Construction of Counting Polynomials

Lemma 4.8. Let k ∈ {0,1, . . . , t−1} and let n = (n0, . . . , nt−1) ∈ N^t. Then n is k- critical if and only if it isk-starting and Sc (n) = [(Tit(n) + 1) (p−1) + 1]p^t−1−p^k. Thus

|n| ≡Sc (n)≡0 (mod p−1) ifn isk-critical. If n is k-critical andnk, . . . , nt−2 < p, then n_k=· · ·=nt−2=p−1 andnt−1= (p−1) (Ti_t(n) + 1).

Proof. Suppose thatn isk-critical. By definitionnisk-starting, and furthermore, we have Ti_t n+e^k

>Ti_t(n), so that Sc n+e^k

≥[(Ti_t(n) + 1) (p−1) + 1]p^t−1. Thus

Sc (n)≥[(Tit(n) + 1) (p−1) + 1]p^t−1−p^k.

On the other hand, by the definition oft-tier,

Sc (n)<[(Tit(n) + 1) (p−1) + 1]p^t−1.

Sc (n) is a multiple of p^k since n is k-starting. But there is only one multiple of p^k that satisfies both our inequalities for Sc (n), namely, Sc (n) = [(Ti_t(n) + 1) (p−1) + 1]p^t−1−p^k. Note that this score is divisible byp−1 and that Sc (n)≡ |n| (mod p−1) by (4.3).

Conversely, suppose n is k-starting and Sc (n) = [(Tit(n) + 1) (p−1) + 1]p^t−1 −p^k. Then Sc n+e^k

= [(Tit(n) + 1) (p−1) + 1]p^t−1, so that Tit n+e^k

= Tit(n) + 1 >

Tit(n). So n is k-critical.

Now suppose n is k-critical and nk, . . . , nt−2 < p. By the first part of the lemma, we know that Sc (n) ≡ p^t−1−p^k (modp^t−1), so that n_kp^k +· · ·+nt−2p^t−2 ≡ p^t−1 −p^k (modp^t−1). By the upper bounds on n_k, . . . , nt−2, we have 0 ≤ n_kp^k+· · ·+nt−2p^t−2 ≤ p^t−1−p^k, which is a range shorter in length that p^t−1. Therefore, our congruence modulo p^t−1 for n_kp^k+· · ·+nt−2p^t−2 exactly determines the value to be p^t−1 −p^k, which forces n_k=· · ·=nt−2=p−1. Thus Sc (n) = (nt−1+ 1)p^t−1−p^k, and the first part of this lemma then tells us thatnt−1= (Tit(n) + 1) (p−1).

The next two lemmas show that the coefficients in the Newton expansion of a (p, t)- periodic function satisfy certain recursion relations. These relations are used to show that the Newton coefficients f_i of a (p, t)-periodic function tend to zero (in thep-adic sense) as Sc (i) tends to infinity.

Lemma 4.9. Let F:Zpt → Zp be (p, t)-periodic with Newton expansion P

i∈N^tf_i ^x_i , let k∈ {0,1, . . . , t−2}, and leth∈N^t withhk ≥p. Then f_h−pe^k_+e^k+1 =fh+Pp−1

j=1 p j

f_h−je^k.

Proof. Since F is (p, t)-periodic, we have T_k^pF =T_k+1F. Then employ Lemma 4.7 on both sides of this equation to compute the Newton coefficient for the (h−pe^k)-term (i.e., the coefficient in front of _h−pe^x k

in the Newton expansion). We obtain

p

X

i=0

p i

f_h−pe^k_+ie^k =

1

X

j=0

1 j

f_h−pe^k_+je^k+1.

The i= 0 term on the left matches thej= 0 term on the right, and thei=pterm on the left is justfh, so we get

fh+

p−1

X

i=1

p i

f_h−pe^k_+ie^k =f_h−pe^k_+e^k+1.

Now re-index the sum on the left withj=p−ito obtain what we were to prove.

Lemma 4.10. Let F:Zpt→Zp be (p, t)-periodic with Newton expansion P

i∈N^tf_i ^x_i ,and let h∈N^t with ht−1 ≥p. Then f_h+Pp−1

j=1 p j

f_h−je^t−1 = 0.

