-k)!_{s}}x^{n-k}.$

This implies that

$\frac{\nabla_{s}^{k}}{k!_{s}}x^{n}=\binom{n}{k}_{s}x^{n-k}.$

The $s$-Pascal identity can therefore be stated as:

$\frac{\nabla_{s}^{k}}{k!_{s}}x^{n}=\frac{\nabla_{s}^{k-1}}{(k-1)!_{s}}x^{n-1}+x\frac{\nabla_{s}^{k}}{k!_{s}}x^{n-1}.$

Rearranging gives

$\frac{\nabla_{s}^{k}}{k!_{s}}x^{n}-x\frac{\nabla_{s}^{k}}{k!_{s}}x^{n-1}=\frac{\nabla_{s}^{k-1}}{(k-1)!_{s}}x^{n-1}.$

Recognizing the left side as a commutator of operators, this can be written as

$\left[\frac{\nabla_{s}^{k}}{k!_{s}},x\right]x^{n-1}=\frac{\nabla_{s}^{k-1}}{(k-1)!_{s}}x^{n-1}.$

This implies the desired equality of operators. ∎

######  Remark 4.1.5.

One can similarly restate the $s$-Lucas theorem (Proposition 2.4.8) via the
$s$-derivative: namely, if the hypotheses of Proposition 2.4.8 are satisfied,
then for any prime $p$ and any nonnegative integers $n_{1},n_{0},k_{1},k_{0}$
such that $n_{0},k_{0}<p$, we have

$\frac{\nabla_{s}^{k_{1}p+k_{0}}}{(k_{1}p+k_{0})!_{s}}(x^{n_{1}p+n_{0}})\equiv\frac{\partial_{x^{p}}^{k_{1}}}{k_{1}!}(x^{n_{1}p})\frac{\nabla_{s}^{k_{0}}}{k_{0}!_{s}}(x^{n_{0}})\pmod{s(p)}.$

######  Proposition 4.1.6 ($s$-product rule).

For a GNS $s$ over $R$, define an $R$-bilinear operator
$\star_{s}:R[x]\otimes_{R}R[x]\to R[x]$ on monomials by

$x^{a}\star_{s}x^{b}=c_{s}(a+b+1,a)x^{a+b}$

for all $a,b\in\mathbf{Z}_{\geq 0}$. Then, for all $f,g\in R[x]$,

$\nabla_{s}(fg)=\nabla_{s}(f)g+f\star_{s}\nabla_{s}(g).$

###### Proof.

Since $\nabla_{s}$ is $R$-linear and $\star_{s}$ is $R$-bilinear, the theorem
follows from the case where $f$ and $g$ are monomials $x^{a}$ and $x^{b}$. In
this case,

$\displaystyle\nabla_{s}(x^{a})x^{b}+x^{a}\star_{s}\nabla_{s}(x^{b})$
$\displaystyle=s(a)x^{a-1}x^{b}+x^{a}\star_{s}s(b)x^{b-1}$
$\displaystyle=s(a)x^{a+b-1}+s(b)c_{s}(a+b,a)x^{a+b-1}$
$\displaystyle=\left(s(a)+s(b)\cdot\frac{s(a+b)-s(a)}{s(b)}\right)x^{a+b-1}$
$\displaystyle=(s(a)+s(a+b)-s(a))x^{a+b-1}$ $\displaystyle=s(a+b)x^{a+b-1}$
$\displaystyle=\nabla_{s}(x^{a+b}).\qed$

######  Example 4.1.7.

Let $s:\mathbf{Z}_{\geq 0}\to\mathbf{Z}[\\![q-1]\\!]$ denote the $q$-integer
GNS from Example 2.1.6. Then

$c_{s}(a+b+1,a)=\frac{[a+b+1]_{q}-[a]_{q}}{[b+1]_{q}}=\frac{q^{a+b+1}-q^{a}}{q^{b+1}-1}=q^{a},$

so that $x^{a}\star_{s}x^{b}=q^{a}x^{a+b}=(qx)^{a}x^{b}$. In particular,

$f(x)\star_{s}\nabla_{s}(g(x))=f(qx)\nabla_{q}(g(x)),$

so that Proposition 4.1.6 reduces to the usual $q$-Leibniz rule.

######  Example 4.1.8.

One can check that the function $s:\mathbf{Z}_{\geq
0}\to\mathbf{Z}[\\![q-1]\\!]$ given by

$s(n)=\frac{1}{q-1}\frac{q^{n}-(2-q)^{n}}{q^{n}+(2-q)^{n}}\in\mathbf{Z}[\\![q-1]\\!]$

defines a GNS; see Example 4.3.2. It follows that

$c_{s}(a+b+1,a)=2q^{a}(2-q)^{a}\frac{q^{b+1}+(2-q)^{b+1}}{(q^{a+b+1}+(2-q)^{a+b+1})(q^{a}+(2-q)^{a})}.$

In particular, unlike for the $q$-integer GNS, there is no simple expression
for $f(x)\star_{s}g(x)$.

######  Corollary 4.1.9.

Fix a GNS $s$ over $R$. Then the complex $s\Omega_{\square,\mathbf{A}^{1}}$
naturally admits the structure of a (noncommutative) differential graded
$R$-algebra.

###### Proof.

Define a left and right $s\Omega_{\square,\mathbf{A}^{1}}^{0}=R[x]$-module
structure on $s\Omega_{\square,\mathbf{A}^{1}}^{1}$ as follows: the right
module structure is the obvious one, and the left module structure is given by
$g(x)\cdot f(x)dx=g\star_{s}f(x)dx$. Then the $s$-Leibniz rule of Proposition
4.1.6 produces a $s\Omega_{\square,\mathbf{A}^{1}}^{0}$-bimodule structure on
$s\Omega_{\square,\mathbf{A}^{1}}^{1}$ such that the $s$-derivative satisfies
the Leibniz rule; this is precisely the structure of a differential graded
$R$-algebra. ∎

###  4.2. $s$-analogues of the Poincaré lemma and Cartier isomorphism

The Poincaré lemma says that over a field $k$ of characteristic zero, the
cohomology of the de Rham complex is concentrated in degree zero (where it is
isomorphic to $k$). A version of this statement is also true over
$\mathbf{Z}$. Namely, if
$\mathbf{Z}\langle{x}\rangle=\mathbf{Z}[x,\frac{x^{n}}{n!}]_{n\geq 0}$ denotes
the divided power envelope of $\mathbf{Z}[x]$, then the cohomology of the
complex
$\Omega^{\bullet}_{\mathbf{Z}[x]/\mathbf{Z}}\otimes_{\mathbf{Z}[x]}\mathbf{Z}\langle{x}\rangle$
is concentrated in degree zero (where it is isomorphic to $\mathbf{Z}$). This
admits a straightforward generalization to the $s$-de Rham complex:

######  Proposition 4.2.1 ($s$-Poincaré lemma).

Let $s$ be a GNS over $R$, and let $R\langle{x}\rangle_{s}$ denote the ring
$R[x,\frac{x^{n}}{n!_{s}}]_{n\geq 0}$. Then the cohomology of the complex
$s\Omega_{\square,\mathbf{A}^{1}}\otimes_{R[x]}R\langle{x}\rangle_{s}$ is
concentrated in degree zero, where it is isomorphic to $R$.

###### Proof.

We need to show that the $R$-linear map

$R\langle{x}\rangle_{s}\xrightarrow{{\nabla_{s}}}R\langle{x}\rangle_{s}dx$

is surjective, and has kernel $R$. Surjectivity follows from the observation
that $\nabla_{s}\frac{x^{n}}{n!_{s}}=\frac{x^{n-1}}{(n-1)!_{s}}$; this also
implies that the kernel of $\nabla_{s}$ is precisely the $R$-submodule of
$R\langle{x}\rangle_{s}$ generated by the constants. ∎

######  Remark 4.2.2.

Note that the ring $R\langle{x}\rangle_{s}$ is nonzero, since the elements
$n!_{s}\in R$ are not zero-divisors; in fact, $R\langle{x}\rangle_{s}$ is a
subring of $R[1/s][x]$.

The $s$-de Rham complex also satisfies an analogue of the Cartier isomorphism.

######  Recollection 4.2.3.

The Cartier isomorphism says that if $A$ is a smooth $\mathbf{F}_{p}$-algebra,
there is a canonical isomorphism
$\Omega^{i}_{A/\mathbf{F}_{p}}\cong\mathrm{H}^{i}(\Omega^{\bullet}_{A/\mathbf{F}_{p}})$.
If $\varphi$ denotes the Frobenius on $R$, this isomorphism is roughly given
by “$\frac{\varphi}{p^{i}}$”. When $A=\mathbf{F}_{p}[x]$, one can interpret
the Cartier isomorphism as giving a canonical isomorphism

$\Omega^{i}_{\mathbf{Z}[x^{p}]/\mathbf{Z}}\otimes_{\mathbf{Z}}\mathbf{F}_{p}\cong\mathrm{H}^{i}(\Omega^{\bullet}_{\mathbf{Z}[x]/\mathbf{Z}}\otimes_{\mathbf{Z}}\mathbf{F}_{p})$

sending $d(x^{p})\mapsto[x^{p-1}dx]$. In [Sch17, Proposition 3.4], Scholze
proves a $q$-analogue of the Cartier isomorphism. Let $\mathbf{Z}[\zeta_{p}]$
denote the quotient $\mathbf{Z}[\\![q-1]\\!]/[p]_{q}$, so that $\zeta_{p}$
denotes a primitive $p$th root of unity. Then there is a canonical isomorphism

$\Omega^{i}_{\mathbf{Z}[x^{p}]/\mathbf{Z}}\otimes_{\mathbf{Z}}\mathbf{Z}[\zeta_{p}]\cong\mathrm{H}^{i}(q\Omega^{\bullet}_{\mathbf{Z}[x]/\mathbf{Z}}\otimes_{\mathbf{Z}[\\![q-1]\\!]}\mathbf{Z}[\zeta_{p}]).$

Both of these results admit an $s$-analogue.

######  Proposition 4.2.4.

Let $s$ be a GNS over $R$ such that $s(1)$ is a unit in $R/s(p)$. Then there
is a canonical isomorphism

$s\Omega_{\square,\mathbf{A}^{1}}^{i}\otimes_{R}R/s(p)\cong\mathrm{H}^{i}(s\Omega_{\square,\mathbf{A}^{1}}\otimes_{R}R/s(p))$

sending $x^{n}\mapsto x^{np}$ in degree zero and
$x^{n}dx\mapsto[x^{np}x^{p-1}dx]$.

