· HYPERDERIVATIVE POWER SUMS, VANDERMONDE MATRICES, AND CARLITZ MULTIPLICATION COEFFICIENTS...

HYPERDERIVATIVE POWER SUMS, VANDERMONDE MATRICES,AND CARLITZ MULTIPLICATION COEFFICIENTS

MATTHEW A. PAPANIKOLAS

In honor and memory of David Goss

Abstract. We investigate interconnected aspects of hyperderivatives of polynomialsover finite fields, q-th powers of polynomials, and specializations of Vandermonde matri-ces. We construct formulas for Carlitz multiplication coefficients using hyperderivativesand symmetric polynomials, and we prove identities for hyperderivative power sums interms of specializations of the inverse of the Vandermonde matrix. As an applicationof these results we give a new proof of a theorem of Thakur on explicit formulas forAnderson’s special polynomials for log-algebraicity on the Carlitz module. Furthermore,by combining results of Pellarin and Perkins with these techniques, we obtain a newproof of Anderson’s theorem in the general case.

1. Introduction

Letting A = Fq[θ] be the polynomial ring in one variable over a finite field and K =Fq(θ) be its fraction field, it is natural to consider polynomial power sums in K,

(1.1) Si(k) =∑a∈Ai+

ak, k ∈ Z,

where Ai+ denotes the finite set of monic polynomials in A of degree i. Vanishing resultsfor these series and special value formulas were obtained early on by Carlitz [9], [10], andLee [25], and they have been studied by many researchers in intervening years for boththeir intrinsic interest and their applications to values of Goss L-functions, multizetavalues, and Anderson’s log-algebraic identities (e.g., see [3], [15], [17], [38], [40], [41]).

As a variant on these types of problems the present paper considers hyperderivativepower sums of the form

(1.2) Hi(j1, . . . , j`; k1, . . . , k`) =∑a∈Ai+

∂j1θ (a)k1 · · · ∂j`θ (a)k` ,

for j1, . . . , j` > 0 and k1, . . . , k` ∈ Z, where ∂jθ(a)k is the k-th power of the j-th hy-perderivative of a with respect to θ (see §2 for the definitions). Hyperderivatives havebecome increasingly important in research in function field arithmetic in recent years(e.g., see [6], [7], [13], [20], [22], [23], [26], [32]). These sums appear in log-algebraicidentities for the Carlitz module (see §6–7), as well as for its tensor powers [31].

Date: July 26, 2020.2010 Mathematics Subject Classification. Primary 11G09; Secondary 05E05, 11M38, 11T55.Key words and phrases. hyperderivatives, polynomial power sums, Carlitz module, symmetric poly-

nomials, log-algebraicity.This project was partially supported by NSF Grant DMS-1501362.

1

2 MATTHEW A. PAPANIKOLAS

In order to investigate these sums and their applications, we first observe that there issubstantial intertwining among q-th powers of polynomials, hyperderivatives, and sym-metric polynomials. For example, in §4 we show that for a ∈ A of degree at most i, if wewrite a(t) for the polynomial obtained by replacing θ by another variable t, then

(1.3)

a(t)a(t)q

...

a(t)qi

= Vi(t− θ, tq − θ, . . . , tqi − θ

)a

∂1θ (a)...

∂iθ(a)

,

where Vi is the Vandermonde matrix in i+ 1 variables. Using this identity, we show thatfor 0 6 j 6 i and k > 0,

(1.4) ∂jθ(a)qk

=i∑

`=0

κij`(t− θqk , tq − θqk , . . . , tqi − θqk

)a(t)q

`

,

where κij` is the entry of V −1i in row j and column ` and is completely explicit (see (4.5)).See Proposition 4.6 for more details. Somewhat surprisingly, as the left-hand side involvesonly θ, the expression on the right is independent of t, and this leads to the followingresult on certain hyperderivative power sums.

Theorem 5.5. Let i > 1, and let 1 6 s 6 q − 1. Then for any µ1, . . . , µs > 0 and any0 6 j1, . . . , js 6 i,∑

a∈Ai+

∂j1θ (a)qµ1 · · · ∂jsθ (a)q

µs

a

=1

Li

s∏r=1

(−1)i−jrei,i−jr(θ − θqµr , θq − θqµr , . . . , θqi−1 − θqµr

).

Here Li = (θ− θq)(θ− θq2) · · · (θ− θqi) ∈ A, and eij denotes the elementary symmetricpolynomial in i variables of degree j.

As pointed out by the referee, one can apply the techniques of Pellarin and Perkins [34]to obtain formulas of the type in Theorem 5.5, without the restriction on s. We explorethe implications of their results to log-algebraic identities in §7.

To demonstrate the connections between hyperderivative power sums and Anderson’slog-algebraicity results, we recall that the Carlitz module C is the A-module structureplaced on any A-algebra R by setting

Cθ(x) = θx+ xq, x ∈ R.As such for any a ∈ A, the Carlitz operation of a on R is given by a polynomial,

Ca(x) =

deg a∑k=0

{a

k

}xq

k ∈ A[x],

where{a0

}= a and

{a

deg a

}∈ Fq is the leading coefficient of a. Carlitz [9] showed that

(1.5)

{a

k

}=

k∑j=0

aqj

DjLqj

k−j

,

HYPERDERIVATIVE POWER SUMS AND CARLITZ MULTIPLICATION COEFFICIENTS 3

where Lk is defined in the previous paragraph, and Dj = (θqj−θ)(θqj−θq) · · · (θqj−θqj−1

).However, by investigating the interplay of the definition of

{ak

}and hyperderivatives, we

prove a new formula for{ak

}in Proposition 3.6,

(1.6)

{a

k

}=

deg a∑j=k

∂jθ(a) · hk,j−k(θq − θ, θq2 − θ, . . . , θqk − θ),

where hkj represents the complete homogeneous symmetric polynomial in k variables ofdegree j (see §2 for details).

In [2], Anderson proved the following power series identity that he termed a log-algebraicity result. For new variables x and z and for β(x) ∈ A[x], we set

(1.7) P(β, z) ..= expC

∑a∈A+

β(Ca(x))

azq

deg a

∈ K[x][[z]],

where expC(z) =∑

i>0 zqi/Di is the Carlitz exponential (see §2).

Theorem 6.1 (Anderson [2, Thm. 3, Prop. 8]). The power series P(β, z) ∈ K[x][[z]] isin fact a polynomial. Moreover,

P(β, z) ∈ A[x, z].

Anderson then used these identities to show that special values at s = 1 of Goss L-functions L(χ, s) for Dirichlet characters could be expressed in a direct way as K-linearcombinations of Carlitz logarithms of values of special polynomials,

Pm(x, z) ..= P(xm, z), m > 0,

at Carlitz torsion points. In [40, §8.10], Thakur discovered exact identities for An-derson’s special polynomials, using Carlitz’s formula (1.5) and formulas for Si(k) (seeTheorem 5.3). His result can be restated in terms of symmetric polynomials as follows.

Theorem 6.3 (Thakur [40, §8.10]).

(a) Let m = qµ for µ > 0. Then

Pqµ(x, z) =

µ∑d=0

(−1)µ−d · Ceµ,µ−d(θ,...,θqµ−1)

(Cθd(x) · z

).

(b) Suppose m is of the form m = qµ1 + · · ·+ qµs for µj > 0 and 1 6 s 6 q− 1. Then

Pm(x, z) =

µ1∑d1=0

· · ·µs∑ds=0

(−1)∑µj−

∑dj · C∏s

r=1 eµr,µr−dr (θ,...,θqµr−1

)

((s∏r=1

Cθdr (x)

)· z

).

In §6 of the present paper we use (1.6) on relating{ak

}to hyperderivatives and The-

orem 5.5 on hyperderivative power sums to devise a new proof of Thakur’s theorem. Inaddition to these ingredients the proof relies heavily on several identities for symmetricpolynomials.

The outline of the paper is as follows. After laying out preliminaries on the Carlitzmodule, hyperderivatives, and symmetric polynomials in §2, we use these objects toconstruct formulas for Carlitz multiplication coefficients

{ak

}in §3. In §4 we investigate

the connections between hyperderivatives and Vandermonde matrices as well as recall


connections with a theorem of Voloch [42]. In §5 we apply the previous techniques toprove formulas for hyperderivative power sums, and then we bring all of these resultstogether in §6.3 to give a new proof of Thakur’s theorem. Finally, in §7 we investigateformulas of Pellarin and Perkins and provide a new proof of Anderson’s theorem.

Acknowledgments. It is with great pleasure that I dedicate this paper to my friendand mentor David Goss. After discussing with him some years ago part of the materialof what would eventually be this paper, David wrote a blog post [19] that outlined theaspects he found most intriguing and included further insights in [20]. I would like tothank him for his immense contributions to function field arithmetic and for the interestand enthusiasm he took to my work throughout my career.

I would like to thank the referee, who made several invaluable suggestions that im-proved the scope of this paper. In particular, the referee pointed out the connectionsbetween Proposition 3.6 and previous work of Jeong [22], and the referee apprised us offormulas of Pellarin and Perkins [34] that would facilitate using the techniques of thispaper to obtain a new proof of Theorem 6.1. I also thank O. Gezmis for helping comparethe contents of [14], [34].

2. Preliminaries

For q a fixed power of a prime p, let A = Fq[θ] be a polynomial ring in one variableover the finite field with q elements, and let K = Fq(θ) be its fraction field. We takeK∞ = Fq((1/θ)) for the completion of K at its infinite place, and we take K for thecompletion of an algebraic closure of K∞. We denote the set of monic elements of A byA+, and for each i > 0, we set

(2.1) Ai+ ..= {a ∈ A+ | deg a = i}.

For i > 0, we define the polynomials [i] = θqi − θ ∈ A, we set L0 = D0 = 1, and we set

(2.2) Di..= [i][i− 1]q · · · [1]q

i−1

, Li ..= (−1)i[i][i− 1] · · · [1], i > 1.

Letting τ : K→ K be the q-th power Frobenius endomorphism, the Carlitz module isthe Drinfeld module C : A→ A[τ ] defined by

Cθ = θ + τ.

The ring A[τ ] is the ring of twisted polynomials in τ with coefficients in A. For a ∈ Aand k ∈ Z+, we define

{ak

}∈ A by

(2.3) Ca =

deg a∑k=0

{a

k

}τ k.