Proof. Since F is (p, t)-periodic, we haveT_t−1^p F =F. Then employ Lemma 4.7 on the left- hand side to compute the Newton coefficient for the (h−pe^t−1)-term (i.e., the coefficient in front of _h−pe^xt−1

in the Newton expansion). We obtain

p

X

i=0

p i

f_h−pe^t−1_+ie^t−1 =f_h−pe^t−1.

The i= 0 term on the left matches the right-hand side, and the i= p term on the left is justf_h, so we get

f_h+

p−1

X

i=1

p i

f_h−pe^t−1_+ie^t−1 = 0.

Now re-index the sum on the left withj=p−ito obtain what we were to prove.

Now we show that the above recursions force the Newton coefficientsf_iof a (p, t)-periodic function F to decay (p-adically) as the t-tier of i increases. Furthermore, we obtain some

additional information on f_i when i is critical. This latter information will be critical in proofs of the sharpness of certain of our bounds on thep-adic valuations of weights in codes.

Theorem 4.11. Let F:Zpt → Zp be (p, t)-periodic with Newton expansion P

n∈N^tf_n ^x_n . Then vp(fn)≥Tit(n) for all n∈N. Furthermore, suppose that

F(r₀, . . . , rt−1) =











1 if r₀+pr₁+· · ·+p^t−1rt−1≡0 (modp^t), 0 otherwise,

(4.4)

for all r0, . . . , rt−1 ∈ Zp. Then f0 = 1, and if n is critical, we have fn ≡ (−p)^Tit(n) (modp^Tit(n)+1).

Proof. First we devise a well-ordering relation on N^t, and then we induct with respect to the ordering. If m,n ∈ N^t with Sc (m) < Sc (n), then m ≺ n. Among the elements of N^t that have the same score, we order them lexicographically, i.e., (m0, . . . , mt−1) ≺ (n0, . . . , nt−1) means that there is some i such that the two t-tuples agree in positions 0 to i−1, but m_i < n_i. There are only finitely many elements of N^t with a given score, so the lexicographic ordering well-orders the elements of like score. Thus≺is a well-ordering relation.

First we prove the lower bound on the p-adic valuation of the Newton coefficients by induction with respect to our ordering. By Proposition 4.5, we know that all Newton coefficients of F are inZp, and there is nothing more to show for those elements, such as (0, . . . ,0), whose t-tier is zero. So assume that n = (n0, . . . , nt−1) with Tit(n) > 0, or equivalently, that Sc (n)≥p^d.

Let us first examine the case whennj ≥pfor some j < t−1. By Lemma 4.9 above, we have

f_n−pej+e^j+1 =f_n+

p−1

X

i=1

p i

f_n−iej.

Then note that n−pe^j +e^j+1 has the same score (and thus the same t-tier) as n, but

n−pe^j +e^j+1 ≺n. Therefore, by inductionvp fn−pe^j+e^j+1

≥Tit(n). So

f_n+

p−1

X

i=1

p i

f_n−iej ≡0 (mod p^Tit(n)). (4.5)

Now note that the t-tuples n −ie^j in the sum over i have strictly lower scores than n, but Sc(n−ie^j) = Sc (n)−ip^j ≥ Sc (n) −(p−1)p^t−1, since i ≤ p−1 and j ≤ t−1.

Thus Ti_t n−ie^j

≥Ti_t(n)−1 fori= 1, . . . , p−1. Then, since n−ie^j ≺n, we can use the induction hypothesis to say that v_p(f_n−iej) ≥ Ti_t(n)−1.On the other hand, all the binomial coefficients ^p_i

with 0< i < pare divisible byp, so the terms of the sum over iin (4.5) all have p-adic valuation at least Tit(n). So fn≡0 (mod p^Tit(n)).