###### Proof.

Let us first compute
$\mathrm{H}^{0}(s\Omega_{\square,\mathbf{A}^{1}}\otimes_{R}R/s(p))$, i.e., the
kernel of $\nabla_{s}$. Observe that if $a\in R$, then $ax^{n}\mapsto
as(n)x^{n-1}dx$. If $p\mid n$, then $s(p)\mid s(n)$, so that
$\nabla_{s}(ax^{n})=0\in R[x]/s(p)$. If $p\nmid n$, it follows from Lemma
2.2.5 that $\nabla_{s}(ax^{n})=0\in R[x]/s(p)$ if and only if $s(p)\mid a$.
This implies that
$\mathrm{H}^{0}(s\Omega_{\square,\mathbf{A}^{1}}\otimes_{R}R/s(p))\cong
R[x^{p}]/s(p)$.

To calculate
$\mathrm{H}^{1}(s\Omega_{\square,\mathbf{A}^{1}}\otimes_{R}R/s(p))$, i.e., the
cokernel of $\nabla_{s}$, we need to determine the image of $\nabla_{s}$. If
$ax^{n}dx$ is in the image of $\nabla_{s}$ for some $a\in R$, then there must
be some $b\in R$ such that $bs(n+1)=a$. If $p\mid n+1$, it follows that
$a=0\in R/s(p)$; if $p\nmid n+1$, then $s(n+1)$ is a unit (by the preceding
discussion), so that $b=\frac{a}{s(n+1)}$. It follows that the image of
$\nabla_{s}$ is precisely $\bigoplus_{p\nmid n+1}R/s(p)\\{x^{n}dx\\}$, so that

$\operatorname{coker}(\nabla_{s})=\bigoplus_{n\geq
1}R/s(p)\\{x^{np-1}dx\\}\cong\bigoplus_{n\geq 1}R/s(p)\\{x^{p(n-1)}\cdot
x^{p-1}dx\\}.$

This implies that
$\mathrm{H}^{1}(s\Omega_{\square,\mathbf{A}^{1}}\otimes_{R}R/s(p))\cong
R[x^{p}]/s(p)d(x^{p})$, as desired. ∎

In fact, the classical and $q$-Cartier isomorphisms are special cases of a
more general result due to Berthelot and Ogus [BO78]. Let us now review this
statement; we will then state and prove the analogue for generalized
$n$-series.

######  Recollection 4.2.5.

Let $R$ be a ring, and let $f\in A$ be a non-zero-divisor. If $M^{\bullet}$ is
a cochain complex of $R$-modules which is termwise $f$-torsionfree, the
décalage $\eta_{f}M^{\bullet}$ is the subcomplex of $M^{\bullet}[1/f]$ defined
via

$(\eta_{f}M)^{i}=\\{x\in f^{i}M^{i}\mid dx\in f^{i+1}M^{i+1}\\}.$

See [BMS18, Section 6] and [BO78]. One basic property of the décalage
construction is the following. Let $\mathrm{H}^{\bullet}(M/f)$ denote the
complex whose underlying graded abelian group is
$\bigoplus_{i\in\mathbf{Z}}\mathrm{H}^{i}(M/f)$, and where the differential is
given by the $f$-Bockstein
$\beta:\mathrm{H}^{i}(M/f)\to\mathrm{H}^{i+1}(M/f)$. Then there is a natural
isomorphism of complexes

(5) $\eta_{f}(M)/f\xrightarrow{{\sim}}\mathrm{H}^{\bullet}(M/f).$

We will only need the case when $M^{\bullet}$ is termwise $f$-torsionfree; but
let us mention that $\eta_{f}$ preserves quasi-isomorphisms, and one can
extend $\eta_{f}$ to a (non-exact) functor $L\eta_{f}:D(R)\to D(R)$ on the the
derived category of $R$.

Let $k$ be a perfect field of characteristic $p>0$. A very special case of a
result of Berthelot and Ogus (in [BO78]) says that if $W(k)$ is the ring of
Witt vectors of $k$ and $A$ is a $W(k)$-algebra, there is a Frobenius333The
existence of a Frobenius on $\Omega_{A}$ is not obvious in general, and
depends on the existence of crystalline cohomology. semilinear isomorphism
$\Omega_{A}\xrightarrow{{\sim}}L\eta_{p}\Omega_{A}$. Suppose for simplicity
that $\Omega_{A}$ is $p$-torsionfree; applying 5 with $f=p$ then defines an
isomorphism of complexes

(6)
$\eta_{p}(\Omega_{A})/p\xrightarrow{{\sim}}\mathrm{H}^{\bullet}(\Omega_{A}/p).$

Note that if we write $A_{0}=A/p$, then $\Omega_{A}/p\cong\Omega_{A_{0}/k}$.
The Frobenius semilinear isomorphism
$\Omega_{A}\xrightarrow{{\sim}}L\eta_{p}\Omega_{A}$ gives a Frobenius
semilinear equivalence
$\Omega_{A_{0}/k}\xrightarrow{{\sim}}\eta_{p}(\Omega_{A_{0}})/p$. Comparing
the left and right-hand sides of 6 recovers the Cartier isomorphism for
$A_{0}$.

A similar result was proved in [BS19, Theorem 1.16(4)] for the $q$-de Rham
complex: namely, if $A$ is a smooth $\mathbf{Z}_{p}[\zeta_{p}]$-algebra, then
there is a Frobenius semilinear equivalence
$q\Omega_{A}\xrightarrow{{\sim}}L\eta_{[p]_{q}}q\Omega_{A}$. As above, using 5
with $f=[p]_{q}$ recovers the $q$-analogue of the Cartier isomorphism.

As one might expect, there is a décalage result for the $s$-de Rham complex,
too:

######  Proposition 4.2.6.

Fix a GNS $s$ over $R$ such that:

  1. (1)

there is a ring endomorphism $\varphi:R\to R$ which sends
$s(n)\mapsto\frac{s(np)}{s(p)}$.

  2. (2)

$s(1)$ is a unit in $R/s(p)$, and $R$ is $s(p)$-adically complete.

Write $\mathbf{A}^{1,(p)}=\operatorname{Spec}R[x^{p}]$, and define a map
$\Phi:s\Omega_{\square,\mathbf{A}^{1,(p)}}\to
s\Omega_{\square,\mathbf{A}^{1}}$ via

$\displaystyle R[x^{p}]$ $\displaystyle\xrightarrow{{\varphi}}R[x],\ \text{in
degree }0,$ $\displaystyle R[x^{p}]d(x^{p})$
$\displaystyle\xrightarrow{{\varphi,d(x^{p})\mapsto s(p)x^{p-1}dx}}R[x]dx,\
\text{in degree }1.$

Then $\Phi$ factors through a quasi-isomorphism
$\varphi^{\ast}s\Omega_{\square,\mathbf{A}^{1,(p)}}\xrightarrow{{\sim}}\eta_{s(p)}s\Omega_{\square,\mathbf{A}^{1}}$.

###### Proof.

Since $s\Omega_{\square,\mathbf{A}^{1}}$ is $s(p)$-torsionfree, we can
directly compute the complex $\eta_{s(p)}s\Omega_{\square,\mathbf{A}^{1}}$. If
$f(x)=\sum_{n\geq 0}a_{n}x^{n}\in R[x]$, then

$\nabla_{s}f(x)=\sum_{n\geq 0}a_{n}s(n)x^{n-1}.$

Since $s(p)\mid s(pj)$ for any $j\geq 0$, we see that $\nabla_{s}f(x)\in
s(p)R[x]$ if and only if $s(p)\mid a_{n}$ for $p\nmid n$. Therefore,

$\displaystyle\eta_{s(p)}s\Omega_{\square,\mathbf{A}^{1}}$
$\displaystyle=\left(R[x^{p}]+s(p)R[x]\xrightarrow{{\nabla_{s}}}s(p)R[x]dx\right).$

The map $\Phi$ clearly factors through the inclusion
$\eta_{s(p)}s\Omega_{\square,\mathbf{A}^{1}}\subseteq
s\Omega_{\square,\mathbf{A}^{1}}$. It remains to check that the map
$\varphi^{\ast}s\Omega_{\square,\mathbf{A}^{1,(p)}}\to\eta_{s(p)}s\Omega_{\square,\mathbf{A}^{1}}$
induces an isomorphism on cohomology. Observe that

$\displaystyle\mathrm{H}^{0}(s\Omega_{\square,\mathbf{A}^{1,(p)}})$
$\displaystyle\cong R,$
$\displaystyle\mathrm{H}^{1}(s\Omega_{\square,\mathbf{A}^{1,(p)}})$
$\displaystyle\cong\bigoplus_{n\geq 1}R/s(n)\\{(x^{p})^{n-1}d(x^{p})\\},$

and similarly

$\displaystyle\mathrm{H}^{0}(\eta_{s(p)}s\Omega_{\square,\mathbf{A}^{1}})$
$\displaystyle\cong R,$
$\displaystyle\mathrm{H}^{1}(\eta_{s(p)}s\Omega_{\square,\mathbf{A}^{1}})$
$\displaystyle\cong\bigoplus_{n\geq
1}(s(p))/(s(np))\\{x^{np-1}dx\\}\oplus\bigoplus_{p\nmid
m}(s(p))/(s(m)s(p))\\{x^{m-1}dx\\}.$

The ring endomorphism $\varphi$ of $R$ sends $s(n)\mapsto\frac{s(np)}{s(p)}$,
and $s(p)$ is a non-zero-divisor in $R$, we see that $\varphi$ descends to an
isomorphism

$\varphi^{\ast}(R/s(n))\xrightarrow{{\sim}}R/\tfrac{s(np)}{s(p)}\xrightarrow{{\sim}}(s(p))/(s(np)).$

If $p\nmid m$, then $s(m)$ is a unit in $R/s(p)$ by Lemma 2.2.5; since $R$ is
$s(p)$-adically complete, this implies that $s(m)$ is a unit in $R$ itself. It
follows that $0\cong R/s(m)\xrightarrow{{\sim}}(s(p))/(s(m)s(p))$. Putting
these together, we see that the map
$\varphi^{\ast}s\Omega_{\square,\mathbf{A}^{1,(p)}}\xrightarrow{{\sim}}\eta_{s(p)}s\Omega_{\square,\mathbf{A}^{1}}$
is a quasi-isomorphism. ∎

In Proposition 4.2.6, the condition that $R$ be $s(p)$-adically complete is
rather powerful. For instance, one of the most basic tools in $p$-adic
mathematics is the Legendre formula; this admits an analogue for generalized
$n$-series, too.