As usual C defines an A-module structure on any A-algebra by way of the commutativepolynomials for a ∈ A,

Ca(x) =

deg a∑k=0

{a

k

}xq

k ∈ A[x].

The Carlitz exponential and logarithm are defined by the infinite series,

expC(z) =∑i>0

zqi

Di

, logC(z) =∑i>0

zqi

Li.


They are mutual inverses of each other and for each a ∈ A, we have expC(az) =Ca(expC(z)). For more information about the Carlitz module, and Drinfeld modulesin general, the reader is directed to [18, Ch. 3–4], [40, Ch. 2–3].

For a field F and a variable θ transcendental over F , the hyperdifferential operators withrespect to θ, ∂jθ : F [θ]→ F [θ], j > 0, are defined F -linearly by setting ∂jθ(θ

n) =(nj

)θn−j.

We note that(nj

)= 0 when n < j, and so these maps are well-defined. Hyperderivatives

then extend uniquely to operators ∂jθ : F (θ)sepv → F (θ)sepv on the separable closure of thecompletion of F (θ) at a place v (see [13, §4], [23, §2]). Hyperderivatives satisfy severalkinds of differentiation rules, such as the product rule,

(2.4) ∂jθ(fg) =

j∑k=0

∂kθ (f)∂j−kθ (f), f, g ∈ F (θ)sepv ,

and the composition rule,

(2.5) ∂jθ(∂kθ (f)) = ∂kθ (∂jθ(f)) =

(j + k

j

)∂j+kθ (f), f ∈ F (θ)sepv .

For various versions of the product rule, quotient rule, power rule, and chain rule, thereader is directed to [23, §2], [31, §2.3].

For a sequence of independent variables θ1, . . . , θm, we can define compatible partialhyperderivatives,

∂jθi : F (θ1, . . . , θm)→ F (θ1, . . . , θm),

in the natural way with the property that for i 6= i′ we have ∂jθi ◦ ∂j′

θi′= ∂j

′

θi′◦ ∂jθi (see [28,

Ch. 2]). Mostly we will focus on the case of two variables, say θ and t, and for functionsf ∈ F (θ, t), we say that f is regular at t = θ if f |t=θ ..= f(θ, θ) is well-defined in F (θ).For f regular at t = θ, it follows from the quotient rule that ∂jt (f) is also regular at t = θfor j > 0. The following standard Taylor series lemma will be used throughout.

Lemma 2.6. For a field F , let f ∈ F (θ, t) be regular at t = θ. Then as an element ofF (θ)[[t− θ]],

f(θ, t) =∞∑j=0

∂jt (f)∣∣t=θ· (t− θ)j.

We further recall definitions of symmetric polynomials. For independent variables t,x1, x2, . . . over Z, the elementary symmetric polynomials eij ∈ Z[x1, . . . , xi] are definedby

(2.7)i∑

j=0

eij(x1, . . . , xi)tj = (1 + x1t)(1 + x2t) · · · (1 + xit),

and the complete homogeneous symmetric polynomials hij ∈ Z[x1, . . . , xi] are defined by

(2.8)∞∑j=0

hij(x1, . . . , xi)tj =

1

(1− x1t)(1− x2t) · · · (1− xit).

It is readily apparent that ei0 = hi0 = 1 and ei1 = hi1 = x1 + · · · + xi for all i > 0. Thepolynomial hij consists of the sum of all monomials in x1, . . . , xi of total degree j, so for


example,

h22 = x21 + x1x2 + x22, h23 = x31 + x21x2 + x1x22 + x32.

By convention we extend eij and hij to all j ∈ Z by setting

(j < 0 or j > i) ⇒ eij = 0,

j < 0 ⇒ hij = 0.

In this notation eij (resp. hij) represents the elementary symmetric polynomial (resp. com-plete homogenous symmetric polynomial) of degree j in i variables. For more detailedinformation on symmetric polynomials, see [37, Ch. 7].

The polynomials eij and hij satisfy several standard recurrence relations through rela-tions on their generating functions. The first is the pair of recursions,

eij(x1, . . . , xi) = ei−1,j(x1, . . . , x`, . . . , xi) + x`ei−1,j−1(x1, . . . , x`, . . . , xi)(2.9)

hij(x1, . . . , xi) = hi−1,j(x1, . . . , x`, . . . , xi) + x`hi,j−1(x1, . . . , xi),(2.10)

where ‘x`’ indicates that the variable x` is omitted. These imply for i > 1 and j > 0,

eij(x1, . . . , xi)|x`=0 = ei−1,j(x1, . . . , x`, . . . , xi),(2.11)

hij(x1, . . . , xi)|x`=0 = hi−1,j(x1, . . . , x`, . . . , xi).(2.12)

A second type of recursive relation holds for eij and hij in tandem. See also [29, Pf. ofThm. 3.2] and [37, §7.6].

Proposition 2.13. For fixed i > 1 and 1 6 k 6 i,

i∑j=k

(−1)i−jei−1,i−j(x1, . . . , xi−1) · hk,j−k(x1, . . . , xk) =

{1 if k = i,

0 if k < i.

Proof. One easily checks the case k = i from the definitions of eij and hij. Thus we cannow assume that k 6 i− 1. From (2.7) and (2.8), we find

i−1∑j=0

∞∑`=0

(−1)jei−1,j(x1, . . . , xi−1) · hk`(x1, . . . , xk)tj+` = (1− xk+1t) · · · (1− xi−1t),

and by reversing the order of the outer sum we have

i∑j=1

∞∑`=0

(−1)i−jei−1,i−j(x1, . . . , xi−1) · hk`(x1, . . . , xk)ti−j+` = (1− xk+1t) · · · (1− xi−1t).

The right-hand side has degree i − k − 1 in t, and so the coefficient of ti−k on the rightis 0. On the other hand, ti−k appears in the left-hand side precisely when ` = j − k,which implies

i∑j=1

(−1)i−jei−1,i−j(x1, . . . , xi−1) · hk,j−k(x1, . . . , xk) = 0.

Since again hk,j−k = 0 when j < k, the result follows. �


Remark 2.14. One useful way of expressing Proposition 2.13 is the following. For d > 1,define lower triangular d× d matrices with entries in Z[x1, . . . , xd−1],

Ed =

1 0 0 · · · 0−e11 1 0 · · · 0e22 −e21 1 · · · 0...

.... . . . . .

...(−1)d−1ed−1,d−1 (−1)d−2ed−1,d−2 · · · −ed−1,1 1

,

Hd =

1 0 0 · · · 0h11 1 0 · · · 0h12 h21 1 · · · 0...

.... . . . . .

...h1,d−1 h2,d−2 · · · hd−1,1 1

.

Then Proposition 2.13 is equivalent to the identity, for i 6 d,

Ed ·Hd = Id,

where Id is the d × d identity matrix. The product Hd · Ed = Id produces a companionformula for Proposition 2.13, which we do not state here but arises in the proof ofProposition 3.10.

These matrices also appear in calculations of the LDU decomposition of the Van-dermonde matrix in x1, . . . , xi (see [29], [30]). In the next section we will use them toderive new formulas for the coefficients

{ak

}of the Carlitz multiplication polynomials Ca

from (2.3).

3. Carlitz multiplication coefficients

Let M = K[t], where t is a variable independent from θ ∈ K. Then we can definea left K[τ ]-module structure on M by setting τm ..= (t − θ)m(1) for m ∈ M . As a leftK[t, τ ]-module, M then has the structure of an Anderson t-motive, in the sense of [1],which is isomorphic to the t-motive of the Carlitz module (see [8, §4.3], [18, §5.8]). Nowdefine polynomials µk(t) ∈ A[t] for k > 0 by setting µ0 = 1, and

(3.1) µk(t) ..= (t− θ)(t− θq) · · · (t− θqk−1

), k > 1.

As degt µk = k, we see that {µk} forms a K-basis for K[t]. Then the following propositionis due to Thakur, based on previous work of Drinfeld and Mumford [27].

Proposition 3.2 (Thakur [39, §0.3.5]). For a = a(θ) ∈ A, the expansion of a(t) ∈ K[t]in terms of the basis {µk} is given by

a(t) =

deg a∑k=0

{a

k

}µk(t).

That is, the Carlitz multiplication coefficients of Ca from (2.3) are also the coefficientsof a(t) in terms of the our K-basis on the t-motive M . See [18, §7.11], [21, §3], [39, §0.3],for additional discussion and details.


In this section we will use Proposition 3.2 to derive a new formula for{ak

}in terms of

hyperderivatives and symmetric polynomials. We first observe from (2.7) that for k > 1,

µk(t) = (t− θ)(t− θ − [1]) · · · (t− θ − [k − 1])(3.3)

=k∑j=1

(−1)k−jek−1,k−j([1], . . . , [k − 1])(t− θ)j.

For d > 0, we take Nd ∈ Matd+1(Z[x1, . . . , xd−1]) to be

Nd = Nd(x1, . . . , xd−1) =

(1 00 Ed

),

where Ed is defined in Remark 2.14, and then (3.3) implies

(3.4)

µ0(t)µ1(t)µ2(t)

...µd(t)

= Nd([1], . . . , [d− 1])

1

t− θ(t− θ)2

...(t− θ)d

.

Now let a ∈ A have degree d in θ. From Lemma 2.6, it follows that

a(t) =d∑j=0

∂jθ(a)(t− θ)j.

Therefore, (3.4) implies that

a(t) =(∂0θ (a), . . . , ∂dθ (a)

)1

t− θ...

(t− θ)d

(3.5)

=(∂0θ (a), . . . , ∂dθ (a)

)Nd([1], . . . , [d− 1])−1

µ0(t)µ1(t)

...µd(t)

.

Now by Remark 2.14, we see that

N−1d =

(1 00 Hd

),

and by comparing entries of Hd with Proposition 3.2 and (3.5), we have proved thefollowing proposition.

Proposition 3.6. Let a ∈ A have degree d > 0. Then for 0 6 k 6 d,{a

k

}=

d∑j=k

∂jθ(a) · hk,j−k([1], . . . , [k]).

Remark 3.7. The proof given above works only for k > 1, but the formula in the propo-sition is valid also when k = 0. In this case

{a0

}= a, whereas the definition of hij implies

that h0j = 1 if j = 0 and that h0j = 0 if j > 0, and the formula holds.