Next we examine the case whennj < pfor allj < t−1. Then Sc (n)< p^t−1+nt−1p^t−1, but since Tit(n)>0, we must havent−1 ≥p. We now apply Lemma 4.10 to obtain

fn+

p−1

X

i=1

p i

fn−ie^t−1 = 0. (4.6)

Now note that the t-tuples n−ie^t−1 in the sum over i have strictly lower scores than n, but Sc(n−ie^t−1)≥Sc (n)−(p−1)p^t−1, sincei≤p−1. Thus Ti_t n−ie^t−1

≥Ti_t(n)−1 fori= 1, . . . , p−1. Then, since n−ie^t−1 ≺n, we can use the induction hypothesis to say that v_p(f_n−ie^t−1) ≥ Ti_t(n)−1.On the other hand, all the binomial coefficients ^p_i

with 0< i < pare divisible byp, so the terms of the sum overiin (4.6) all havep-adic valuation at least Tit(n). So fn ≡0 (mod p^Tit(n)). This completes the induction proof of the lower bound onp-adic valuations of coefficients.

Now suppose that F satisfies (4.4). Then f0 = (∆⁰F)(0) = F(0) = 1. We prove the congruence for Newton coefficients with critical indices by induction with respect to our ordering. For the base of our induction, we shall prove the congruence for all critical n = (n₀, . . . , nt−1) with n_i < p for all i. So suppose that k ∈ {0,1, . . . , t−1} and n = (n₀, . . . , nt−1) is k-critical with n_i < p for all i. Then by Lemma 4.8 above, we have n₀=· · ·=nk−1 = 0,n_k=· · ·=nt−2 =p−1, andnt−1 = (p−1) (Ti_t(n) + 1). This forces Ti_t(n) = 0 and nt−1=p−1. So n₀ =· · ·=nk−1= 0 and n_k =· · ·=nt−1 =p−1 and we

must showfn ≡1 (modp). LetS ={0,1, . . . , p−1}and consider truncation

f(x₀, . . . , xt−1) = X

ik,...,it−1∈S

f_{0,...,0,ik,...,it−1} x_k

ik

. . .

xt−1

it−1

(4.7)

of the Newton expansion forF. By Corollary 4.4, this polynomial agrees withF(x₁, . . . , x_t) on the setU ={(r₀, . . . , rt) :r0=· · ·=rk−1= 0, r_k ∈S, . . . , rt−1 ∈S}. That is,

f(r0, . . . , rt−1) =











1 ifr0+· · ·+p^t−1rt−1 ≡0 (modp^t), 0 otherwise,

for all r∈U. This is equivalent to saying that

f(r₀, . . . , rt−1) =











1 ifr_k+· · ·+p^t−k−1rt−1≡0 (modp^t−k), 0 otherwise,

for all r∈ U. But the only way that r_k+· · ·+p^t−k−1rt−1 can vanish modulo p^t−k as we varyr_k, . . . , rt−1 overS ={0,1, . . . , p−1}is for r_k=· · ·=rt−1 = 0. That is

f(r0, . . . , rt−1) =











1 ifrk=· · ·=rt−1 = 0, 0 otherwise,

for all r ∈ U. Lemma 4.4 also tells us that f(x₁, . . . , x_t) is the unique polynomial in Qp[x_k, . . . , xt−1] of degree at most p−1 in each indeterminate that takes these values on U. So we know that

f(x₁, . . . , x_t) =

t−1

Y

i=k

(−1)^p−1

x_i−1 p−1

.

Note that the coefficient of the monomial x^p−1_k · · ·x^p−1_t−1 in f(x₁, . . . , xt−1) is ₍₋₁₎p−1

(p−1)!

t−k

. But the only term in the definition (4.7) of f that can give rise to a monomial of this degree is the term with i_k =· · ·=it−1 =p−1, and so, matching coefficients, we see that fn= (−1)^{(p−1)(t−k)}≡1 (modp).

Now the induction step. We suppose thatn= (n0, . . . , nt−1) isk-critical and hasnj ≥p

for some j. We first examine the case when j < t−1. Of course j ≥ k since ni = 0 for i < k. In this case, consider the t-tuplem=n−pe^j+e^j+1. This newt-tuple has the same score (and therefore the same t-tier) as n, but it is lexicographically lower, hence m≺n.