######  Proposition 4.2.7 ($s$-analogue of Legendre formula).

Let $s$ be a GNS over $R$ satisfying the hypotheses of Proposition 4.2.6. For
any $n\geq 0$, we have

$\displaystyle n!_{s}$ $\displaystyle=u\prod_{j\geq
1}\varphi^{j-1}(s(p))^{\lfloor n/p^{j}\rfloor}$

for some unit $u\in R^{\times}$.

###### Proof.

The argument is a straightforward adaptation of [BS19, Lemma 12.6]. Since $R$
is $s(p)$-adically complete, we know that if $p\nmid m$, then $s(m)$ is a unit
in $R$. This implies that

$\displaystyle(np)!_{s}$
$\displaystyle=u\prod_{i=1}^{n}s(np)=u\prod_{i=1}^{n}\left(\frac{s(np)}{s(p)}\cdot
s(p)\right)$
$\displaystyle=us(p)^{n}\prod_{i=1}^{n}\varphi(s(n))=us(p)^{n}\varphi(n!_{s}),$

where $u=\prod_{1\leq j\leq np,p\nmid j}s(j)$ is a unit in $R$. Using the
above identity to inductively strip powers of $p$ off of $n$ produces the
desired claim. ∎

This implies the following analogue of [BS19, Lemma 12.5], which gives a
criterion for admitting “$s$-divided powers”.

######  Corollary 4.2.8.

Let $s$ be a GNS over $R$ satisfying the hypotheses of Proposition 4.2.6, and
suppose that for any $n\geq 0$, $\varphi(n!_{s})$ is a non-zero-divisor in
$R/s(p)$. Let $A$ be an $s(p)$-completely flat $R$-algebra equipped with a
$R$-linear multiplicative map $\phi:\varphi^{\ast}A\to A$ and an element $x\in
A$ such that $\varphi(x)=x^{p}$, and $\varphi(x)$ is divisible by $s(p)$. Then
$n!_{s}\mid x^{n}$, i.e., $\frac{x^{n}}{n!_{s}}$ is well-defined in $A$.

###### Proof.

Because $s(j)$ is a unit in $R$ is $p\nmid j$, it suffices to show: for any
$n\geq 0$, if $n!_{s}\mid x^{n}$, then $(np)!_{s}\mid x^{np}$. By (the proof
of) Proposition 4.2.7, it suffices to show that $\varphi(n!_{s})s(p)^{n}\mid
x^{np}$. Because $\varphi(n!_{s})$ is a non-zero-divisor in $R/s(p)$ (by
assumption), it is also a non-zero-divisor in $A/s(p)$ by flatness of $A$. It
therefore suffices to show that $\varphi(n!_{s})$ and $s(p)^{n}$ each
individually divide $x^{np}$. Since $n!_{s}\mid x^{n}$, it is clear that
$\varphi(n!_{s})\mid\varphi(x^{n})=x^{np}$. Since $s(p)\mid x^{p}$ by
assumption, we also see that $s(p)^{n}\mid x^{np}$, as desired. ∎

###  4.3. Formal group law $n$-series and the $s$-derivative

Some of the most important (and accessible) examples of generalized $n$-series
come from formal group laws. Throughout this section, we will fix a base
commutative ring $R$.

######  Recollection 4.3.1.

A ($1$-dimensional) formal group law over a commutative ring $R$ is a two-
variable power series $F(x,y)\in R[\\![x,y]\\!]$ such that
$F(F(x,y),z)=F(x,F(y,z))$ and $F(x,y)\equiv x+y\pmod{(x,y)^{2}}$. It will
sometimes be convenient to denote $F(x,y)$ by $x+_{F}y$. A morphism $f:F\to G$
of formal group laws is a power series $f(x)\in R[\\![x]\\!]$ such that
$f(F(x,y))=G(f(x),f(y))$; a morphism is called an isomorphism if it admits a
compositional inverse.

If $n\geq 0$ is an integer, the $n$-series of $F$ is defined via the formula

$[n]_{F}(t)=\overbrace{F(t,F(t,F(t,\dots
F(t,t)\dots)))}^{n}=\overbrace{t+_{F}t+_{F}\cdots+_{F}t}^{n}.$

This can be extended to all integers $n\in\mathbf{Z}$ by using the existence
of inverses for the formal group law. The $n$-series $[n]_{F}(t)$ is an
element of $R[\\![t]\\!]$, and it is always divisible by $t$. We will define
$\langle n\rangle_{F}(t):=[n]_{F}(t)/t$ (where we agree that $\langle
0\rangle_{F}(t)=0$). Sometimes, it will be notationally convenient to simply
write these as $[n]_{F}$ and $\langle n\rangle_{F}$ (it being implicit that
these are functions of $t$). If $R$ is an $\mathbf{F}_{p}$-algebra, then
either $[p]_{F}(t)=0$ or $[p]_{F}(t)=\lambda t^{p^{h}}+O(t^{p^{h}+1})$ for
some $h>0$. If $v_{j}$ denotes the coefficient of $t^{p^{j}}$ in $[p]_{F}(t)$,
then $F$ is said to be of height $\geq n$ if $v_{j}=0$ for $j<n$; if $v_{n}$
is a unit, then $F$ is said to be of height $n$.

If $\mathbf{Q}\subseteq R$, then every formal group law $F(x,y)$ is isomorphic
to the additive formal group law via the logarithm. Let
$F_{y}(x,y)=\partial_{y}F(x,y)$; then, the logarithm is given by the integral

$\ell_{F}(x):=\int^{x}_{0}\frac{dt}{F_{y}(t,0)}.$

We will write $\mathscr{E}_{F}(x)$ to denote its compositional inverse, so
that $F(x,y)=\mathscr{E}_{F}(\ell_{F}(x)+\ell_{F}(y))$. Observe that
$[n]_{F}(t)=\mathscr{E}_{F}(n\ell_{F}(t))$ for any $n\in\mathbf{Z}$.

######  Example 4.3.2.

Fix a base commutative ring $R$. The polynomial $F(x,y)=x+y$ is known as the
additive formal group law, and $\langle n\rangle_{F}(t)=n$. The polynomial
$F(x,y)=x+y+xy$ is known as the multiplicative formal group law, and

$\langle n\rangle_{F}(t)=\frac{(1+t)^{n}-1}{t}\in R[\\![t]\\!].$

Note that $\langle{n}\rangle_{F}=[n]_{q}$, where we set $q=t+1$. The power
series $F(x,y)=\frac{x+y}{1+xy}$ is known as the hyperbolic formal group law
(since it describes the addition law for $\tanh$), and a simple induction on
$n$ shows that

$\langle
n\rangle_{F}(t)=\frac{1}{t}\frac{(1+t)^{n}-(1-t)^{n}}{(1+t)^{n}+(1-t)^{n}}=\frac{1}{q-1}\frac{q^{n}-(2-q)^{n}}{q^{n}+(2-q)^{n}}\in
R[\\![q-1]\\!].$

######  Remark 4.3.3.

Given a formal group law $F(x,y)$ over a (torsionfree, say) commutative ring
$R$, one can define a “rescaled” formal group law $\widetilde{F}(x,y)$ over
$R[\\![t]\\!]$ which is characterized by the property that
$\widetilde{F}(x,y)=\frac{1}{t}F(xt,yt)$. Observe that over the special fiber
(i.e., $t=0$), $\widetilde{F}(x,y)$ degenerates to the additive formal group
law. Over $(R\otimes\mathbf{Q})[\\![t]\\!]$, the logarithm
$\widetilde{\ell}_{F}(x)$ is given by $\frac{1}{t}\ell_{F}(tx)$; note that
this power series does not have polar terms in $t$, since $x\mid\ell_{F}(x)$
(and hence $tx\mid\ell_{F}(tx)$).

We begin by showing that the map $n\mapsto[n]_{F}(t)$ is a GNS over
$R[\\![t]\\!]$, as long as $F$ satisfies a mild condition. The existence of
the power series $\ell_{F}$ is the main reason that GNS arising via formal
group laws are particularly well-behaved.

######  Proposition 4.3.4.

Let $F$ be a formal group law over a ring $R$, and suppose that $[n]_{F}(t)\in
R[\\![t]\\!]$ is not a zero-divisor for any $n>0$. Define $s:\mathbf{Z}_{\geq
0}\to R[\\![t]\\!]$ by $s(n)=[n]_{F}(t)$. Then, $s$ is a GNS over
$R[\\![t]\\!]$. Similarly, the function $s_{F}:\mathbf{Z}_{\geq 0}\to
R[\\![t]\\!]$ sending $s_{F}(n)=\langle n\rangle_{F}(t)$ is a GNS over
$R[\\![t]\\!]$.

###### Proof.

It is easy to see that if $s$ is a GNS, the same is true of $s_{F}$. Let us
now show that $s$ is a GNS by checking the conditions of Definition 2.1.4.
Condition (1) is clear, since $[0]_{F}=0$. Condition (2) is already assumed in
the theorem statement.

For condition (3), let $G(x)\in R[\\![t]\\!][\\![x]\\!]$ be the power series
$F(t,x)$. Then,

$[n+1]_{F}-[k+1]_{F}=G([n]_{F})-G([k]_{F}).$

Since $x-y$ divides $x^{j}-y^{j}$ for each $j\geq 0$, and $G$ is a power
series, $x-y$ also divides $G(x)-G(y)$. In particular,
$s(n)-s(k)=[n]_{F}-[k]_{F}$ divides $G([n]_{F})-G([k]_{F})$; but
$G([n]_{F})-G([k]_{F})=[n+1]_{F}-[k+1]_{F}$ is precisely $s(n+1)-s(k+1)$, so
$s(n)-s(k)$ divides $s(n+1)-s(k+1)$. Inducting, we conclude that $s(n)-s(k)$
divides $s(n+j)-s(k+j)$ for all $n,k,j\in\mathbf{Z}_{\geq 0}$. For $k=0$, this
gives $s(n)\mid s(n+j)-s(j)$ for all $n,j\in\mathbf{Z}_{\geq 0}$. After
relabeling the indices, this becomes $s(n-k)\mid s(n)-s(k)$, which proves
condition (3). ∎

######  Lemma 4.3.5.

Let $F$ be a formal group law over a ring $R$. If $n\in\mathbf{Z}_{>0}$ is not
a zero-divisor in $R$ (e.g., $R$ is torsionfree), then $[n]_{F}$ and $\langle
n\rangle_{F}$ are not zero-divisors in $R[\\![t]\\!]$.