Remark 3.8. It is worth comparing the formula in Proposition 3.6 with other formulasfor{ak

}. For example, there is the formula due to Carlitz [9, Thm. 2.1] that

(3.9) Ek(z) ..=k∑j=0

zqj

DjLqj

k−j

∈ K[z] ⇒{a

k

}= Ek(a).

By its definition as a coefficient of the polynomial Ca ∈ A[τ ], we know that{ak

}∈ A, but

in contrast to Proposition 3.6 this is not particularly clear from Carlitz’s formula on itsown. Jeong [22] obtained additional formulas for

{ak

}through hyperdifferential identities

involving Ek(z). See Proposition 3.11 and the subsequent discussion for connections withProposition 3.6.

In [40, §8.10], Thakur uses (3.9) to give explicit formulas for special polynomials fromAnderson’s log-algebraicity theorem for the Carlitz module [2, Prop. 8]. In §6, we willreformulate Thakur’s results (see Theorem 6.3) and use Proposition 3.6 to design a newproof.

We can also write{ak

}in terms of symmetric polynomials in other ways that we will

need. The following proposition provides a different formula for{θm

k

}.

Proposition 3.10. Let m > 0 and 0 6 k 6 m. Then{θm

k

}= hk+1,m−k

(θ, θq, . . . , θq

k).

Proof. The essential argument is to expand tm in terms of the polynomials µk(t) and useProposition 3.2. To do this we consider

m∑k=0

hk+1,m−k(θ, . . . , θq

k)µk(t) =

m∑k=0

hk+1,m−k(θ, . . . , θq

k)(t− θ

)· · ·(t− θqk−1)

=m∑k=0

hk+1,m−k(θ, . . . , θq

k)×

k∑j=0

(−1)k−jek,k−j(θ, . . . , θq

k−1)tj

=m∑j=0

tjm∑k=j

(−1)k−jhk+1,m−k · ek,k−j.

Now the inner sum is the same as the entry in row m+ 1 and column j+ 1 in the matrixproduct Hd · Ed = Id from Remark 2.14 (with any d > m). Thus the inner sum is 1 ifj = m and 0 otherwise, so

m∑k=0

hk+1,m−k(θ, . . . , θq

k)µk(t) = hm+1,0

(θ, . . . , θq

m) · em,0(θ, . . . , θqm−1) · tm = tm,

and the result follows from Proposition 3.2. �

As was pointed out by the referee, Proposition 3.6 is similar to work of Jeong [22,Cor. 1], and below we give another proof of Proposition 3.6 that follows from a combina-tion of Jeong’s results and Proposition 3.10. The proof below requires also Lemma 6.6


from later in the paper, whose proof is independent of any of these considerations. How-ever, for the sake of exposition we have left Lemma 6.6 where it was in the original versionof this paper, and we refer the reader to its proof in §6. Recalling Ek(z) from (3.9), Jeongproved the following.

Proposition 3.11 (Jeong [22, Cor. 1]). For j, k > 0, set Bj,k..=∑j

i=0(−1)j−i∂iθ(θj)Ek(θ

i).Then for u ∈ Fq[[θ]] we have

Ek(u) =∞∑j=0

Bj,k∂jθ(u).

Second proof of Proposition 3.6. Because{ak

}= Ek(a) by (3.9) and since hij = 0 for

j < 0, it suffices by Proposition 3.11 to show that Bj,k = hk,j−k([1], . . . , [k]) for k 6 j 6 d.Combining the definition of Bj,k with (3.9), we see

Bj,k =

j∑i=k

(−1)j−i(j

i

)θj−i

{θi

k

}=

j∑i=k

(−1)j−i(j

i

)θj−ihk+1,i−k

(θ, θq, . . . , θq

k),

where the second equality follows from Proposition 3.10. First, taking i ← i + k toreindex the sum and then using d← k + 1, j ← i, k ← j − k in Lemma 6.6(b), we find

Bj,k =

j−k∑i=0

(−1)j−i−k(

j

i+ k

)θj−i−khk+1,i

(θ, θq, . . . , θq

k)= (−1)j−khk+1,j−k

(T − θ, . . . , T − θqk

)∣∣T=θ

.

Finally from (2.12) we see that Bj,k = hk,j−k([1], . . . , [k]). �

Remark 3.12. It is possible to extend Propositions 3.6 and 3.10 to multiplication coeffi-cients of tensor powers of the Carlitz module, and the reader is directed to [31, Ch. 3]for further details.

4. Vandermonde matrices and a theorem of Voloch

In [42], Voloch proved the following proposition that relates q-th powers of power seriesover Fq to their hyperderivatives. Voloch’s proof is straightforward, but below we giveanother proof that then lends itself well to generalizations for our purposes.

Proposition 4.1 (Voloch [42]). Let g ∈ Fq[[θ]]. Then for k > 0,

gqk

=∞∑j=0

∂jθ(g) · [k]j,

where by convention we set [0]0 = 1.

Proof. By substituting θ ← t into g, the definition of hyperderivatives yields

g(t) =∞∑j=0

∂jθ(g)(t− θ)j.

Then for k > 0,

g(t)qk

= g(tqk)

=∞∑j=0

∂jθ(g)(tqk − θ)j.


Upon substituting t← θ, we obtain the result. �

Fixing i > 0, for variables x0, . . . , xi we define the (i+1)×(i+1) Vandermonde matrix,

(4.2) Vi(x0, . . . , xi) =

1 x0 · · · xi01 x1 · · · xi1...

......

1 xi · · · xii

.

For a polynomial a ∈ A of degree at most i, as in the proof of Proposition 4.1, we haveidentities for k > 0,

a(t)qk

= a(tqk)

=i∑

j=0

∂jθ(a)(tqk − θ

)j.

This yields a matrix identity

(4.3)

a(t)a(t)q

...

a(t)qi

= Vi(t− θ, tq − θ, . . . , tqi − θ

)a

∂1θ (a)...

∂iθ(a)

.

Inverting the Vandermonde matrix, we have

(4.4)

a

∂1θ (a)...

∂iθ(a)

= Vi(t− θ, tq − θ, . . . , tqi − θ

)−1a(t)a(t)q

...

a(t)qi

.

Notably the left-hand side involves only the variable θ, implying that the expression onthe right-hand side is independent of t. The inverse of Vi can be obtained through variousmeans (e.g., see [24, p. 36], [29, Thm. 3.2]), and setting

Vi(x0, . . . , xi)−1 =

(κij`(x0, . . . , xi)

)06j6i, 06`6i

,

the entries κij` can be expressed in terms of elementary symmetric functions,

(4.5) κij`(x0, . . . , xi) =(−1)i−jei,i−j(x0, . . . , x`, . . . , xi)∏

m 6=`(x` − xm).

The following proposition is then immediate.

Proposition 4.6. For i > 0, let a ∈ A satisfy deg a 6 i.

(a) For 0 6 j 6 i,

∂jθ(a) =i∑

`=0

κij`(t− θ, tq − θ, . . . , tqi − θ

)a(t)q

`

.

(b) More generally, for 0 6 j 6 i and k > 0,

∂jθ(a)qk

=i∑

`=0

κij`(t− θqk , tq − θqk , . . . , tqi − θqk

)a(t)q

`

.


Proof. The proof of (a) has already been shown. For (b), we apply a Frobenius twist.

That is, we need only observe that ∂jθ(a)qk, as it is a polynomial in Fq[θ], is obtained

from ∂jθ(a) by replacing θ ← θqk. Thus making the same replacement on the right-hand

side of (a) will yield the desired result. �

Remark 4.7. Of note in part (b) of the proposition is that the degree in t on the right-handside is bounded independently from k.

Remark 4.8. We see from (4.5) that

(4.9) κij`(t− θqk , tq − θqk , . . . , tqi − θqk

)=

(−1)i−jei,i−j(t− θqk , . . . , tq`−1 − θqk , tq`+1 − θqk , . . . , tqi − θqk

)(tq` − t

)· · ·(tq` − tq`−1

)(tq` − tq`+1

)· · ·(tq` − tqi

) ,

which will be important for calculations in §5–6.

5. Hyperderivative power sums

Power sums of polynomials in A are defined for i > 0 and k ∈ Z by setting

(5.1) Si(k) =∑a∈Ai+

ak.

Research on vanishing criteria and precise formulas for these sums goes back at least toCarlitz [9], [10], and Lee [25], and it has been continued by several authors, [3], [15], [17],[36], [40, Ch. 5], [41]. More generally, one can ask for similar results about hyperderivativepower sums of the form

Hi(j1, . . . , j`; k1, . . . , k`) =∑a∈Ai+

∂j1θ (a)k1 · · · ∂j`θ (a)k` ,

for j1, . . . , j` > 0 and k1, . . . , k` ∈ Z. We will explore some results for reasonably simplehyperderivative power sums in this section, but they are studied in more depth in [31,Ch. 5, 8–9].

In this section the following results on Si(k) for k > 0 are fundamental. For k > 0, ifwe write k =

∑sj=0 kjq

j with 0 6 kj 6 q − 1, then we let

σq(k) ..= k0 + · · ·+ ks

be the sum of its digits base q.

Theorem 5.2 (Carlitz [9, Thm. 9.5], [10, p. 497]; Lee [25, Lem. 7.1]; see also Gekeler [15,Cor. 2.12] and Thakur [40, Thm. 5.1.2]). Let i > 0. For k > 0, if σq(k) < i(q − 1), thenSi(k) = 0. Moreover, if k < qi − 1, then Si(k) = 0.

Theorem 5.3 (Carlitz [11, p. 941]; Lee [25, Thm. 4.1]; see also Gekeler [15, Thm. 4.1,Rmk. 6.6]). Let i > 0. Suppose 1 6 s 6 q − 1 and k = q`1 + · · ·+ q`s − 1 > 0.

(a) If `r < i for some r = 1, . . . , s, then Si(k) = 0.(b) If `r > i for all r = 1, . . . , s, then

Si(k) =1

Li

s∏r=1

D`r

Dqi

`r−i

.


(c) In particular,

Si(q`+i − 1) =

D`+i

LiDqi

`

.