By Lemma 4.9 above, we have

f_m=f_n+

p−1

X

i=1

p i

f_n−iej. (4.8)

Now note that the t-tuples n −ie^j in the sum over i have strictly lower scores than n, but Sc(n−ie^j) = Sc (n) −ip^j ≥ Sc (n) −(p−1)p^j, since i ≤ p−1. By Lemma 4.8, Sc (n) = [(Ti_t(n) + 1) (p−1) + 1]p^t−1−p^k, so that

Sc(n−ie^j)≥[(Tit(n) + 1) (p−1) + 1]p^t−1−p^k−(p−1)p^j

fori= 1, . . . , p−1. Thus Ti_t n−ie^j

≥j

Ti_t(n) + 1−^p_(p−1)p^k+(p−1)p_t−1^jk

fori= 1, . . . , p−1. But recall thatk≤j < t−1, so thatp^k+ (p−1)p^j ≤p^j+1≤p^t−1, so that Ti_t n−ie^j

= Ti_t(n) for i= 1, . . . , p−1. Thus, by the first part of the theorem, we have v_p(f_n−iej) ≥Ti_t(n) for i= 1, . . . , p−1. Since the binomial coefficients ^p_i

with 0 < i < p are divisible by p, this means that all the terms in the sum over iin (4.8) vanish modulop^Tit(n)+1. So

f_m≡f_n (modp^Tit(n)+1) (4.9)

Now we claim thatm=n−pe^j+e^j+1 isk-starting. Since n isk-starting, this is obvious if j > k. Otherwise we would have k = j < t −1, and then note that Sc (n) ≡ −p^k (modp^k+1) by Lemma 4.8, which implies that nk ≡ −1 (modp); since p ≤nj =nk, this means that n_k≥2p−1, so thatm_k ≥p−1>0. Thus mis indeed k-starting. Also recall that Sc (m) = Sc (n), so Lemma 4.8 tells us that mis k-critical like n. Since m≺n with Ti_t(m) = Ti_t(n), the induction hypothesis tells us that f_m ≡ (−p)^Tit(n) (modp^Tit(n)+1).

This, combined with (4.9), completes the induction step in the case where n_j ≥pfor some j < t−1.

So we are left to analyze the case where n is a k-critical t-tuple with nt−1 ≥ p and

nj < pfor all j < t−1. We now apply Lemma 4.10 to obtain

f_n+

p−1

X

i=1

p i

f_n−ie^t−1 = 0. (4.10)

Now note that the t-tuples n−ie^t−1 in the sum over i have strictly lower scores than n.

If 0 < i ≤ p−2, then Sc(n−ie^t−1) ≥ Sc (n)−(p−2)p^t−1. By Lemma 4.8, Sc (n) ≥ (Ti_t(n) + 1) (p−1)p^t−1, so that Sc(n−ie^t−1)≥[Ti_t(n)(p−1) + 1]p^t−1 for 0< i≤p−2.

Therefore, Ti_t n−ie^t−1

≥Ti_t(n) for 0 < i≤p−2. Then the first part of the theorem tells us that vp(f_n−ie^t−1) ≥ Tit(n) for such i. Since the binomial coefficients ^p_i

with 0< i < p−1 are divisible byp, this means that thei= 1, . . . , p−2 terms in the sum over iin (4.10) vanish modulo p^Tit(n)+1. So

f_n≡ −pf_n−(p−1)et−1 (modp^Tit(n)+1). (4.11)

Let m =n−(p−1)e^t−1. Note that since nt−1 ≥ p, we have Sc (n) ≥ p^t, so Tit(n) ≥1.

Then note that Sc (m) = Sc (n)−(p−1)p^t−1 ≥ p^t−1, so that Ti_t(m) = Ti_t(n)−1. By Lemma 4.8, Sc (n) = [(Ti_t(n) + 1) (p−1) + 1]p^t−1−p^k, so that

Sc (m) = [Tit(n)(p−1) + 1]p^t−1−p^k

= [(Ti_t(m) + 1) (p−1) + 1]p^t−1−p^k.

Note thatmis k-starting just liken, so that Lemma 4.8 tells us thatmis k-critical. Fur- thermore,m≺n, so we may apply the induction hypothesis to it to obtainfm≡(−p)^Tit(m) (modp^Tit(m)+1). Combining this with (4.11), and recalling that Ti_t(n) = Ti_t(m) + 1, we obtain f_n≡(−p)^Tit(n) (modp^Tit(n)+1). This completes the induction proof.

With this knowledge of the Newton expansions of (p, t)-periodic functions, we now construct polynomials thatp-adically approximate such functions.