###### Proof.

Suppose for the sake of contradiction that $[n]_{F}(t)$ is a zero-divisor for
some $n$ (the same argument works for $\langle n\rangle_{F}$). Then there
exists a power series $f(t)\in R[\\![t]\\!]$ such that

$f(t)\cdot[n]_{F}(t)=0.$

It follows that the product of the coefficients of the lowest-degree terms of
$f(t)$ and $[n]_{F}(t)$ must be $0$. Since the lowest-degree term of
$[n]_{F}(t)$ is $nt$, this implies that $n$ times the lowest-degree
coefficient of $f(t)$ is $0$. In particular, $n$ is a zero-divisor in $R$. ∎

######  Definition 4.3.6.

Let $F$ be a formal group law over $R$ such that $[n]_{F}(t)\in R[\\![t]\\!]$
is not a zero-divisor for any $n>0$.444Many of the results below do not rely
on this assumption; but we keep it nonetheless, since Proposition 4.3.4 allows
to immediately transport many results about generalized $n$-series obtained
above. The interested reader should have no trouble removing this condition as
necessary in the results below. Let $F\Omega_{\square,\mathbf{A}^{1}}$ denote
the differential graded $R[\\![t]\\!]$-algebra given by the $s$-de Rham
complex associated to the GNS $s:\mathbf{Z}_{\geq 0}\to R$ sending
$n\mapsto\langle n\rangle_{F}$. We will refer to
$F\Omega_{\square,\mathbf{A}^{1}}$ as the $F$-de Rham complex of the affine
line $\mathbf{A}^{1}=\operatorname{Spec}R[x]$; we will abusively also refer to
the differential $\nabla_{F}$ as the $F$-derivative.

######  Example 4.3.7.

  * •

For the additive formal group law, $\langle n\rangle_{F}=n$; so the resulting
$F$-de Rham complex is simply the usual de Rham complex.

  * •

For the multiplicative formal group law, $\langle n\rangle_{F}=[n]_{q}$; so
the resulting $F$-de Rham complex is simply the $q$-de Rham complex. Note that
the $F$-derivative can be defined directly on polynomials (instead of only on
monomials) via $f(x)\mapsto\frac{f(qx)-f(x)}{(q-1)x}$. This can be seen
directly from Proposition 4.3.9: indeed,

$\mathscr{E}_{F}(\ell_{F}(t)z)=\frac{(1+t)^{z}-1}{t}=\frac{q^{z}-1}{q-1},$

and the operator $q^{x\partial_{x}}$ sends $f(x)\mapsto f(qx)$.

######  Remark 4.3.8.

One can consider a slight variant of the $F$-de Rham complex, given by the
complex

$C^{\bullet}:=\left(R[\\![t]\\!][x]\to R[\\![t]\\!][x]dx\right),\
x^{n}\mapsto[n]_{F}x^{n-1}dx.$

This is the $s$-de Rham complex for the function $s:\mathbf{Z}_{\geq 0}\to R$
sending $n\mapsto[n]_{F}$. Then, there is a quasi-isomorphism
$F\Omega_{\square,\mathbf{A}^{1}}\simeq\eta_{t}C^{\bullet}$.

When $\mathbf{Q}\subseteq R$, there is a general formula for the
$F$-derivative.

######  Proposition 4.3.9.

Suppose that $\mathbf{Q}\subseteq R$, and let $F$ be a formal group law over
$R$. Then there is an equality of $R[\\![t]\\!]$-linear operators on
$R[\\![t]\\!][x]$:

$\nabla_{F}=\frac{1}{tx}\mathscr{E}_{F}(\ell_{F}(t)x\partial_{x}).$

In particular, there is a canonical isomorphism
$F\Omega_{\square,\mathbf{A}^{1}}\cong\Omega_{\mathbf{Q}[x]/\mathbf{Q}}\otimes_{\mathbf{Q}}R[\\![t]\\!]$.

###### Proof.

Let $\nabla_{F}^{\prime}$ denote the expression on the right-hand side. By
definition of $\nabla_{F}$, it suffices to check that
$\nabla_{F}^{\prime}(x^{m})=\langle{m}\rangle x^{m-1}$ for every $m\geq 1$.
Write $\mathscr{E}_{F}(t)=\sum_{n}a_{n}t^{n}$; then

$\displaystyle\nabla^{\prime}_{F}(x^{m})$
$\displaystyle=\frac{1}{xt}\sum_{n}a_{n}\ell_{F}(t)^{n}(x\partial_{x})^{n}(x^{m})$
$\displaystyle=\frac{1}{xt}\sum_{n}a_{n}(m\ell_{F}(t))^{n}x^{m}$
$\displaystyle=\frac{1}{t}\mathscr{E}_{F}(m\ell_{F}(t))x^{m-1}=\langle{m}\rangle
x^{m-1},$

as desired. ∎

######  Remark 4.3.10.

Since every formal group law over a $\mathbf{Q}$-algebra is isomorphic to the
additive formal group law, the final statement of Proposition 4.3.9 is a
special case of the following more general observation: if $F_{1}$ and $F_{2}$
are isomorphic formal group laws, then the associated de Rham complexes are
also isomorphic.

######  Example 4.3.11.

Using Proposition 4.3.9, we can make the $F$-derivative explicit for the
hyperbolic formal group law. In this case, $\mathscr{E}_{F}(t)=\tanh(t)$, so
that $\ell_{F}(t)=\tanh^{-1}(t)$, and

$\mathscr{E}_{F}(\ell_{F}(t)z)=\tanh(z\tanh^{-1}(t))=\frac{q^{z}-(2-q)^{z}}{q^{z}+(2-q)^{z}},$

where $q-1=t$. Since the operator $q^{x\partial_{x}}$ sends $f(x)\mapsto
f(qx)$, the $F$-derivative can be expressed as

$\nabla_{F}:f(x)\mapsto\frac{1}{(q-1)x}\frac{f(qx)-f((2-q)x)}{f(qx)+f((2-q)x)}.$

######  Example 4.3.12.

Let $R_{0}=\mathbf{F}_{p}[v_{n}]$. Then, there is a unique formal group law
(known as the Honda formal group law; see [Hon70]) over $R_{0}$ which is
characterized by the property that its $p$-series is given by
$[p]_{F}(t)=v_{n}t^{p^{n}}$. This implies that up to a unit in
$R_{0}[\\![t]\\!]$, we have $[m]_{F}(t)=v_{n}^{v_{p}(m)}t^{m^{n}}$, where
$v_{p}(m)$ denotes the $p$-adic valuation of $m$. The formal group law over
$R_{0}$ lifts to a formal group law over $R=\mathbf{Z}_{p}[v_{n}]$ such that
over $R\otimes\mathbf{Q}\cong\mathbf{Q}_{p}[v_{n}]$, its logarithm is given by

$\ell_{F}(x)=\sum_{j\geq
0}v_{n}^{\frac{p^{jn}-1}{p^{n}-1}}\frac{x^{p^{jn}}}{p^{j}}.$

For example, if $n=1$ and we adjoin a $(p-1)$st root $\beta$ of $v_{1}$, this
is essentially the logarithm of the Artin-Hasse exponential, so that

$\ell_{F}(x)=-\frac{1}{\beta}\sum_{p\nmid
d}\frac{\mu(d)}{d}\mathrm{log}(1-(\beta x)^{d}).$

One can say something similar for general $n$. Recall that the polylogarithm
is defined by $\mathrm{Li}_{s}(x)=\sum_{j\geq 1}\frac{x^{j}}{j^{s}}$, so that
$\mathrm{Li}_{1}(x)=-\mathrm{log}(1-x)$. If $\beta$ denotes a $(p^{n}-1)$st
root of $v_{n}$, then $\ell_{F}(x)$ can be understood as a “$p^{n}$-typical”
version of $\frac{1}{\beta}\mathrm{Li}_{1/n}(\beta x)$.

When $n=1$, the resulting $F$-de Rham complex over $\mathbf{F}_{p}[\\![t]\\!]$
is closely related to the mod $p$ reduction of the $q$-de Rham complex:
indeed, observe that the $p$-series of the multiplicative formal group law is
congruent to $t^{p}\pmod{p}$. The resulting lifted formal group law over
$\mathbf{Z}_{p}[v_{1}]$ is the $p$-typification of the multiplicative formal
group law (see [Rav86, Appendix 2]). For higher $n$, the resulting $F$-de Rham
complex behaves qualitatively similar to the case $n=1$. For instance, we have

$\mathrm{H}^{0}(F\Omega_{\square,\mathbf{A}^{1}})\cong\mathbf{F}_{p^{n}}[v_{n}][\\![t]\\!],\
\mathrm{H}^{1}(F\Omega_{\square,\mathbf{A}^{1}})\cong\bigoplus_{j\geq
1}\mathbf{F}_{p^{n}}[v_{n},t]/(v_{n}^{v_{p}(j)}t^{j^{n}})\\{x^{j-1}dx\\}.$

######  Example 4.3.13.

Let $n\geq 1$, let $k$ be an algebraically closed field of characteristic
$p>0$, and let $R=W(k)[\\![u_{1},\cdots,u_{n-1}]\\!]$ denote the Lubin-Tate
ring. Let $F$ denote the formal group law associated to the universal
deformation of a chosen formal group law of height $n$ over $k$. Then, the
resulting $F$-de Rham complex specializes to the $q$-de Rham complex when
$n=1$. For general $n$, this $F$-de Rham complex is closely related to deep
phenomena in chromatic homotopy theory (see Remark 4.3.24).

######  Lemma 4.3.14 ($F$-Taylor expansion).

Let $F$ be a formal group law over $R$ such that $[n]_{F}(t)\in R[\\![t]\\!]$
is not a zero-divisor for any $n>0$. If
$f(x)\in(R\otimes\mathbf{Q})[\\![t,x-1]\\!]$, there is a Taylor expansion

$f(x)=\sum_{n\geq 0}\nabla_{F}^{n}(f(x))|_{x=1}\frac{(x-1)^{n}_{s}}{n!_{F}}.$

Here, $(x-1)^{n}_{s}$ denotes the symbol from Definition 2.3.5 with $y=-1$.

###### Proof.

This is the same argument as in [AL20, Proposition 4.4]. First, observe that
if $g(x)\in(R\otimes\mathbf{Q})[\\![t,x-1]\\!]$ is a function such that
$\nabla_{F}^{n}(g(x))|_{x=1}=0$ for all $n\geq 0$, then $g=0$. Indeed, since
$\nabla_{F}$ is simply the usual derivative modulo $t$, we see that $g(x)$ is
divisible by $t$. Write $g(x)=tg_{1}(x)$; then,
$\nabla_{F}^{n}(g_{1}(x))|_{x=1}=0$ for all $n\geq 0$, so $t\mid g_{1}(x)$.
Continuing, we see that $g(x)$ is infinitely $t$-divisible, and hence is zero
(since $t$ is topologically nilpotent).