The second theorem can be proved especially cleanly by way of results of Angles andPellarin [3], who adapted techniques of Simon [3, Lem. 4]. In particular, Theorem 5.3

can be obtained by specializing the following result at t1 = θq`1 , . . . , ts = θq

`s.

Proposition 5.4 (Angles-Pellarin [3, Prop. 10]). Let i > 0. For 0 6 s 6 q − 1,∑a∈Ai+

a(t1) · · · a(ts)

a=

1

Li

s∏r=1

i−1∏ν=0

(tr − θq

ν).

In this proposition the coefficient of the top degree term in t1, . . . , ts is Si(−1) =∑a∈Ai+ 1/a = 1/Li (see [9, Eq. (9.09)]). The main result of this section is the following.

Theorem 5.5. Let i > 1, and let 1 6 s 6 q − 1. Then for any µ1, . . . , µs > 0 and any0 6 j1, . . . , js 6 i,∑

a∈Ai+

∂j1θ (a)qµ1 · · · ∂jsθ (a)q

µs

a

=1

Li

s∏r=1

(−1)i−jrei,i−jr(θ − θqµr , θq − θqµr , . . . , θqi−1 − θqµr

).

Proof. By Proposition 4.6, we see that∑a∈Ai+

∂j1θ (a)qµ1 · · · ∂jsθ (a)q

µs

a=∑a∈Ai+

1

a·

s∏r=1

i∑`r=0

κijr`r(t− θqµr , . . . , tqi − θqµr

)a(t)q

`r.

If we substitute t = θ, then we have∑a∈Ai+

∂j1θ (a)qµ1 · · · ∂jsθ (a)q

µs

a=∑a∈Ai+

1

a·

s∏r=1

i∑`r=0

κijr`r(θ − θqµr , . . . , θqi − θqµr

)aq

`r

=i∑

`1=0

· · ·i∑

`s=0

(s∏r=1

κijr`r(θ − θqµr , . . . , θqi − θqµr

))×∑a∈Ai+

aq`1+···+q`s−1.

Thus for fixed `1, . . . , `s, each term is multiplied by Si(q`1 +· · ·+q`s−1). By Theorem 5.3,

this sum is 0, unless `1 = · · · = `s = i, in which case

Si(q`1 + · · ·+ q`s − 1) = Si(sq

i − 1) =Dsi

Li.

Therefore,

(5.6)∑a∈Ai+

∂j1θ (a)qµ1 · · · ∂jsθ (a)q

µs

a=Dsi

Li

s∏r=1

κijri(θ − θqµr , . . . , θqi − θqµr

).


Now for 0 6 j 6 i and µ > 0, we see from Remark 4.8 that

κiji(θ − θqµ , . . . , θqi − θqµ

)=

(−1)i−jei,i−j(θ − θqµ , . . . , θqi−1 − θqµ

)Di

,

and the result follows by substituting into (5.6). �

Remark 5.7. As pointed out by the referee, Theorem 5.5 can be obtained from Proposi-tion 5.4 by appropriately taking hyperderivatives with respect to the variables t1, . . . , tsand specializing. Furthermore, Pellarin and Perkins [34] have more general formulas forthe sums in Proposition 5.4, which can be turned into formulas for the hyperderivativesums in Theorem 5.5 that are valid for all s > 1. We explore these ideas in §7.

Example 5.8. We will see applications of Theorem 5.5 to log-algebraicity calculationsin §6, but for the moment there are some interesting cases to observe. For s 6 q − 1, ifwe take j1 = · · · = js = j, with 0 6 j 6 i, and µ1 = · · · = µs = 0, then we have

∑a∈Ai+

∂jθ(a)s

a=

1

Li· (−1)s(i−j)ei,i−j

(0, θq − θ, . . . , θqi−1 − θ

)s=

(−1)s(i−j)

Li· ei−1,i−j([1], [2], . . . , [i− 1])s.

Example 5.9. Similarly, if we take s = 1, 0 6 j 6 i, and µ > 0, then

∑a∈Ai+

∂jθ(a)qµ

a=

(−1)i−j

Li· ei,i−j

(θ − θqµ , θq − θqµ , . . . , θqi−1 − θqµ

).

(a) If µ > i, then we have

∑a∈Ai+

∂jθ(a)qµ

a=

(−1)i−j

Li· ei,i−j

(−[µ],−[µ− 1]q, . . . ,−[µ− i+ 1]q

i−1).

(b) If µ < i, then the simplification, using (2.11), is slightly different with

∑a∈Ai+

∂jθ(a)qµ

a

=(−1)i−j

Li· ei,i−j

(−[µ],−[µ− 1]q, . . . ,−[1]q

µ−1

, 0, [1]qµ

, [2]qµ

, . . . , [i− µ− 1]qµ)

=(−1)i−j

Li· ei−1,i−j

(−[µ],−[µ− 1]q, . . . ,−[1]q

µ−1

, [1]qµ

, [2]qµ

, . . . , [i− µ− 1]qµ).

These cases will arise in the next section on log-algebraicity formulas.

Remark 5.10. More general hyperderivative power sum formulas can be used to analyzelog-algebraicity formulas on tensor powers of the Carlitz module. For more details onsuch formulas, see [31, Ch. 5, 9].


6. Thakur’s method for log-algebraicity on the Carlitz module

In [2], Anderson developed the notion of log-algebraic power series identities for rank 1Drinfeld modules based on previous special cases of Thakur [38]. For the Carlitz module,Anderson’s main theorem was the following. We take x, z, and θ to be independentvariables over Fq.

Theorem 6.1 (Anderson [2, Thm. 3, Prop. 8]). For β(x) ∈ A[x], let

(6.2) P(β, z) ..= expC

∑a∈A+

β(Ca(x))

azq

deg a

∈ K[x][[z]].

Then in fact P(β, z) ∈ A[x, z].

By using the Carlitz A-module operation, it is evident that P(β, z) is A-linear in β, andso the values of P(β, z) are completely determined by Anderson’s special polynomials,

Pm(x, z) ..= P(xm, z), m > 0.

Anderson [2, Prop. 8] derives several properties of special polynomials, including boundsfor their degrees in x, z, and θ. However, Anderson’s proof was indirect, and so exceptin certain cases he does not provide exact formulas for Pm(x, z). In [40, §8.10], Thakurconstructed a new method for proving Anderson’s theorem using the power sum formulasfor Si(k) (Theorem 5.3 and others) and Carlitz’s formulas for

{ak

}in (3.9). One benefit

of Thakur’s method was that he derived exact formulas for Pm(x, z) for a large class ofvalues of m. In this section we restate Thakur’s results using symmetric polynomials andprovide a proof using the techniques of sections §3–5.

Theorem 6.3 (Thakur [40, §8.10]).

(a) Let m = qµ for µ > 0. Then

Pqµ(x, z) =

µ∑d=0

(−1)µ−d · Ceµ,µ−d(θ,...,θqµ−1)

(Cθd(x) · z

).

(b) Suppose m is of the form m = qµ1 + · · ·+ qµs for µj > 0 and 1 6 s 6 q− 1. Then

Pm(x, z) =

µ1∑d1=0

· · ·µs∑ds=0

(−1)∑µj−

∑dj · C∏s

r=1 eµr,µr−dr (θ,...,θqµr−1

)

((s∏r=1

Cθdr (x)

)· z

).

Remark 6.4. (a) Also in [40, §8.10], Thakur gives a proof that Pm(x, z) ∈ K[x, z] forgeneral m using the same methods, though the intricate calculations present difficultiesto conclude that the coefficients are in A. Following suggestions of the referee, we proveTheorem 6.1 and thus that Pm(x, z) ∈ A[x, z] for all m, though at the expense of theexplicitness of Theorem 6.3. (b) Multivariable versions of Theorem 6.1 and parts ofTheorem 6.3 were worked out by Angles, Pellarin, and Tavares Ribeiro [4] based on earlierwork of Pellarin [33], and it would be interesting to investigate their constructions usingthe techniques in this section. (c) Further results on connections between polynomialpower sums and log-algebraicity results can be found in [5], [12], [14], [21], [35], [36].(d) While Thakur’s methods do not readily transfer to the setting of tensor powers of theCarlitz module, generalizations of our proof here of Theorem 6.3 to higher tensor powerswill be the subject of [31, Ch. 9].


Example 6.5. As observed by Thakur [40, Rmk. 8.10.1], we have P1(x, z) = xz,Pq(x, z) = xqz − xqzq, and (for q > 2) Pq+1(x, z) = xq+1z − x2qzq. If we considerm = 2q for q > 2, then Theorem 6.3(b) implies

P2q(x, z) = Ce11(θ)e11(θ)(x2z)− 2Ce11(θ)

(Cθ(x) · xz

)+ Cθ(x)2 · z

= Cθ2(x2z)− 2Cθ

((θx2 + xq+1)z

)+ (θx+ xq)2z

= x2qz −((θq − θ)x2q + 2xq

2+q)zq + x2q

2

zq2

,

which agrees with [2, Eq. (27)]. For other examples using Theorem 6.3, we find for q > 2,

Pq2(x, z) = Cθ2(x) · z − Cθq+θ(Cθ(x) · z

)+ Cθq+1(xz),

Pq2+1(x, z) = Cθ2(x) · xz − Cθq+θ(Cθ(x) · xz

)+ Cθq+1(x2z),

Pq2+q(x, z) = Cθ(x) · Cθ2(x) · z − Cθ(Cθ2(x) · xz

)− Cθq+θ

(Cθ(x)2 · z

)+ Cθq+1+θ2

(Cθ(x) · xz

)+ Cθq+1

(Cθ(x) · xz

)− Cθq+2(x2z),

P2q2(x, z) = Cθ2(x)2 · z − Cθq+θ(Cθ2(x) · Cθ(x) · z

)+ Cθ2q+2θq+1+θ2

(Cθ(x)2 · z

)+ Cθq+1

(Cθ2(x) · z

)− Cθ2q+1+θq+2

(Cθ(x) · xz

)+ Cθ2q+2(xz),

Pq3(x, z) = Cθ3(x) · z − Cθq2+θq+θ(Cθ2(x) · z

)+ Cθq2+q+θq2+1+θq+1

(Cθ(x) · z

)− Cθq2+q+1(xz).

The remainder of this section is devoted to a new proof of Theorem 6.3. We first needsome lemmas on symmetric polynomials.