Theorem 4.12. Let m ≥ 1 and let F: Zpt → Zp be (p, t)-periodic. There exists a poly-

nomial

f(x) = X

n∈N^t Tit(n)<m

fn

x n

, (4.12)

with all f_n ∈ Zp, such that f(r) ≡F(r) (mod p^m) for all r ∈ Zpt. For each n ∈ N^t with Ti_t(n)< m, we have v_p(f_n)≥Ti_t(n).

Suppose further that

F(r) =











1 if r0+pr1+· · ·+p^t−1rt−1 ≡0 (modp^t), 0 otherwise,

(4.13)

for all r ∈ Zpt. Then f₀ = 1 and if n ∈ N^t is critical with Ti_t(n) < m, we also have f_n ≡ (−p)^Tit(n) (modp^Tit(n)+1). Furthermore, if F is as given in (4.13), there exists a polynomial

g(x) = X

n∈N^t,Tit(n)<m

|n|≡0 (modp−1)

gnxⁿ

in Qp[x] such that (i) g(r) ≡F(r) (mod p^m) for all r∈ Zpt, (ii) g0 = 1, and (iii) if n is critical with Ti_t(n) =m−1, then n!g_n≡(−p)^m−1 (mod p^m).

Proof. Recall that any (p, t)-periodic function is p-adically continuous. Therefore, the ex- istence of the polynomial f for a generic (p, t)-periodic function F is guaranteed by ap- plying Theorem 4.11 and Lemma 4.6, with the set S in Lemma 4.6 equal to the set {n ∈ N^t : Ti_t(n) < m}. (This set is finite because only finitely many n have a certain t-tier.) Indeed, the coefficientf_n of our polynomialf is precisely the Newton coefficient for the term _n^x

in the Newton expansion ofF . Thus the bounds on valuations of coefficients and the congruences for certain coefficients (in the special case when F is given by (4.13)) follow from the bounds and congruences given in Theorem 4.11. Furthermore, ifF is given by (4.13), we have f0= 1.

Now suppose that F is given by (4.13). Given the polynomialf described in the first

half of the theorem, consider the polynomial

g(x) = 1 p−1

p−2

X

h=0

f(ζ_p−1^h x).

Since ζp−1 is a unit in Zp, we have F(ζ_p−1^h r) = F(r) for all h ∈ Z and r ∈ Zpt. Thus f(ζ_p−1^h r) ≡ f(r) (modp^m) for all h ∈ Z and r ∈ Zpt. Therefore g(r) ≡ f(r) ≡ F(r) (modp^m) for all r ∈ Zpt. Note that if we expand out the terms ^x_n

to write f(x) = P

n∈N^t Tit(n)<m

c_nxⁿ, then

g(x) = X

n∈N^t,Tit(n)<m

|n|≡0 (modp−1)

c_nxⁿ. (4.14)

In particular, g(x) has the same constant term asf(x), namely, 1. Supposeh isk-critical with Tit(h) =m−1. We want to calculate the coefficient ofx^h ing(x). Since his critical,

|h| ≡0 (mod p−1) by Lemma 4.8. Thusg(x) andf(x) have the same coefficient for x^h, namely, c_h. We want to relate c_h to the coefficients f_n in (4.12). Thus, we want to see which terms in the sum in (4.12) might involve the monomial x^h. Note that the monomial x^h can occur in ^x_n

only ifn_j ≥h_j for allj and only ifn_j = 0 for allj such thath_j = 0. So the monomial x^h can occur in _n^x

only ifn isk-starting and Sc (n) ≥Sc (h). But since h isk-critical, Lemma 4.8 shows that anyk-starting t-tuple with a strictly higher score than Sc (h) will be in a highert-tier; sucht-tuples are not included in our sum. So we need only consider thoset-tuplesn in our sum that arek-starting, that have the same score ash, and that have n_j ≥h_j for all j. But the second condition forces equality in all the inequalities of the third condition, showing that the only term in the sum in (4.12) that can involve the monomial x^h is the term with n =h. So the coefficient of x^h in f(x) is c_h = ^f_h!^h. This is also the coefficient of x^h ing(x), as we noted above. Now h!c_h=f_h, and recall thatf_h is the Newton coefficient for the _h^x

-term in the Newton expansion of F. So we know from Theorem 4.11 thatf_h ≡(−p)^m−1 (modp^m).