We can now apply the above observation to

$g(x):=f(x)-\sum_{n\geq
0}\nabla_{F}^{n}(f(x))|_{x=1}\frac{(x-1)^{n}_{s}}{n!_{F}}.$

By definition of $(x-1)^{n}_{s}$, we know that
$\nabla_{F}(\frac{(x-1)^{n}_{s}}{n!_{F}})=\frac{(x-1)^{n-1}_{s}}{(n-1)!_{s}}$;
so $\nabla_{F}^{n}(g(x))|_{x=1}=0$ for all $n\geq 0$, and hence $g=0$, as
desired. ∎

######  Corollary 4.3.15 ($F$-logarithm).

Let $F$ be a formal group law over $R$ such that $[n]_{F}(t)\in R[\\![t]\\!]$
is not a zero-divisor for any $n>0$. Consider the function
$F\mathrm{log}(x)\in(R\otimes\mathbf{Q})[\\![t,x-1]\\!]$ given by
$\frac{t}{\ell_{F}(t)}\mathrm{log}(x)$. Then, we have:

  1. (1)

$\nabla_{F}(F\mathrm{log}(x))=\frac{1}{x}$.

  2. (2)

$F\mathrm{log}(xy)=F\mathrm{log}(x)+F\mathrm{log}(y)$.

  3. (3)

There is a series expansion

$F\mathrm{log}(x)=\sum_{n\geq
1}\frac{\langle{-n+1}\rangle_{F}\cdots\langle{-1}\rangle_{F}}{n!_{F}}(x-1)^{n}_{s}.$

###### Proof.

The first statement follows from Proposition 4.3.9. Indeed, write
$\mathscr{E}_{F}(y)=\sum_{n\geq 1}a_{n}y^{n}$; the condition that
$F(x,y)\equiv x+y\pmod{(x,y)^{2}}$ forces $a_{1}=1$. Since

$(x\partial_{x})(F\mathrm{log}(x))=\frac{t}{\ell_{F}(t)}(x\partial_{x})\mathrm{log}(x)=\frac{t}{\ell_{F}(t)},$

we see that

$\displaystyle x\nabla_{F}(F\mathrm{log}(x))$
$\displaystyle=\frac{1}{t}\sum_{n\geq
1}a_{n}\ell_{F}(t)^{n}(x\partial_{x})^{n}(F\mathrm{log}(x))$
$\displaystyle=\frac{1}{t}\left(\ell_{F}(t)\cdot\frac{t}{\ell_{F}(t)}+\sum_{n\geq
2}a_{n}\ell_{F}(t)^{n}(x\partial_{x})^{n}(F\mathrm{log}(x))\right).$

The second sum vanishes, since $(x\partial_{x})^{n}(F\mathrm{log}(x))=0$ for
$n\geq 2$. The first term cancels out to give
$x\nabla_{F}(F\mathrm{log}(x))=1$, as desired.

The second statement is clear. For the third statement, observe that

$\nabla_{F}^{n}(F\mathrm{log}(x))=\nabla_{F}^{n-1}(1/x)=\langle{-n+1}\rangle_{F}\cdots\langle{-1}\rangle_{F}x^{-n}.$

Evaluating at $x=1$ and using Lemma 4.3.14 gives the desired claim. ∎

######  Warning 4.3.16.

The $F$-logarithm $F\mathrm{log}(x)$ is not the same as the logarithm
$\ell_{F}(x)$ associated to the formal group law. This unfortunate terminology
stems from attempting to simultaneously emulate the standard terminology
“$q$-logarithm” and the “logarithm of the multiplicative formal group law”.

######  Remark 4.3.17.

Corollary 4.3.15 implies that $F\mathrm{log}(x)$ is a well-defined class in
the ring $R[\\![t]\\!]\left[x^{\pm 1},\frac{(x-1)^{n}_{s}}{n!_{F}}\right]$;
this is the ring of functions on an $F$-analogue of the divided power
completion of the identity section of $(\mathbf{G}_{m})_{R[\\![t]\\!]}$.

######  Remark 4.3.18.

The formal series in Corollary 4.3.15(3) can be written for arbitry GNS $s$;
when it exists and converges, its $s$-derivative will formally be $1/x$.
However, we have chosen to state Corollary 4.3.15 only in the case of GNS
arising via formal group laws, since it is otherwise difficult to get a
computational grip on the resulting formal series.

######  Example 4.3.19.

When $F$ is the multiplicative formal group law over $\mathbf{Z}$, the
function $F\mathrm{log}(x)$ can be identified with the $q$-logarithm

$\mathrm{log}_{q}(x)=\sum_{n\geq
1}(-1)^{n+1}q^{-\binom{n}{2}}\frac{(x-1)(x-q)\cdots(x-q^{n-1})}{[n]_{q}}\in\mathbf{Q}[\\![q-1,x-1]\\!].$

Indeed, this follows from Corollary 4.3.15(3) and the observation that
$\langle{-j}\rangle_{F}=[-j]_{q}=-q^{-j}[j]_{q}$. See [AL20, Section 4] for
more on the $q$-logarithm.

Let us summarize some of the results from the previous section upon
specialization to the $F$-de Rham complex:

######  Theorem 4.3.20.

Let $R$ be a commutative ring, and let $F$ be a formal group law over $R$ such
that $[n]_{F}(t)\in R[\\![t]\\!]$ is not a zero-divisor for any $n>0$. Let
$\hat{\mathbf{G}}=\operatorname{Spf}R[\\![t]\\!]$ denote the formal group over
$R$. Then:

  1. (1)

Let $R[\\![t]\\!]\langle{x}\rangle_{F}$ denote the ring
$R[\\![t]\\!][x,\frac{x^{n}}{[n]_{F}!}]_{n\geq 0}$. Then the Poincaré lemma
holds: the cohomology of the complex
$F\Omega_{\square,\mathbf{A}^{1}}\otimes_{R[\\![t]\\!][x]}R[\\![t]\\!]\langle{x}\rangle_{F}$
is concentrated in degree zero, where it is isomorphic to $R[\\![t]\\!]$.

  2. (2)

The Cartier isomorphism holds: there is a canonical isomorphism

$F\Omega_{\square,\mathbf{A}^{1,(p)}}^{i}\otimes_{R[\\![t]\\!]}R[\\![t]\\!]/\langle{p}\rangle_{F}\cong\mathrm{H}^{i}(F\Omega_{\square,\mathbf{A}^{1}}\otimes_{R[\\![t]\\!]}R[\\![t]\\!]/\langle{p}\rangle_{F})$

sending $(x^{p})^{n}\mapsto x^{np}$ in degree zero and
$(x^{p})^{n}d(x^{p})\mapsto[x^{np}x^{p-1}dx]$. Note that
$\operatorname{Spf}R[\\![t]\\!]/[p]_{F}\cong\hat{\mathbf{G}}[p]$.

  3. (3)

The décalage isomorphism holds: replace $R[\\![t]\\!]$ with its
$\langle{p}\rangle_{F}$-adic completion. Let $\varphi:R[\\![t]\\!]\to
R[\\![t]\\!]$ denote the $R$-algebra map sending $t\mapsto[p]_{F}(t)$, i.e.,
the map induced on rings by the multiplication-by-$p$ map
$\hat{\mathbf{G}}\to\hat{\mathbf{G}}$. Then, there is a quasi-isomorphism
$\varphi^{\ast}F\Omega_{\square,\mathbf{A}^{1,(p)}}\xrightarrow{{\sim}}\eta_{\langle{p}\rangle_{F}}F\Omega_{\square,\mathbf{A}^{1}}$.

  4. (4)

The hypotheses of Corollary 4.2.8 are satisfied, so that there is a criterion
for admitting “$F$-divided powers”. Namely, replace $R[\\![t]\\!]$ by its
$(p,\langle{p}\rangle_{F})$-adic completion, and suppose that for any $n\geq
0$, $\varphi(n!_{F})$ is a non-zero-divisor in
$R[\\![t]\\!]/\langle{p}\rangle_{F}$. Let $A$ be a
$\langle{p}\rangle_{F}$-completely flat $R[\\![t]\\!]$-algebra equipped with a
$R[\\![t]\\!]$-linear multiplicative map $\phi:\varphi^{\ast}A\to A$. If $x\in
A$ is an element such that $\varphi(x)=x^{p}$ and
$\langle{p}\rangle_{F}\mid\varphi(x)$, then $\frac{x^{n}}{n!_{s}}\in A$.

###### Proof.

The first part is Proposition 4.2.1. The second part is Proposition 4.2.4,
where the hypothesis in the proposition holds because
$\langle{1}\rangle_{F}=1$. The third (and fourth) part is an application of
Proposition 4.2.6. Note that the first hypothesis holds by construction of
$\varphi:R[\\![t]\\!]\to R[\\![t]\\!]$: indeed, $\varphi$ sends

$s(n)=\frac{[n]_{F}(t)}{t}\mapsto\frac{[n]_{F}([p]_{F}(t))}{[p]_{F}(t)}=\frac{[np]_{F}(t)}{[p]_{F}(t)}=\frac{s(np)}{s(p)}.$

The second hypothesis follows from the assumption on $R[\\![t]\\!]$. ∎

######  Remark 4.3.21.

Using Remark 4.3.10, one can upgrade Theorem 4.3.20 to the case when $R$
(rather, $\operatorname{Spec}R$) is replaced by the moduli stack of formal
groups. However, we will not discuss this further in this article.

Motivated by [Sch17, Conjecture 3.1], we propose the following conjecture. In
fact, considerations with ring stacks following [Dri22] strongly indicate that
the conjecture is false, but we have stated it nonetheless in the hopes that
it might spur investigation into these sort of questions. Arpon Raksit has
informed the first author that he is currently working on some variant of this
conjecture.