Lemma 6.6. Let d > 1, and let T be a variable independent from x1, . . . , xd.

(a) For 0 6 k 6 d,

ed,d−k(T − x1, . . . , T − xd) =d∑j=k

(−1)d−j(j

k

)ed,d−j(x1, . . . , xd)T

j−k.

(b) For k > 0,

hd,k(T − x1, . . . , T − xd) =k∑j=0

(−1)j(d+ k − 1

k − j

)hd,j(x1, . . . , xd)T

k−j.

Proof. Both identities are proved by taking hyperderivatives with respect to T . For (a),consider the polynomial

P (T ) = (T − x1) · · · (T − xd) =d∑j=0

(−1)d−jed,d−j(x1, . . . , xd)Tj.

By the product rule (see [23, §2.2] or [31, §2.3])

∂kT (P (x)) =∑

k1,...,kd>0k1+···+kd=k

∂k1T (T − x1) · · · ∂kdT (T − xd)

= ed,d−k(T − x1, . . . , T − xd).


On the other hand,

∂kT (P (x)) =d∑j=k

(−1)d−j(j

k

)ed,d−j(x1, . . . , xd)T

j−k.

For (b) we proceed by a double induction. We first note that the result holds for allk > 0 for h1,k = (T−x1)k by the binomial theorem. Now suppose for some d0 > 2 that theresult holds for all (d, k) in the set {(d, k) | d 6 d0−1, k > 0}. Now hd0,0(x1, . . . , xd) = 1,so the result also holds for (d, k) = (d0, 0), so suppose further that there is some k0 > 1so that the result holds for all (d0, k) with k 6 k0 − 1. If we let xi denote the tuple(x1, . . . , xi), then using (2.10) the induction hypothesis implies

hd0,k0(T − x1, . . . , T − xd) = hd0−1,k0(T − x1, . . . , T − xd0−1)+ (T − xd0)hd0,k0−1(T − x1, . . . , T − xd0)

=

k0∑j=0

(−1)j(d0 + k0 − 2

k0 − j

)hd0−1,j(xd0−1)T

k0−j

+ (T − xd0)k0∑j=1

(−1)j−1(d0 + k0 − 2

k0 − j

)hd0,j−1(xd0)T

k0−j.

Collecting terms and using (2.10) again, we find that

hd0,k0(T − x1, . . . , T − xd) =

k0∑j=0

(−1)j((

d0 + k0 − 2

k0 − j

)+

(d0 + k0 − 2

k0 − j − 1

))× hd0,j(xd0)T k0−j,

and the result follows in the case (d, k) = (d0, k0). �

Now for i > 1 and 1 6 ` 6 k 6 i− 1, define

(6.7) Gi,k,` =i∑

j=k

(−1)i−jei−1,i−j(y1, . . . , y`, x`+1, . . . , xi−1) · hk,j−k(x1, . . . , xk),

where x1, . . . , xi−1, y1, . . . , y` are independent variables over Z. To emphasize the orderof the variables when making substitutions, we will write

Gi,k,` = Gi,k,`(x1, . . . , xk;xk+1, . . . , xi−1; y1, . . . , y`).

Ostensibly these polynomials are in Z[x1, . . . , xi−1, y1, . . . , y`], but in fact they do notcontain the variables x`+1, . . . , xk, as shown in the following lemma.

Lemma 6.8. For i > 1 and 1 6 ` 6 k 6 i− 1,

Gi,k,`(x1, . . . , xk;xk+1, . . . , xi−1; y1, . . . , y`)

= Gi−(k−`),`,`(x1, . . . , x`;xk+1, . . . , xi−1; y1, . . . , y`).

Proof. A short calculation reveals that the right-hand side of this formula is obtainedfrom the left by substituting in x`+1 = · · · = xk = 0, so the proof reduces to showingthat Gi,k,` does not actually contain x`+1, . . . , xk. As the defining expression for Gi,k,` is


symmetric in x`+1, . . . , xk, it suffices to show that it does not contain xk when ` < k.(When ` = k, there is nothing to prove.) Define the following tuples:

y` = (y1, . . . , y`), x`,k = (x`+1, . . . , xk, . . . , xi−1), xj = (x1, . . . , xj).

Then, since hkj(x1, . . . , xk) =∑j

n=0 hk−1,j−n(xk−1)xnk (by (2.10)),

∂

∂xk(Gi,k,`) =

i∑j=k

(−1)i−j(ei−2,i−j−1(y`, x`,k) · hk,j−k(xk)

+ ei−1,i−j(y`, x`+1, . . . , xi−1) ·j−k∑n=0

n · hk−1,j−k−n(xk−1)xn−1k

)

=i∑

j=k

(−1)i−j(ei−2,i−j−1(y`, x`,k)

j−k∑n=0

hk−1,j−k−n(xk−1)xnk

+(ei−2,i−j(y`, x`,k) + xkei−2,i−j−1(y`, x`,k))

×j−k∑n=0

n · hk−1,j−k−n(xk−1)xn−1k

),

after applying (2.9). By rearranging terms and reordering the sums we finally obtain

∂

∂xk(Gi,k,`) =

i−k∑n=0

(i∑

j=n+k

(−1)i−j(ei−2,i−j−1(y`, x`,k)hk−1,j−k−n(xk−1)

+ ei−2,i−j(y`, x`,k)hk−1,j−k−n−1(xk−1)))

(n+ 1)xnk .

The inner sum telescopes leaving only

∂

∂xk(Gi,k,`) = (−1)i−n−kei−2,i−n−k(y`, x`,k) · hk−1,−1(xk−1)

+ ei−2,−1(y`, x`,k) · hk−1,i−k−n(xk−1),

and since both of these terms are zero, as they have terms with negative indices, we seethat (∂/∂xk)(Gi,k,`) ≡ 0 identically. Thus Gi,k,` does not contain the variable xk. �

Proof of Theorem 6.3(a). Although part (a) of Theorem 6.3 is a special case of part (b),once complete the argument for (a) will simplify the argument for (b). For i > 0, we let

(6.9) λi(qµ) =

∑a∈Ai+

Ca(x)qµ

a,

so that

(6.10) Pqµ(x, z) = expC

( ∞∑i=0

λi(qµ)zq

i

).


The major focus of the proof is to find a simplification for λi(qµ). Using Proposition 3.6,

we see that

λi(qµ) =

∑a∈Ai+

1

a

i∑k=0

{a

k

}qµxq

k+µ

=∑a∈Ai+

1

a

i∑k=0

i∑j=k

∂jθ(a)qµ · hk,j−k([1], . . . , [k])q

µ · xqk+µ .

By reordering the sum and applying the calculation from Example 5.9, we see that

λi(qµ) =

i∑k=0

i∑j=k

hk,j−k([1]q

µ

, . . . , [k]qµ) · xqk+µ ∑

a∈Ai+

∂jθ(a)qµ

a(6.11)

=i∑

k=0

i∑j=k

hk,j−k([1]q

µ

, . . . , [k]qµ) · xqk+µ

× (−1)i−j

Li· ei,i−j

(θ − θqµ , θq − θqµ , . . . , θqi−1 − θqµ

).

Just as in Example 5.9, there are the two cases i 6 µ and i > µ. We will consider herethe case i > µ. The case where i 6 µ is similar with the same resulting formula, andfor the sake of space we leave it to the reader. Using the formula in Example 5.9(b), wethen see that

(6.12) λi(qµ) =

i∑k=0

i∑j=k

(−1)i−jei−1,i−j(−[µ], . . . ,−[1]q

µ−1

, [1]qµ

, . . . , [i− µ− 1]qµ)

× hk,j−k([1]q

µ

, . . . , [k]qµ) · xqk+µ

Li.

By Proposition 2.13, the inner sum vanishes when k 6 i − µ − 1, and so using (6.7) wehave

λi(qµ) =

i∑k=i−µ

Gi,k,k−i+µ+1

([k]q

µ

, . . . , [1]qµ

;−[1]qµ−1

, . . . ,−[i− k − 1]qµ+k−i+1

;

− [i− k]qµ+k−i

, . . . ,−[µ])· x

qk+µ

Li.

Lemma 6.8 then implies that if we set `k ..= k − i+ µ+ 1, then

λi(qµ) =

i∑k=i−µ

Gµ+1,`k,`k

([k]q

µ

, . . . , [i− µ]qµ

;−[1]qµ−1

, . . . ,−[i− k − 1]qµ+k−i+1

;

− [i− k]qµ+k−i

, . . . ,−[µ])· x

qk+µ

Li.


=

µ+1∑`=1

Gµ+1,`,`

([i− µ+ `− 1]q

µ

, . . . ,[i− µ]qµ

;−[1]qµ−1

, . . . ,−[µ− `]q` ;

− [µ− `+ 1]q`−1

, . . . ,−[µ])· x

qi+`−1

Li.

Although this looks complicated, the important thing is that the number of variablesin Gµ+1,`,` is bounded independent of i. Moreover, applying the definition of Gµ+1,`,`

from (6.7) and reordering the sum, we find

(6.13) λi(qµ) =

µ+1∑j=1

j∑`=1

(−1)µ+1−jeµ,µ+1−j(−[1]q

µ−1

,−[2]qµ−2

, . . . ,−[µ])

× h`,j−`([i− µ+ `− 1]q

µ

, . . . , [i− µ]qµ) · xqi+`−1

Li.

Now by Proposition 3.10 and Lemma 6.6(b), we see that

h`,j−`([i− µ+ `− 1]q

µ

, . . . , [i− µ]qµ)

= h`,j−`(θq

i+`−1 − θqµ , . . . , θqi − θqµ)

= (−1)j−`j−`∑n=0

(−1)n(

j − 1

j − `− n

)h`,n(θ, θq, . . . , θq

`−1)qi · θ(j−`−n)qµ=

j−`∑n=0

(−1)j−`+n(

j − 1

j − `− n

)θ(j−`−n)q

µ ·{θ`+n−1

`− 1

}qi.