In later chapters, we shall also want single-variable polynomialsp-adically approximating certain functions that are constant on cosets ofp^tZp inZp. Such approximations as we need can be obtained easily from the polynomials we just constructed.

Corollary 4.13. Let m≥1 and let F:Zp →Zp be a function withF(a) =F(b) whenever a≡b (modp^t). Let dj = [j(p−1) + 1]p^t−1−1 for j∈N. There exists a polynomial

f(x) =

dm

X

n=0

f_n x

n

,

with allf_n∈Zp such thatf(r)≡F(r) (mod p^m)for all r∈Zp. Ifn > d_j thenv_p(f_n)≥j.

Suppose further that

F(r) =











1 if r ≡0 (mod p^t), 0 otherwise,

(4.15)

for all r ∈ Zp. Then f₀ = 1 and if n = d_j for some j = 1,2, . . . , m, then we have f_n≡(−p)^j−1 (modp^j). Furthermore, if F is as in (4.15), there exists a polynomial

g(x) = X

0≤n≤d_m p−1|n

gnxⁿ

of degree d_m in Qp[x] such that (i) g(r) ≡F(r) (mod p^m) for all r ∈Zp, (ii) g₀ = 1, and (iii) d_m!g_dm ≡(−p)^m−1 (modp^m).

Proof. We set F^∗(x) =F(x0+px1+· · ·+p^t−1xt−1), and it is not hard to check thatF^∗ is (p, t)-periodic. So Theorem 4.12 provides us with a polynomial

f^∗(x) = X

n∈N^t Tit(n)<m

f_n^∗ x

n

in Qp[x] that agrees with F^∗(x) modulo p^m everywhere on Zpt. In particular, we have f^∗(r,0,0, . . . ,0)≡F(r) (mod p^m) for all r ∈Zp. So if we set f(x) =f^∗(x,0,0, . . . ,0), we obtain a polynomial inQp[x] of the form

f(x) = X

n∈N Tit(n,0,0,...,0)<m

f(n,0,0,...,0)^∗

x n

,

which agrees with F(x) modulo p^m on Zp. So f(x) = Pdm

n=0f(n,0,0,...,0)^∗ x n

. If n > dj, then

Ti_t(n,0,0, . . . ,0)≥j, so we know thatv_p

f(n,0,0,...,0)^∗

≥j from Theorem 4.12.

Now suppose that F(x) is given by (4.15). Then F^∗(x) is given by (4.13), and so Theorem 4.12 gives us additional information. If n = dj for some j ∈ {1, . . . , m}, then (n,0,0, . . . ,0) is 0-starting and Sc ((n,0,0, . . . ,0)) = dj = [j(p−1) + 1]p^t−1 −1, so that (n,0,0, . . . ,0) is of t-tier j−1, and by Lemma 4.8 we see that (n,0,0, . . . ,0) is critical.

Thus, from Theorem 4.12, we know thatf(n,0,0,...,0)^∗ ≡(−p)^j−1 (modp^j). Also f_(0,...,0)^∗ = 1 from Theorem 4.12.

Furthermore, Theorem 4.12 furnishes a polynomial

g^∗(x) = X

n∈N^t,Tit(n)<m

|n|≡0 (modp−1)

g_n^∗xⁿ

inQp[x] that agrees withF^∗(x) modulo p^m onZpt and has g_(0,...,0) = 1. Furthermore, if n is critical and oft-tierm−1, thenn!g_n ≡(−p)^m−1 (modp^m). Setg(x) =g^∗(x,0,0, . . . ,0), so that

g(x) = X

n∈N Tit(n,0,0,...,0)<m

p−1-n

g(n,0,0,...,0)^∗ xⁿ,

and g(x) agrees with F(x) modulo p^m on Zp. Nowg(x) = P

0≤n≤dm

p−1-n

g(n,0,0,...,0)^∗ xⁿ and has constant term 1. Furthermore, ifn=dm, then (n,0,0, . . . ,0) is 0-critical and oft-tierm−1 by the calculation in the previous paragraph, so that we know dm!g^∗_(d

m,0,0,...,0) ≡(−p)^m−1 (modp^m) from Theorem 4.12.