######  Conjecture 4.3.22.

Let $\mathrm{Poly}_{R}$ denote the category of polynomial $R$-algebras and
$R$-algebra maps between them. Let $F$ be a formal group law over $R$ such
that $[n]_{F}(t)\in R[\\![t]\\!]$ is not a zero-divisor for any $n>0$. Then,
there is a functor
$\Gamma_{F\text{-}\mathrm{dR}}({-}):\mathrm{Poly}_{R}\to\mathrm{CAlg}(R[\\![t]\\!])$
landing in the $\infty$-category of
${\mathbf{E}_{\infty}}$-$R[\\![t]\\!]$-algebras which sends
$R[x_{1},\cdots,x_{n}]\mapsto F\Omega_{\square,\mathbf{A}^{n}}$.555In other
words, the assignment $R[x_{1},\cdots,x_{n}]\mapsto
F\Omega_{\square,\mathbf{A}^{n}}$ is functorial in $R$-algebra maps of
polynomial $R$-algebras. Furthermore, each part of Theorem 4.3.20 admits a
generalization to $\Gamma_{F\text{-}\mathrm{dR}}({-})$.

######  Remark 4.3.23.

If the preceding conjecture is true, then the construction
$F\Omega_{\square,-}$ can be extended to all (animated) $R$-schemes by left
Kan extension:

$\textstyle{\mathrm{Poly}_{R}^{\mathrm{op}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\Gamma_{F\text{-}\mathrm{dR}}({-})}$$\textstyle{\mathrm{CAlg}(R[\\![t]\\!])^{\mathrm{op}}}$$\textstyle{\mathrm{Sch}_{R}^{\mathrm{op}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\Gamma_{F\text{-}\mathrm{dR}}({-})}$

We expect that the resulting functor
$X\mapsto\Gamma_{F\text{-}\mathrm{dR}}({X})$, if it exists, should be rather
interesting. When $F$ is the multiplicative formal group law, 4.3.22 is true,
and the resulting assignment $X\mapsto\Gamma_{F\text{-}\mathrm{dR}}({X})$ is
the $q$-de Rham cohomology of [BS19].

Let us end this section by discussing the motivation behind the construction
of the $F$-de Rham complex. We will necessarily be brief, since this is not
the main subject of the present article.

######  Remark 4.3.24.

Let $A$ be an even-periodic ${\mathbf{E}_{\infty}}$-ring equipped with a
complex orientation. Then, Quillen defined a canonical formal group law
$F(x,y)$ over $R:=\pi_{0}(A)$, whose underlying formal group is
$\operatorname{Spf}\pi_{0}(A^{hS^{1}})$. Let $\tau_{\geq 0}A$ denote the
connective cover of $A$, and let
$\mathrm{F}^{\star}_{\mathrm{ev}}\mathrm{HP}(\tau_{\geq 0}A[x]/\tau_{\geq
0}A)$ denote the even filtration on the periodic cyclic homology of
$\tau_{\geq 0}A[x]$ (defined in [HRW22]). The assignment $A\mapsto(\tau_{\geq
0}A)^{tS^{1}}$ is the homotopical analogue of the construction of the
“rescaled” formal group law from Remark 4.3.3; in other words,
$\operatorname{Spf}\pi_{0}((\tau_{\geq 0}A)^{tS^{1}})$ is the rescaled
analogue of the Quillen formal group $\operatorname{Spf}\pi_{0}(A^{hS^{1}})$.

Unpublished work of Arpon Raksit shows that the $F$-de Rham complex
$F\Omega_{\square,\mathbf{A}^{1}}$ arises as the zeroth associated graded
piece $\mathrm{gr}^{0}_{\mathrm{ev}}\mathrm{HP}(\tau_{\geq 0}A[x]/\tau_{\geq
0}A)$.666Using similar methods, one can also show that the variant of the
$F$-de Rham complex from Remark 4.3.8 arises as the zeroth associated graded
piece $\mathrm{gr}^{0}_{\mathrm{ev}}\mathrm{HC}^{-}(A[x]/A)$ in the negative
cyclic homology of $A[x]$. In particular, in this case, the $F$-de Rham
complex admits the structure of an
${\mathbf{E}_{\infty}}$-$R[\\![t]\\!]$-algebra. There are homotopical
analogues of each part of Theorem 4.3.20: for example, the décalage
isomorphism of Theorem 4.3.20(3) is proved as [Dev23b, Proposition 3.5.3].

When $A=\mathrm{KU}$ is periodic complex K-theory (so $\tau_{\geq
0}A=\mathrm{ku}$ is connective complex K-theory), the formal group law over
$\pi_{0}(A)$ is precisely the multiplicative one; so the $q$-de Rham complex
arises as
$\mathrm{gr}^{0}_{\mathrm{ev}}\mathrm{HP}(\mathrm{ku}[x]/\mathrm{ku})$. As
explained in [DR23], the $p$-completion of
$\mathrm{HP}(\mathrm{ku}[x]/\mathrm{ku})$ can be understood via the
topological negative cyclic homology of $\mathbf{Z}_{p}[\zeta_{p}][x]$; this
is a homotopical analogue of the Bhatt-Scholze construction [BS19] of $q$-de
Rham cohomology via prismatic cohomology.

When $A=E_{n}$ is the Morava E-theory associated to the Lubin-Tate formal
group (see Example 4.3.13) and $\tau_{\geq 0}A=e_{n}$ is its connective cover,
the $F$-de Rham complex of Example 4.3.13 arises as
$\mathrm{gr}^{0}_{\mathrm{ev}}\mathrm{HP}(e_{n}[x]/e_{n})$. The periodic
cyclic homology $\mathrm{HP}(e_{n}[x]/e_{n})$ plays an important role in
higher chromatic analogues of the work of Bhatt-Morrow-Scholze [BMS19], and
will be explored in future work.

Although this does not quite fall into the above framework, the first author
hopes to show in future work that when $A=L^{s}(\mathbf{Z})$ is the symmetric
L-theory of the integers (see [HLN21]), the $F$-de Rham complex associated to
the hyperbolic formal group law is closely related to
$\mathrm{gr}^{0}_{\mathrm{ev}}\mathrm{HP}(L^{s}(\mathbf{Z})[x]/L^{s}(\mathbf{Z}))$.
This is a manifestation of the observation (going back to the Hirzebruch
signature theorem) that the logarithm of the formal group law associated to
the complex orientation on $L^{s}(\mathbf{Z})$ is given by the hyperbolic
tangent function $\tanh(x)$.

###  4.4. A variant of the Weyl algebra

It is well-known that the classical de Rham complex over a base commutative
ring $R$ is Koszul dual to the usual Weyl algebra of differential operators on
the affine line:

$\mathscr{D}_{\mathbf{A}^{1}}=R\langle{x,\partial_{x}}\rangle/([\partial_{x},x]=1).$

The complex of Remark 4.3.8 for the additive formal group is also Koszul dual
to a rescaled analogue $R[\\![t]\\!]\langle{x,D}\rangle/([D,x]=t)$ of this
Weyl algebra; this rescaling amounts to replacing $\partial_{x}$ by
$t\partial_{x}$. This fact has an analogue for arbitrary formal group laws, as
we now explain. It turns out to be significantly more convenient to study the
Weyl algebra of $\mathbf{G}_{m}$ instead, so we will restrict to that case.
Some of the discussion in this section appears briefly in [Dev23a, Section
3.3]. As usual, fix a base commutative ring $R$ and a formal group law
$F(x,y)$ over $R$.

######  Definition 4.4.1.

The $F$-Weyl algebra of $\mathbf{G}_{m}=\operatorname{Spec}R[x^{\pm 1}]$ is
defined to be the associative $R[\\![t]\\!]$-algebra given by

$F\mathscr{D}_{\square,\mathbf{G}_{m}}:=R[\\![t]\\!]\langle{x^{\pm
1},y}\rangle^{\wedge}_{y}/(yx=xF(y,t)).$

######  Example 4.4.2.

For the additive formal group law, the relation imposed in Definition 4.4.1 is
just $yx=x(y+t)$, or equivalently that $[y,x]=tx$. This defines an isomorphism
between $F\mathscr{D}_{\square,\mathbf{G}_{m}}$ and the rescaled Weyl algebra
for $\mathbf{G}_{m}$ by sending $y$ to the rescaled vector field
$tx\partial_{x}$.

######  Example 4.4.3.

For the multiplicative formal group law, let us write $q=1+t$. If we define
$\widetilde{y}=1+y$, the relation imposed in Definition 4.4.1 is just

$(\widetilde{y}-1)x=x(q\widetilde{y}-1),$

or equivalently that $\widetilde{y}x=qx\widetilde{y}$. Observe that
$\widetilde{y}$ acts as the operator $q^{x\partial_{x}}$, so that we obtain an
isomorphism between $F\mathscr{D}_{\square,\mathbf{G}_{m}}$ and the completion
of $R[\\![q-1]\\!]\langle{x^{\pm
1},q^{x\partial_{x}}}\rangle/(q^{x\partial_{x}}x=qxq^{x\partial_{x}})$ at the
ideal $(q^{x\partial_{x}}-1)$. This algebra is essentially the $q$-Weyl
algebra of $\mathbf{G}_{m}$, and is sometimes known as the “quantum torus”, as
well as the algebra of $q$-difference operators on the torus.

######  Lemma 4.4.4.

The algebra $R[\\![t]\\!][x^{\pm 1}]$ is canonically a left
$F\mathscr{D}_{\square,\mathbf{G}_{m}}$-module, where the action of $y$ sends
$x^{n}\mapsto[n]_{F}x^{n}$.

###### Proof.

Observe that

$(yx)(x^{n})=[n+1]_{F}x^{n+1}=xF([n]_{F},t)x^{n}=xF(y,t)x^{n},$

so that the action prescribed above satisfies the relation in Definition
4.4.1. ∎

The following result describing the complex from Remark 4.3.8 as Koszul dual
to the $F$-Weyl algebra is sketched in [Dev23a, Proposition 3.3.9]. We will
not need this result below, so we only state it for completeness.

######  Proposition 4.4.5.

The derived tensor product $R[\\![t]\\!][x^{\pm
1}]\otimes_{F\mathscr{D}_{\square,\mathbf{G}_{m}}}R[\\![t]\\!][x^{\pm 1}]$ can
be identified with the complex

$C^{\bullet}\otimes_{R[x]}R[x^{\pm 1}]\cong\left(R[\\![t]\\!][x^{\pm 1}]\to
R[\\![t]\\!][x^{\pm 1}]d\mathrm{log}(x)\right),\
x^{n}\mapsto[n]_{F}x^{n}d\mathrm{log}(x).$

######  Remark 4.4.6.

In [Dev23a, Section 3.3], we show that if $R$ is a complex-oriented even-
periodic ${\mathbf{E}_{\infty}}$-ring and $F(x,y)$ is the formal group law
over $\pi_{0}(R)$, then $F\mathscr{D}_{\square,\mathbf{G}_{m}}$ arises as the
loop-rotation equivariant homology $\pi_{\ast}(R[\Omega T]^{h(T\times
S^{1}_{\mathrm{rot}})})$ where $T$ is a compact torus of rank $1$ (i.e., a
circle, not to be confused with the loop-rotation circle).777In fact, the
first author initially came across the $F$-Weyl algebra in this manner, and
this article was originally intended to be about this algebra. However, upon
learning of the $F$-de Rham complex from Arpon Raksit, it became clear that
the $F$-de Rham complex was a much simpler object to work with; hence the
present form of the article. We also explained that, under the discussion of
Remark 4.3.24, the Koszul duality of Proposition 4.4.5 is a manifestation of
the Koszul duality between $R[\Omega T]^{h(T\times S^{1}_{\mathrm{rot}})}$ and
the negative cyclic homology
$R[\mathscr{L}T_{+}]^{hS^{1}_{\mathrm{rot}}}=\mathrm{HC}^{-}(R[\Omega T]/R)$.