Therefore, after some reordering and reindexing of sums,

j∑`=1

h`,j−`([i− µ+ `− 1]q

µ

, . . . , [i− µ]qµ) · xqi+`−1

Li(6.14)

=

j−1∑n=0

j−n∑`=1

(−1)j−`+n(

j − 1

j − `− n

)θ(j−`−n)q

µ ·{θ`+n−1

`− 1

}qi· x

qi+`−1

Li

=

j−1∑n=0

j−1∑d=n

(−1)j−d−1(

j − 1

j − d− 1

)θ(j−d−1)q

µ ·{

θd

d− n

}qi· x

qi+d−n

Li

=

j−1∑d=0

(−1)j−d−1(j − 1

d

)θ(j−d−1)q

µd∑

n=0

{θd

d− n

}qi·(xq

d−n)qiLi

.

=

j−1∑d=0

(−1)j−d−1(j − 1

d

)θ(j−d−1)q

µ · Cθd(x)qi

Li.


We then substitute (6.14) into (6.13), and find

λi(qµ) =

µ+1∑j=1

eµ,µ+1−j([1]qµ−1

, . . . , [µ])

×j−1∑d=0

(−1)j−d−1(j − 1

d

)θ(j−d−1)q

µ · Cθd(x)qi

Li

(6.15)

=

µ∑j=0

j∑d=0

(−1)j−deµ,µ−j([1]qµ−1

, . . . , [µ])

(j

d

)θ(j−d)q

µ · Cθd(x)qi

Li

=

µ∑d=0

(−1)µ−deµ,µ−d(θq

µ − [1]qµ−1

, θqµ − [2]q

µ−2

, . . . , θqµ − [µ]

)· Cθd(x)q

i

Li

=

µ∑d=0

(−1)µ−deµ,µ−d(θq

µ−1

, θqµ−2

, . . . , θ) · Cθd(x)qi

Li,

where in the third equality we have applied Lemma 6.6(a). From this we see that

(6.16)∞∑i=0

λi(qµ)zq

i

=

µ∑d=0

(−1)µ−deµ,µ−d(θ, θq, . . . , θq

µ−1

) · logC(Cθd(x)

),

which after applying expC(z) to both sides yields the desired result. �

Proof of theorem 6.3(b). The proof of part (b) runs along the same lines as part (a),but with more bookkeeping. Nevertheless, we have carried out much of the difficultcalculations in (a). As in (6.9), we set

λi(m) =∑a∈Ai+

Ca(x)m

a,

so that Pm(x, z) = expC(∑∞

i=0 λi(m)zqi)

. Now since m = qµ1 + · · ·+ qµs , we see that

λi(m) =∑a∈Ai+

(i∑

k1=0

{a

k1

}qµ1xq

k1+µ1

)· · ·

(i∑

ks=0

{a

ks

}qµsxq

ks+µs

)1

a(6.17)

=i∑

k1=0

· · ·i∑

ks=0

(∑a∈Ai+

{a

k1

}qµ1· · ·{a

ks

}qµs· 1

a

)xq

k1+µ1+···+qks+µs .

We apply Proposition 3.6 to the inner sum and find as in (6.11),∑a∈Ai+

{a

k1

}qµ1· · ·{a

ks

}qµs· 1

a(6.18)

=i∑

j1=k1

· · ·i∑

js=ks

hk1,j1−k1([1], . . . , [k1]

)qµ1 · · ·hks,js−ks([1], . . . , [ks])qµs

×∑a∈Ai+

∂j1θ (a)qµ1 · · · ∂jsθ (a)q

µs

a.


Now the value of the final hyperderivative sum has been obtained in Theorem 5.5, andwe can use the methods of Example 5.9 to simplify it as we did in (6.11) and (6.12).Again the cases where i 6 µr are similar with the same resulting formula, but for thepurpose of space we leave the details to the reader. We assume then that i > µr for eachr = 1, . . . , s, and obtain that

λi(m) =i∑

k1=0

· · ·i∑

ks=0

xqk1+µ1+···+qks+µs

Li

(k1∑j1=0

· · ·ks∑js=0

·s∏r=1

(−1)i−jr

× ei−1,i−jr(−[µr], . . . ,−[1]q

µr−1

, [1]qµr, . . . , [i− µr − 1]q

µr )× hkr,jr−kr

([1]q

µr, . . . , [kr]

qµr))

=1

Li

s∏r=1

(i∑

kr=0

i∑jr=kr

(−1)i−jrei−1,i−jr(−[µr], . . . , [i− µr − 1]q

µr )× hkr,jr−kr

([1]q

µr, . . . , [kr]

qµr)· xqkr+µr

).

The inner double sum has already been evaluated in the proof of part (a), startingwith (6.12). Therefore, using (6.15), we obtain

(6.19) λi(m) =1

Li

s∏r=1

µr∑dr=0

(−1)µr−dreµr,µr−dr(θq

µr−1

, θqµr−2

, . . . , θ)· Cθdr (x)q

i

,

and so

(6.20)∞∑i=0

λi(m)zqi

=

µ1∑d1=0

· · ·µs∑ds=0

(−1)∑µr−

∑dr

×s∏r=1

eµr,µr−dr(θ, θq, . . . , θq

µr−1) · logC

((s∏r=1

Cθdr (x)

)· z

),

which yields the desired result after exponentiation. �

7. Proof of Anderson’s theorem

At the suggestion of the referee, in this section we demonstrate how the techniques inthis paper in conjunction with results of Pellarin and Perkins [34] can be used to proveAnderson’s Theorem 6.1. In particular we show that Pm(x, z) = P(xm, z) ∈ A[x, z] forall m > 0.

For s, i > 0, we let

(7.1) Ss,i ..=∑a∈Ai+

a(t1) · · · a(ts)

a

and

(7.2) Fs,i+1..=

i∑j=0

Ss,i.


The proof of Theorem 6.3 hinged on the hyperderivative formulas in Theorem 5.5, whichwere restricted to 1 6 s 6 q − 1. However, as mentioned in Remark 5.7 it is possibleto derive Theorem 5.5 from Proposition 5.4. More specifically, in the notation of thestatement of Theorem 5.5, one finds from Lemma 2.6 that

(7.3)∑a∈Ai+

∂j1θ (a)qµ1 · · · ∂jsθ (a)q

µs

a=(∂j1t1 ◦ · · · ◦ ∂

jsts

)(Ss,i)∣∣t1=θq

µ1 ,...,ts=θqµs .

However, Proposition 5.4 with 1 6 s 6 q − 1, combined with the product rule, yields

(7.4)(∂j1t1 ◦ · · · ◦ ∂

jsts

)(Ss,i)

=1

Li

s∏r=1

(−1)i−jrei,i−jr(θ − t, θq − t, . . . , θqi−1 − t

),

from which Theorem 5.5 follows after substitution.Now the formula in (7.3) holds for any s > 1, and it is the subsequent formula in (7.4)

for 1 6 s 6 q − 1 that allows for the proof of Theorem 6.3. There are formulas for Ss,ithat hold for arbitrary s in [34] that lose some of the precision of Proposition 5.4 but thatcan still be used to derive formulas like (7.4) in these more general cases. Because of thisdiminished precision the resulting identities for Pm(x, z) are not as explicit as in Thakur’sTheorem 6.3, but we retain enough information to prove Anderson’s Theorem 6.1.

Now as in Theorem 6.3, we can express m = qµ1 + · · · + qµs , but since we will allowarbitrary s, it will be simpler to take µ1 = · · · = µs = 0. Thus, m = s throughoutthis section. (This also bypasses the need for the Gi,k,` polynomials of §6.) BecauseTheorem 6.3 already covers the case s 6 q − 1, we will also assume that s > q.

Before presenting the proof of Theorem 6.1, we focus on identities for Ss,i and Fs,i+1

derived from [34]. Indeed [34, Thm. 7] provides a formula for Ss,i when i > b(s−1)/(q−1)cand arbitrary s, but in order to show that Ps(x, z) ∈ A[x, z] we need somewhat morerefined information about the coefficients that make up the identities for Ss,i as well asidentities that are valid for all i > 0. Thus, instead of [34, Thm. 7] we build on identitiesand techniques from the proof of [34, Thm. 2].

Remark 7.5. Demeslay [14, Thm. 3.3.6] has derived remarkable identities for Ss,i for alls > 1, which could shed light on the developments here. However, Demeslay’s resultsare derived from multivariable log-algebraicity identities of Angles, Pellarin, and TavaresRibeiro [4], which themselves are generalizations of Anderson’s Theorem 6.1. Becauseour purpose in this section is to give an independent proof of Theorem 6.1, we do notexplore Demeslay’s results here. For more information about Demeslay’s identities, thereader is also directed to Gezmis [16, §2.1–2].

We review some of the essential notation from Pellarin and Perkins [34], but for expedi-ence we will assume that the reader has some facility with the notation and constructionsthere. When possible we have tried to use their notation, unless we have already definednotation for the same object (e.g., their bi(t) is µi(t) of (3.1) in the present paper).Recalling the polynomials Ei(z) ∈ K[z] from (3.9), for j = j0 + j1q + · · · + juq

u with0 6 jk 6 q − 1, we set

Gj(z) ..= Ej00 · · ·Eju

u ∈ K[z],


with G0 = 1. Since degz(Gj) = j, the polynomials {Gj} form a K-basis of K[z]. For` = (`1, . . . , `ν) ∈ Zν>0, we can uniquely define a sequence {ε`,j} ⊆ K so that

(7.6) E`1 · · ·E`ν =∑j>0

ε`,jGj.

We note that ε`,j = 0 for j > degz(E`1 · · ·E`ν ) = q`1 + · · ·+ q`ν . As necessary we allow νto vary.

Now fix σ ∈ Z with s < σ so that σ = 1 + ρ(q − 1) with ρ > 2, and let i > 0. Asin the proof of [34, Thm. 7], we observe that Ss,i is the coefficient of tis+1 · · · tiσ in Fσ,i+1.Moreover, from [34, §2.1.2], Pellarin and Perkins prove

(7.7) Fσ,i+1 = −blogq(σ)c−1∑

h=0

1

Li+1+h

∑(`1,...,`σ)6i

ε(`1,...,`σ),qi+1+hµ`1(t1) · · ·µ`σ(tσ), i > 0,

where ` 6 i indicates that each entry of ` is at most i. If i > ρ− 1, then as Pellarin andPerkins observe, Fσ,i+1|ti=θqν = 0 for 0 6 ν 6 i− ρ, and so this identity simplifies as

Fσ,i+1 = −blogq(σ)c−1∑

h=0

1

Li+1+h

(7.8)

×∑

i+1−ρ6(`1,...,`σ)6i

ε(`1,...,`σ),qi+1+hµ`1(t1) · · ·µ`σ(tσ), i > ρ− 1.

Our preliminary goal is to build on [34, Lem. 9] to obtain more detailed informationabout the coefficients ε(`1,...,`σ),qk .

To this end we develop a kind of combinatorial game for working with decompositionsof products of the form E`1 · · ·E`ν in (7.6). Let

S ..={

(w0, w1, . . .) ∈ Z∞>0 | ∀ k � 0, wk = 0},

and for a new variable Y , let W be the free A[Y ]-module on S. For w = (wk) ∈ S, wesay w is reduced if wk 6 q − 1 for all k. Otherwise, we let κ = κ(w) denote the smallestindex such that wκ > q. We then define functions u, v : S→ S and m : S→W by

u(w) ..=

{w if w is reduced,

(w0, . . . , wκ−1, wκ − (q − 1), wκ+1, . . .) otherwise,

v(w) ..=

{w if w is reduced,

(w0, . . . , wκ−1, wκ − q, wκ+1 + 1, wκ+2, . . .) otherwise,

m(w) ..=

{w if w is reduced,

u(w) +(Y qκ+1 − θ

)v(w) otherwise.

We extend A[Y ]-linearly to obtain maps u, v, m : W→W. For w = (wk) ∈ S we furtherdefine the weight and size of w to be the non-negative integers

wt(w) =∑k>0

wkqk, sz(w) =

∑k>0

wk.

It is readily apparent that if w is not reduced, then

wt(u(w)) = wt(w)− (q − 1)qκ sz(u(w)) = sz(w)− (q − 1),(7.9)


wt(v(w)) = wt(w) sz(v(w)) = sz(w)− (q − 1).

We note that for B > 0,#{w ∈ S | wt(w) < B} <∞.

For arbitrary x ∈ W, we set wt(x) and sz(x) to be the maximum weight and sizerespectively of the generators from S in the support of x, and sets of bounded weight inW are supported on a finite set of generators in S.

We think of m as being the primary ‘move’ in our game. For any x ∈ W, if we letmj(x) ..= (m ◦ · · · ◦m)(x) be the j-fold iterate of m applied to x, the sequence

m(x),m2(x),m3(x), . . .

is a sequence of bounded weight. Moreover, it eventually stabilizes, since after finitelymany applications of m the size of mj(x) is necessarily strictly less than the size of x butthe weight of mj(x) is no more than the weight of x. Thus eventually each generatorfrom S in the support of mj(x) is reduced, and we let M(x) ∈W denote this stabilizationof this sequence.

Example 7.10. The following calculations hold. We let ‘0’ denote a sequence of infinitelymany 0’s.

M(q, 0) = (1, 0) +(Y q − θ

)(0, 1, 0),

M(2q − 1, 0) = (1, 0) +(Y q − θ

)(0, 1, 0) + (Y q − θ)(q − 1, 1, 0),

M(0, q, 0) = (0, 1, 0) +(Y q2 − θ

)(0, 0, 1, 0),

M(0, 2q − 1, 0) = (0, 1, 0) +(Y q2 − θ

)(0, 0, 1, 0) + (Y q − θ)(0, q − 1, 1, 0),

M(q2, 0) = (1, 0) +((Y q − θ

)+(Y q − θ

)q)(0, 1, 0)

+(Y q − θ

)q(Y q2 − θ

)(0, 0, 1, 0),

M(0, q2, 0) = (0, 1, 0) +((Y q2 − θ

)+(Y q2 − θ

)q)(0, 0, 1, 0)

+(Y q2 − θ

)q(Y q3 − θ

)(0, 0, 0, 1, 0).

For n > 0,

M(qn, 0) = (1, 0) +n∑k=1

( ∑06jk<···<j16n

(Y q − θ

)qj1 · · · (Y qk − θ)qjk)

(0, . . . , 0︸︷︷︸k

, 1, 0).

For w = (wk) ∈ S, we set

(7.11) e(w) ..=∏k>0

Ewkk ,

and we extend it A[Y ]-linearly to a map e : W → K[z] by setting also Y = θ. We notethat if w ∈ S is reduced, then e(w) = Gwt(w). Thus for x =

∑w∈S aww ∈ W, with

aw ∈ A[Y ], if each w in the support of x is reduced, then

(7.12) e(x) =∑w∈S

aw(θ)Gwt(w).

The motivation for this entire construction is to obtain identities for ε`,j (see (7.16))by combining (7.12) with the following lemma, which itself follows directly from [34,Prop. 8(3)].


Lemma 7.13. For any x ∈W, we have

e(x) = e(m(x)) = e(M(x)).

It will be useful to have some additional operators, inspired by the calculations inExample 7.10. We define shift operators α, β : S→ S so that

α(w0, w1, . . .) ..= (0, w0, w1, . . .), β(w0, w1, . . .) ..= (w1, w2, . . .),

and then extend A[Y ]-linearly to α, β : W → W. Certainly, for all n > 0, we haveβn ◦ αn = IdW, and if for w ∈ S, w0 = · · · = wn−1 = 0, then αn ◦ βn(w) = w. We notethat for w ∈ S and n > 0,

(7.14) M(αn(w)) = αn(M(w)

∣∣Y←Y qn

)= αn

(M(w)

)∣∣Y←Y qn .

A word of caution is that this identity does not extend A[Y ]-linearly to all of W, thoughit will not impact us here. If w0 = · · · = wn−1 = 0, then it also follows that

(7.15) M(βn(w)) = βn(M(w)

∣∣Y←Y 1/qn

)= βn

(M(w)

)∣∣Y←Y 1/qn .

Furthermore, we let I ..= ∪σ>1Zσ>0, taken as a disjoint union, and define ξ : I → S bysetting

ξ(`1, . . . , `σ) ..= (ξ0, ξ1, . . .),

where ξk ..= #{i | `i = k}. (Previously we had assumed σ ≡ 1 (mod q − 1), but we donot need to continue with that assumption until later.) Our main computational toolis then the following. For ` ∈ I and for j > 0, expressed as j = j0 + j1q + · · · + juq

u,0 6 ji 6 q − 1, we set

m`,j(Y ) ..= coefficient of (j0, j1, . . .) in M(ξ(`)),

which is an element of A[Y ]. It then follows from (7.12) and Lemma 7.13 that

(7.16) ε`,j = m`,j(θ).

As it turns out, the polynomials m`,j(Y ) are the same as another class of polynomialsc`,j(Y ) ∈ A[Y ] defined in [34].

Lemma 7.17 (Pellarin-Perkins [34, Lem. 9]). Let ` ∈ I. For each j > 0, there existsa polynomial c`,j(Y ) ∈ A[Y ], all but finitely many of which are non-zero, so that for alln > 0,

e(αn(ξ(`))) =∑j>0

c`,j(θq

n)Gjqn .

Corollary 7.18. With notation as above, m`,j(Y ) = c`,j(Y ).

Proof. It suffices to show that m`,j(Y ) and c`,j(Y ) agree at Y = θqn

for all n > 0. By (7.6)and (7.16), we see that m`,j(θ) = c`,j(θ). Moreover, for n > 1, letting ` + n denote thetuple (`1 + n, . . . , `σ + n) and using (7.14) and Lemma 7.17, we have

c`,j(θq

n)= ε`+n,jqn(7.19)

= coefficient of αn(j1, . . . , ju, 0) in M(αn(ξ(`))) with Y = θ,

= coefficient of (j0, . . . , ju, 0) in M(ξ(`)) with Y = θqn

,

= m`,j

(θq

n).

�


Henceforth we will use c`,j(Y ) exclusively to denote m`,j(Y ). To complete our calcula-tions for (7.7)–(7.8), we have the following lemma, whose proof is a variant of the proofof Lemma 7.17 in [34].

Lemma 7.20. Let ` ∈ I, and suppose ξ(`) = (ξ0, ξ1, . . .) with ξ0 = · · · = ξk0−1 = 0 andξk0 6= 0 for k0 > 0. Then for any k > k0,(

Y qk0+1 − θ)(Y qk0+2 − θ

)· · ·(Y qk − θ

)divides c`,qk(Y )

in A[Y ].

Proof. For k > 0, let sk ∈ S be given by sk ..= (0, . . . , 0, 1, 0) where the 1 is in the k-thentry of sk. The key thing to keep to in mind is that by the definition of m`,qk , it followsthat c`,qk(Y ) is the coefficient of sk in M(ξ(`)). We proceed by induction on the sizesz(ξ(`)) =

∑i ξi. If sz(ξ(`)) = 1, then by the definition of k0 we have ξ(`) = sk0 , and also

M(ξ(`)) = ξ(`). Thus c`,qk(Y ) = 0 identically for every k > k0.Now suppose sz(ξ(`)) > 2 and that the statement holds for elements of size < sz(ξ(`)).

If 1 6 ξk0 6 q − 1, then by the moves of our game, every element w ∈ S in the supportof M(ξ(`)) starts as

w = (0, . . . , 0︸︷︷︸k0

, ξk0 , . . .),

and so we have c`,qk(Y ) = 0 identically for all k > k0.Therefore, suppose that ξk0 > q, and apply m to ξ(`):

m(ξ(`)) =(0, . . . , 0︸︷︷︸

k0

, ξk0 − (q − 1), ξk0+1, . . .)

+(Y qk0+1 − θ

)(0, . . . , 0︸︷︷︸

k0

, ξk0 − q, ξk0+1 + 1, . . .)

(7.21)

=.. y +(Y qk0+1 − θ

)z.