######  Remark 4.4.7 (Mellin transform).

Let $\mathbf{A}^{1}_{R[\\![t]\\!]}$ denote the affine line over $R[\\![t]\\!]$
with coordinate $y$, so that $\mathbf{Z}$ acts on
$\mathbf{A}^{1}_{R[\\![t]\\!]}$ via the map $y\mapsto F(y,t)$. It follows from
Definition 4.4.1 and Morita theory that the category of
$F\mathscr{D}_{\square,\mathbf{G}_{m}}$-modules is equivalent to the category
of quasicoherent sheaves on $\mathbf{A}^{1}_{R[\\![t]\\!]}/\mathbf{Z}$ which
are $y$-complete. In the case of the additive formal group law, this is a
$t$-deformation of the Mellin transform, which gives an equivalence between
the category of $\mathscr{D}_{\mathbf{G}_{m}}$-modules and quasicoherent
sheaves on $\mathbf{A}^{1}/\mathbf{Z}$ (where $\mathbf{Z}$ acts by $y\mapsto
y+1$).

Let us now describe some special properties of the center of
$F\mathscr{D}_{\square,\mathbf{G}_{m}}$.

######  Recollection 4.4.8.

One property satisfied by the ordinary Weyl algebra in characteristic $p>0$ is
that it has a large center: namely, if $R$ is an $\mathbf{F}_{p}$-algebra,
there is an isomorphism $Z(\mathscr{D}_{\mathbf{G}_{m}})\cong R[x^{\pm
p},x^{p}\partial_{x}^{p}]$ which identifies $Z(\mathscr{D}_{\mathbf{G}_{m}})$
with the ring of functions on the cotangent bundle of the Frobenius twist
$(\mathbf{G}_{m})^{(p)}$. Note that
$x^{p}\partial_{x}^{p}\equiv(x\partial_{x})^{p}-x\partial_{x}\pmod{p}$. Under
the Koszul duality between $\mathscr{D}_{\mathbf{G}_{m}}$ and the de Rham
complex, this identification of $Z(\mathscr{D}_{\mathbf{G}_{m}})$ is in fact
Koszul dual to the Cartier isomorphism
$\mathrm{H}^{\ast}(\Omega^{\bullet}_{\mathbf{G}_{m}/R})\cong\Omega^{\ast}_{(\mathbf{G}_{m})^{(p)}/R}$.

It is therefore natural to ask for a description of the center of
$F\mathscr{D}_{\square,\mathbf{G}_{m}}$; this leads to the following result,
which is Koszul dual to the Cartier isomorphism of Theorem 4.3.20(b).

######  Theorem 4.4.9.

The center of
$F\mathscr{D}_{\square,\mathbf{G}_{m}}\otimes_{R[\\![t]\\!]}R[\\![t]\\!]/\langle{p}\rangle_{F}$
can be identified as follows:

$Z(F\mathscr{D}_{\square,\mathbf{G}_{m}}\otimes_{R[\\![t]\\!]}R[\\![t]\\!]/\langle{p}\rangle_{F})\cong
R[\\![t]\\!]\left[x^{\pm
p},\prod_{j=0}^{p-1}(y+_{F}[j]_{F})\right]/\langle{p}\rangle_{F}.$

###### Proof.

Observe that

$yx^{p}=x^{p}F(y,[p]_{F})\equiv x^{p}y\pmod{\langle{p}\rangle_{F}},$

so that $x^{p}$ is in the center of
$F\mathscr{D}_{\square,\mathbf{G}_{m}}\otimes_{R[\\![t]\\!]}R[\\![t]\\!]/\langle{p}\rangle_{F}$.
Similarly, since
$\prod_{j=0}^{p-1}(y+_{F}[j]_{F})\equiv\prod_{j=0}^{p-1}(y+_{F}[j]_{F})\pmod{\langle{p}\rangle_{F}}$,
we have

$\prod_{j=0}^{p-1}(y+_{F}[j]_{F})x=x\prod_{j=1}^{p}(y+_{F}[j]_{F})\equiv
x\prod_{j=0}^{p-1}(y+_{F}[j]_{F})\pmod{\langle{p}\rangle_{F}},$

and hence $\prod_{j=0}^{p-1}(y+_{F}[j]_{F})$ is in the center of
$F\mathscr{D}_{\square,\mathbf{G}_{m}}\otimes_{R[\\![t]\\!]}R[\\![t]\\!]/\langle{p}\rangle_{F}$.
This defines an inclusion

$R[\\![t]\\!]\left[x^{p},\prod_{j=0}^{p-1}(y+_{F}[j]_{F})\right]/\langle{p}\rangle_{F}\subseteq
Z(F\mathscr{D}_{\square,\mathbf{G}_{m}}\otimes_{R[\\![t]\\!]}R[\\![t]\\!]/\langle{p}\rangle_{F}),$

which can be checked to be an isomorphism. ∎

######  Example 4.4.10.

For the additive formal group law over $\mathbf{Z}$ (say), we have

$\prod_{j=0}^{p-1}(y+jt)\equiv y^{p}-t^{p-1}y\pmod{p},$

so that upon identifying $y=x\partial_{x}$, Theorem 4.4.9 implies the
following isomorphism for the rescaled Weyl algebra:

$Z(F\mathscr{D}_{\square,\mathbf{G}_{m}}/p)\cong\mathbf{F}_{p}[\\![t]\\!][x^{\pm
p},(x\partial_{x})^{p}-t^{p-1}x\partial_{x}].$

Regarding the above algebra as a graded ring with both $t$ and $x\partial_{x}$
in weight $1$ allows us to replace $t$ by a polynomial generator (instead of a
power series variable). Inverting $t$ and taking $\mathbf{G}_{m}$-invariants
(which is to be thought of as setting $t=1$) then produces the ring
$\mathbf{F}_{p}[x^{\pm
p},t^{-p}x^{p}\partial_{x}^{p}]=\mathscr{O}_{T^{\ast}\mathbf{G}_{m}^{(1)}}$.

######  Example 4.4.11.

For the multiplicative formal group law over $\mathbf{Z}$ (say), if we define
$\widetilde{y}=y+1$ as above (so that $\widetilde{y}=q^{x\partial_{x}}$), we
have

$\prod_{j=0}^{p-1}(y+_{F}(q^{j}-1))=\prod_{j=0}^{p-1}(q^{j}\widetilde{y}-1)\equiv
q^{p(p-1)/2}(\widetilde{y}^{p}-1)\pmod{[p]_{q}}.$

Note that $q^{p(p-1)/2}$ is $1\pmod{[p]_{q}}$ for $p>2$, but is
$-1\pmod{[2]_{q}}$. In either case, $q^{p(p-1)/2}$ is a unit, so Theorem 4.4.9
implies the following isomorphism:

$Z(F\mathscr{D}_{\square,\mathbf{G}_{m}}/[p]_{q})\cong\mathbf{Z}[\zeta_{p}][x^{\pm
p},q^{px\partial_{x}}].$

######  Remark 4.4.12.

Since $F\mathscr{D}_{\square,\mathbf{G}_{m}}$ can be recovered from the
equivariant homology $\pi_{\ast}(R[\Omega T]^{h(T\times
S^{1}_{\mathrm{rot}})})$ for a compact torus $T$ of rank $1$ (see Remark
4.4.6), it is natural to wonder whether there is an explanation of Theorem
4.4.9 from the perspective of the affine Grassmannian. In the case when $R$ is
ordinary (integral) homology or K-theory, this has been answered in [Lon17] —
in fact, the methods there are sufficiently geometric that they work even for
more general $R$ (and for $T$ replaced by a more general connected compact Lie
group!), so we refer the reader to loc. cit. for further discussion of this
question.

###  4.5. An analogue of the Bhatt-Lurie Cartesian square

Throughout this section, we will fix a $p$-completely flat
$\mathbf{Z}_{p}$-algebra $R$ and a formal group law $F(x,y)$ over $R$ (so $R$
is torsionfree). The symbol $R[\\![t]\\!]$ will always denote the $p$-adic
completion of the formal power series ring, and all constructions will be done
internal to the category of $(p,t)$-adically $R[\\![t]\\!]$-schemes. (We have
omitted the completion from the notation for readability.) Let
$\hat{\mathbf{G}}$ denote the associated formal group, so that its underlying
formal scheme is $\operatorname{Spf}R[\\![t]\\!]$.

In [BL22, Lemma 3.5.18], Bhatt-Lurie showed that there is a Cartesian square
of group schemes over $R$:

(7)
$\textstyle{\mathbf{G}_{m}^{\sharp}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\mathrm{log}}$$\textstyle{\mathbf{G}_{a}^{\sharp}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{x\mapsto\exp(px)}$$\textstyle{\mathbf{G}_{m}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{x\mapsto
x^{p}}$$\textstyle{\mathbf{G}_{m}^{(1)}.}$

Taking vertical quotients, we obtain an isomorphism between
$\mathbf{G}_{m}/\mathbf{G}_{m}^{\sharp}$ and
$\mathbf{G}_{m}^{(1)}/\mathbf{G}_{a}^{\sharp}$; one can identify
$\mathbf{G}_{m}/\mathbf{G}_{m}^{\sharp}$ with the “de Rham stack”
$\mathbf{G}_{m}^{\mathrm{dR}}$ of $\mathbf{G}_{m}$ (see, e.g., [Bha22]), so
that this isomorphism describes $\mathbf{G}_{m}^{\mathrm{dR}}$ in terms of the
group scheme $\mathbf{G}_{a}^{\sharp}$.