As noted in (7.9), the sizes of each term on the right is sz(ξ(`))− (q−1), so we can applyour induction hypothesis. Since ξk0 − (q − 1) > 1, the induction hypothesis applied to y

implies that for k > k0, the coefficient of sk inM(y) is divisible by (Y qk0+1−θ) · · · (Y qk−θ).We consider M(z). If ξk0 > q, then the argument in the previous paragraph holds as

well for M(z). So finally we suppose that ξk0 = q. If k = k0 + 1, then there is nothing

to show since the Y qk0+1 − θ in front in (7.21) is the divisor we want and it persiststhroughout all calculations of M(ξ(`)) = M(m(ξ(`))). On the other hand, if k > k0 + 1,since ξk0+1+1 6= 0 we can apply induction with k0 replaced by k0+1, and so the coefficient

of sk in M(z) is divisible by (Y qk0+2 − θ) · · · (Y qk − θ). Again adding on the Y qk0+1 − θterm from (7.21), we are done. �

Using Lemma 7.20 together with (7.7)–(7.8), we now derive identities for Ss,i for i > 0.As earlier in the section we let σ = 1 + ρ(q− 1) with ρ > 2 and assume s < σ. Since Ss,iis the coefficient of tis+1 · · · tiσ in Fσ,i+1, we see from (7.7) that

Ss,i = −blogq(σ)c−1∑

h=0

1

Li+1+h

∑`6i

ε(`,ı),qi+1+hµ`1(t1) · · ·µ`s(ts),


where ` ∈ Zs>0 and ı represents a string of i’s repeated σ − s times (so (`, ı) ∈ Zσ>0). Aswe saw in (7.8), if i − ρ + 1 66 `, then ε(`,ı),qi+1+h = 0. Thus if we let δ(`) denote theminimum of the entries of `, then for all i > 0,

(7.22) Ss,i = −blogq(σ)c−1∑

h=0

1

Li+1+h

ρ∑ν=max(1,ρ−i)

∑i−ρ+ν6`6iδ(`)=i−ρ+ν

ε(`,ı),qi+1+hµ`1(t1) · · ·µ`s(ts).

If δ(`) = i − ρ + ν > 0, then Lemma 7.17 with n = i − ρ + ν yields (as in [34, §2.1.2],and see also (7.19) above)

(7.23) ε(`,ı),qi+1+h = c(`1−i+ρ−ν,...,`s−i+ρ−ν,ρ−ν,...,ρ−ν),qh+ρ−ν+1(Y )|Y=θqi−ρ+ν .

This prompts the following definition. For 1 6 ν 6 ρ, suppose η ∈ Zs>0 satisfies 0 6 η 6ρ− ν with δ(η) = 0. Then for 0 6 h 6 blogq(s)c − 1, Lemma 7.20 enables us to define apolynomial in A[Y ],

(7.24) ∆η,ν,h(Y ) ..=c(η,ρ−ν,...ρ−ν),qh+ρ−ν+1(Y )

(Y q − θ) · · · (Y qh+ρ−ν+1 − θ)∈ A[Y ].

Moreover, for ` = (`1, . . . , `s) 6 i, for 1 6 ν 6 ρ with δ(`) = i − ρ + ν > 0, and for0 6 h 6 blogq(σ)c−1, we let ηi(`) ..= `− (i−ρ+ ν) = (`1− i+ρ− ν, . . . , `s− i+ρ− ν) ∈{0, . . . , ρ− ν}s. We then find from (7.23) that

−ε(`,ı),qi+1+h

Li+1+h

= (−1)h+ρ−ν∆ηi(`),ν,h

(θq

i−ρ+ν)Li−ρ+ν

.

Combining this identity with (7.22) we prove the following lemma, after which we canfinally prove Theorem 6.1.

Lemma 7.25. Let σ = 1 + ρ(q − 1) with ρ > 2, and let s < σ. Then for i > 0,

Ss,i =

blogq(σ)c−1∑h=0


(−1)h+ρ−ν∑

i−ρ+ν6`6iδ(`)=i−ρ+ν

∆ηi(`),ν,h

(θq


· µ`1(t1) · · ·µ`s(ts),

=



(−1)h+ρ−ν∑

06η6ρ−νδ(η)=0

∆η,ν,h

(θq


× µη1+i−ρ+ν(t1) · · ·µηs+i−ρ+ν(ts).

Proof of Theorem 6.1. As in the beginning of the present section, we take m = s, andfor i > 0, we consider λi(s) with µ1 = · · · = µs = 0 in (6.17) so that Ps(x, z) =

expC(∑∞

i=0 λi(s)zqi). The formula in (6.18) remains valid in this more general case, and

we see from (7.3) and the argument in (7.4) that∑a∈Ai+

∂j1θ (a) · · · ∂jsθ (a)

a=(∂j1t1 ◦ · · · ◦ ∂

jsts

)(Ss,i)∣∣t1=θ,...,ts=θ


=



(−1)h+ρ−ν∑


∆ηi(`),ν,h

(θq


×s∏r=1

(−1)`r−jre`r−1,`r−jr([1], . . . , [`r − 1]).

We note that e`r−1,`r−jr = 0 unless jr 6 `r 6 i, and so when we substitute this expressioninto (6.18) and reorder the sum, we obtain an interior sum of the form

s∏r=1

`r∑jr=kr

(−1)`r−jre`r−1,`r−jr([1], . . . , [`r − 1]) · hkr,jr−kr([1], . . . , [kr])

=

{1 if k1 = `1, . . . , ks = `s,

0 otherwise,

where the equality follows from Proposition 2.13. Combining all of this with (6.17) andthe definition of ηi(`), we obtain

λi(s) =



(−1)h+ρ−ν∑


∆ηi(`),ν,h

(θq


· xq`1+···+q`s

=



(−1)h+ρ−ν∑

06η6ρ−νδ(η)=0

∆η,ν,h

(θq


·(xq

η1+···+qηs)qi−ρ+ν .By reordering the sum and making the substitution i← i+ ρ− ν, we find

∞∑i=0

λi(s)zqi =


ρ∑ν=1

(−1)h+ρ−ν∑

06η6ρ−νδ(η)=0

∑i>0

∆η,ν,h

(θq

i)Li

·(xq

η1+···+qηs)qi(zqρ−ν)qi .If ∆η,ν,h(Y ) =

∑dk=0 aη,ν,h,kY

k with aη,ν,h,k ∈ A, then this becomes

∞∑i=0

λi(s)zqi =


ρ∑ν=1

(−1)h+ρ−ν∑

06η6ρ−νδ(η)=0

(7.26)

×d∑

k=0

aη,ν,h,k logC

(θk · xqη1+···+qηs · zqρ−ν

),

and the result follows upon exponentiation. �

Example 7.27. We demonstrate elements of the proof of Theorem 6.1 in the case s =q + 1 for q > 2. We take σ = 2q − 1, and so ρ = 2. We note that h = 0 throughoutthis calculation. For ν = 2 and for η = (0, . . . , 0) ∈ Zq+1

>0 , we see that ξ(η, 0, . . . , 0) =

(2q − 1, 0) ∈ S, and the calculations in Example 7.10 together with (7.24) show that

∆(0,...,0),2,0(Y ) = 1.


Likewise for ν = 1 and for η = (0, . . . , 0, 1) ∈ Zq+1>0 , we see that ξ(η, 1, . . . , 1) = (q, q −

1, 0) ∈ S. Similar to the calculations in Example 7.10, we find

M(q, q − 1, 0) = (1, q − 1, 0) + (Y q − θ)(0, 1, 0) + (Y q − θ)(Y q2 − θ)(0, 0, 1, 0),

from which we find using the coefficient of (0, 0, 1, 0) and (7.24) that also

∆(0,...,0,1),1,0(Y ) = 1.

We can permute (0, . . . , 0, 1) in q + 1 ways and obtain the same polynomial, but otherthan that there are no other non-zero ∆η,ν,0(Y ) in this case. Then (7.26) becomes

∞∑i=0

λi(q + 1)zqi

= logC(xq+1z

)− (q + 1) logC

(xq·1+qzq

),

and so Pq+1(x, z) = xq+1z − x2qzq, which agrees with Example 6.5.

Example 7.28. We also consider the case s = 2q for q > 2. We take σ = 3q− 2, and soρ = 3. As in the previous example h = 0 throughout. There are four types of choices ofη ∈ Z2q

>0 that produce non-zero polynomials ∆η,ν,0(Y ). For ν = 3, we take η0

= (0, . . . , 0),

for which ξ(η, 0, . . . , 0) = (3q − 2, 0) ∈ S. We calculate

M(3q − 2, 0) = (1, 0) + (Y q − θ)(0, 1, 0) + 2(Y q − θ)(q − 1, 1, 0) + (Y q − θ)2(q − 2, 2, 0),

and using the coefficient of (0, 1, 0) in (7.24) we see that ∆η0,3,0(Y ) = 1.For ν = 2, we let ηa,b = (0, . . . , 0, 1, . . . , 1), with 0 given a times and 1 given b times

and a+ b = 2q. For ξ(ηq,q) = (q, 2q − 2, 0), we find

M(q, 2q − 2, 0) = (1, q − 1, 0) + (Y q2 − θ)(1, q − 2, 1, 0) + (Y q − θ)(0, 1, 0)

+ (Y q − θ)(Y q2 − θ)(0, 0, 1, 0) + (Y q − θ)2(0, q − 1, 1, 0),

and using the coefficient of (0, 0, 1, 0) in (7.24) we have ∆ηq,q ,2,0(Y ) = 1. Similarly, wehave ∆η2q−1,1,2,0(Y ) = 1 and ∆η2q,0,2,0(Y ) = Y q − θ, but we omit the details.

Each of these four types of ∆η,ν,0(Y ) polynomials are unchanged by permuting the

entries of η, but beyond that all other ∆η,ν,0(Y ) = 0, including for all η with ν =

1. Taking into account the possible permutations, ∆η0,3,0(Y ) occurs once; ∆ηq,q ,2,0(Y )

occurs(2qq

)≡ 2 (mod p) times; ∆η2q−1,1,2,0(Y ) occurs

(2q1

)≡ 0 (mod p) times; and finally,

∆η2q,0,2,0(Y ) occurs once. Assembling all of this information into (7.26), we have

∞∑i=0

λi(2q)zqi = logC

(x2qz

)− 2 logC

(xq

2+qzq)

+ 0 + θ logC(x2qzq

)− logC

(θqx2qzq

).

After exponentiating and simplifying, we find

P2q(x, z) = x2qz − 2xq2+qzq + Cθ

(x2qzq

)− θqx2qzq

= x2qz −(2xq

2+q − θx2q + θqx2q)zq + x2q

2

zq2

,

which agrees with Example 6.5.


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Department of Mathematics, Texas A&M University, College Station, TX 77843,U.S.A.

Email address: [email protected]

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