The proof of 7 in loc. cit. used the relationship between the group schemes
appearing in the square and the ring scheme of Witt vectors. In this section,
we prove an $F$-analogue of this result (see Theorem 4.5.10); in the case of
the additive formal group law, this reproves 7. Let us state at the outset
that in the case when $F$ is the multiplicative formal group law, this result
was obtained in a discussion between the first author and Michael Kural.
Moreover, the argument in this section rests crucially on 12, the $q$-analogue
of which (Example 4.5.8) was proved by Michael Kural. Any errors below are
solely the fault of the first author!

######  Definition 4.5.1.

Remark 4.3.3 gives a formal group $\hat{\mathbf{G}}_{t}$ over
$\operatorname{Spf}R[\\![t]\\!]$ whose logarithm is
$\widetilde{\ell}_{F}(x)=\frac{1}{t}\ell_{F}(tx)$. Let $x$ denote the
coordinate on $\hat{\mathbf{G}}_{t}$, so that its underlying formal scheme is
$\operatorname{Spf}R[\\![t,x]\\!]$. Let $\hat{\mathbf{G}}_{t}^{\vee}$ denote
the Cartier dual
$\operatorname{Hom}(\hat{\mathbf{G}}_{t},(\mathbf{G}_{m})_{R[\\![t]\\!]})$ of
$\hat{\mathbf{G}}_{t}$; see [Dri21, Section 3] for some generalities on
Cartier duals of formal groups. The element
$x\in\mathscr{O}_{\hat{\mathbf{G}}_{t}}$ defines a homomorphism
$\tau:\hat{\mathbf{G}}_{t}^{\vee}\to(\mathbf{G}_{a})_{R[\\![t]\\!]}$.

######  Observation 4.5.2.

Over $(R\otimes\mathbf{Q})[\\![t]\\!]$, the rescaled logarithm
$\widetilde{\ell}_{F}$ of Remark 4.3.3 defines an isomorphism
$\widetilde{\ell}_{F}:\hat{\mathbf{G}}_{t}\xrightarrow{{\sim}}(\hat{\mathbf{G}}_{a})_{(R\otimes\mathbf{Q})[\\![t]\\!]}$
of formal groups. Therefore, the canonical pairing
$\hat{\mathbf{G}}_{t}\times_{R[\\![t]\\!]}\hat{\mathbf{G}}_{t}^{\vee}\to(\mathbf{G}_{m})_{R[\\![t]\\!]}$
fits into a diagram

$\textstyle{\hat{\mathbf{G}}_{t}\times_{R[\\![t]\\!]}\hat{\mathbf{G}}_{t}^{\vee}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\mu}$$\scriptstyle{\sim}$$\scriptstyle{\widetilde{\ell}_{F}\times\mathrm{id}}$$\textstyle{(\hat{\mathbf{G}}_{a})_{(R\otimes\mathbf{Q})[\\![t]\\!]}\times_{R[\\![t]\\!]}\hat{\mathbf{G}}_{t}^{\vee}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\nu}$$\textstyle{(\mathbf{G}_{m})_{(R\otimes\mathbf{Q})[\\![t]\\!]}.}$

Since $R[\\![t]\\!]$ is $(p,t)$-adically complete, the Cartier dual of
$(\hat{\mathbf{G}}_{a})_{R[\\![t]\\!]}$ can be identified with the divided
power completion $(\mathbf{G}_{a}^{\sharp})_{R[\\![t]\\!]}$. The pairing $\nu$
is base-changed from $R[\\![t]\\!]$ itself, where it is given by the formula

$\nu:(x,y)\mapsto\exp(xy).$

It follows that the pairing $\mu$ is given by

$\mu(x,y)=\exp(\widetilde{\ell}_{F}(x)y).$

This can be expanded as a power series in $x$:

$\mu(x,y)=\sum_{n\geq 0}\beta_{n}(y)x^{n}.$

Unwinding the definition of the Cartier dual, and using that $R[\\![t]\\!]$ is
$p$-torsionfree (using our assumption that $R$ is a $p$-completely flat
$\mathbf{Z}_{p}$-algebra), we see that the ring of functions on
$\hat{\mathbf{G}}_{t}^{\vee}$ has underlying $R[\\![t]\\!]$-module given by
(the $(p,t)$-adic completion of)

$\mathscr{O}_{\hat{\mathbf{G}}_{t}^{\vee}}=R[\\![t]\\!]\\{\beta_{n}(y)\\}_{n\geq
0}.$

######  Example 4.5.3.

When $F$ is the multiplicative formal group law, the function $\mu$ is simply

$\mu(x,y)=\exp\left(\frac{y}{q-1}\mathrm{log}(1+(q-1)x)\right)=(1+(q-1)x)^{y/(q-1)};$

its power series expansion is given by

$\mu(x,y)=\sum_{n\geq 0}\frac{\prod_{j=0}^{n-1}(y-j(q-1))}{n!}x^{n}.$

This expression plays an important role in [Dri21].

######  Example 4.5.4.

When $F$ is the hyperbolic formal group law (so that
$\ell_{F}(x)=\tanh^{-1}(x)=\frac{1}{2}\mathrm{log}\left(\frac{1+x}{1-x}\right)$),
the function $\mu$ is

$\mu(x,y)=\exp\left(\frac{y}{2(q-1)}\mathrm{log}\left(\frac{1+(q-1)x}{1-(q-1)x}\right)\right)=\left(\frac{1+(q-1)x}{1-(q-1)x}\right)^{y/2(q-1)}.$

The power series expansion of this function is somewhat complicated: one can
show that upon writing $\mu(x,y)=\sum_{n\geq 0}\beta_{n}(y)x^{n}$, we have
$\beta_{0}=1$, $\beta_{1}=y$, and there is a recurrence

$\beta_{n+2}(y)=\frac{y\beta_{n+1}(y)+n(q-1)^{2}\beta_{n}(y)}{n+2}.$

Using a computer, one can compute that the first few terms of this expansion
are

$\displaystyle\mu(x,y)$
$\displaystyle=1+yx+\frac{y^{2}}{2}x^{2}+\frac{2(q-1)^{2}y+y^{3}}{3!}x^{3}+\frac{8(q-1)^{2}y^{2}+y^{4}}{4!}x^{4}$
$\displaystyle+\frac{24(q-1)^{4}y+20(q-1)^{2}y^{3}+y^{5}}{5!}x^{5}+\cdots.$

Observe that $\beta_{n}(y)\equiv\frac{y^{n}}{n!}\pmod{2}$, reflecting the fact
that the base-change of the hyperbolic formal group to $\mathbf{F}_{2}$ is
isomorphic to the additive formal group.

######  Definition 4.5.5.

Let $\mathbf{G}_{m}^{\sharp,F}$ denote the formal scheme over
$\operatorname{Spf}R[\\![t]\\!]$ given by (the $(p,t)$-adic completion of)

$\mathbf{G}_{m}^{\sharp,F}=\operatorname{Spf}R[\\![t]\\!]\left[y^{\pm
1},\frac{(y-1)^{n}_{s}}{n!_{F}}\right]_{n\geq 0}.$

This can be viewed as the “$F$-divided power hull” of the identity section of
$(\mathbf{G}_{m})_{R[\\![t]\\!]}$. Equip $\mathbf{G}_{m}^{\sharp,F}$ with the
structure of a group scheme where the coproduct sends $y\mapsto y\otimes y$.
It is not immediate that this is well-defined, but we will prove this below in
Corollary 4.5.9. There is a canonical homomorphism
$\mathrm{can}:\mathbf{G}_{m}^{\sharp,F}\to(\mathbf{G}_{m})_{R[\\![t]\\!]}$.

Note that Remark 4.3.17 implies that $F\mathrm{log}(y)$ defines an element of
the coordinate ring of $\mathbf{G}_{m}^{\sharp,F}$, i.e., it defines a map
$F\mathrm{log}:\mathbf{G}_{m}^{\sharp,F}\to(\mathbf{G}_{a})_{R[\\![t]\\!]}$.
This is in fact a homomorphism, since
$F\mathrm{log}(y_{1}y_{2})=F\mathrm{log}(y_{1})+F\mathrm{log}(y_{2})$.

######  Proposition 4.5.6.

Work over the base $(R\otimes\mathbf{Q})[\\![t]\\!]$. Then, the iterated
$F$-derivative of $\mu(x,F\mathrm{log}(y))$ with respect to the variable $y$
is given by

(8)
$\nabla_{F,y}^{n}\mu(x,F\mathrm{log}(y))=\frac{x(x+_{\widetilde{F}}\langle{-1}\rangle_{F}(t))\cdots(x+_{\widetilde{F}}\langle{-n+1}\rangle_{F}(t))}{y^{n}}\mu(x,F\mathrm{log}(y)).$

###### Proof.

Observe that:

$\displaystyle\mu(x,F\mathrm{log}(y))$ $\displaystyle=\sum_{n\geq
0}\beta_{n}(F\mathrm{log}(y))x^{n}=\exp(F\mathrm{log}(y)\widetilde{\ell}_{F}(x))$
$\displaystyle=\exp\left(\frac{t}{\ell_{F}(t)}\mathrm{log}(y)\cdot\frac{\ell_{F}(tx)}{t}\right)=\exp\left(\mathrm{log}(y)\frac{\ell_{F}(tx)}{\ell_{F}(t)}\right)=y^{\frac{\ell_{F}(tx)}{\ell_{F}(t)}};$

the third equality used the definition of $F\mathrm{log}(y)$ via Corollary
4.3.15 and the definition of $\widetilde{\ell}_{F}(x)$ via Remark 4.3.3. One
can deduce 8 from this; let us illustrate this rather inefficiently. Let us
write $a=\frac{\ell_{F}(tx)}{\ell_{F}(t)}$ for notational simplicity, so that

$y^{a}=\sum_{m\geq 0}\frac{a(a-1)\cdots(a-(m-1))}{m!}(y-1)^{m},$

and $\partial_{y}y^{a}=ay^{a-1}$.

We can now inductively compute the iterated $F$-derivative using Proposition
4.3.9. We begin with the base case $n=1$. Note that
$(y\partial_{y})y^{a}=ay^{a}$, so that

(9) $(y\partial_{y})^{m}y^{a}=a^{m}y^{a}$

by an easy induction on $m$. Write $\mathscr{E}_{F}(z)=\sum_{m\geq
0}b_{m}z^{m}$; then using 9, we have:

$\displaystyle\nabla_{F,y}\mu(x,F\mathrm{log}(y))$
$\displaystyle=\frac{1}{yt}\sum_{m\geq
0}b_{m}\ell_{F}(t)^{m}(y\partial_{y})^{m}y^{a}=\frac{1}{yt}\sum_{m\geq