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ON THE RING OF NEWTON INTERPOLATING SERIES OVER A LOCAL FIELD

GHIOCEL GROZA and AZEEM HAIDER

Newton interpolating polynomials at m distinct nodes x0, x1, . . . , xm−1 of mul- tiplicities ni, i= 0,1, . . . , m1,have been used to prove the transcendence of the value of the exponential function at an algebraic number [6]. In this paper, by using these polynomials, we construct a Banach algebra over a local non- archimedean field K which contains the Tate algebra in n indeterminates over K. Many properties of this Banach algebra are analogues of those of the Tate algebras.

AMS 2000 Subject Classification: 13J05, 12J25, 12J27, 32P05, 46J05.

Key words: formal power series, noetherian ring, Tate algebra.

1. INTRODUCTION

The Tate algebrasTn over a complete non-archimedean fieldK,consist- ing of all power series innvariables converging on the “closed” unit polydisc in Kn, lead to the definition of affinoid algebras ([2], Part B or [3], Ch. 2).

Motivated by the question how to characterize elliptic curves with bad reduc- tion this category of analytic-algebraic objects was born in 1961 when J. Tate gave a seminar at Harvard entitled “Rigid Analytic Spaces”. This category of objects make possible analytic continuation over totally disconnected ground fields. A generalization of Tate algebras by using power series is given in [5].

In this paper, we consider a commutativeK-algebraRover an arbitrary fieldK and distinct elementsx0, x1, . . . , xm−1 ofK.In Section 2 we construct a commutative R-algebra Rx0,...,xm−1[[X]] of formal series in nindeterminates which is noetherian whenRis noetherian (see Theorem 2.3). Then we describe the units of this algebra which in the case m = 1 and x0 = 0 becomes the R-algebra of formal power series. In Section 3, we consider a local fieldR=K and define aK-subalgebraT Kx0,...,xm−1[[X]] ofKx0,...,xm−1[[X]] including Tate algebra Tn which contains series converging in the “closed” unit polydisk.

The corresponding residue algebra is isomorphic to K(x0,...,xm−1)[[X]], where K is the residue field of K(see Proposition 3.2) and each series gives rise to

MATH. REPORTS9(59),4 (2007), 343–356

(2)

a continuous function. We describe the units of T Kx0,...,xm−1[[X]] and prove that it is a noetherian domain (see Theorem 3.6).

2. THE ALGEBRA OF NEWTON INTERPOLATING SERIES

Let K be a commutative field, m 1 an integer and x0, x1, . . . , xm−1

distinct elements of K.Ifn is a nonnegative integer we consider the quotient q(n) and the remainderr(n) obtained whenn is divided bym.Construct the polynomials

(1) un=

m−1 k=0

(X−xk)q(n)+σ(r(n)−k−1)∈K[X], n= 0,1,2, . . . ,

where σ(x) is equal to 0 if x < 0 and 1 otherwise. Thus, u0 = 1, u1 = X−x0, . . . , um = (X−x0). . .(X−xm−1) and

(2) ui+jm=uiujm.

IfP ∈K[X],we denote by Pt(z1;z2;. . .;zt) the divided difference of P with respect tot distinct elements z1, z2, . . . , zt∈K. Thus,P1(z1) =P(z1) and (3) Ps(z1;z2;. . .;zs) = Ps−1(z2;. . .;zs)−Ps−1(z1;. . .;zs−1)

zs−z1 , for 2≤s ≤t. Letd < tbe the degree ofP. Then, for everyx∈K,by Newton interpolating formula, we have

(4) P(x) =

d+1

s=1

Ps(z1;z2;. . .;zs)us−1(x).

Lemma 2.1. Let K be a commutative field and x0, x1, . . . , xm−1 dis- tinct elements of K. Then there exist elements ck(s, t) K, where s, t {0,1, . . . , m1}, max{s, t} ≤ k s+t, such that in the ring K[X], for all nonnegative integers iand j,

(5) uiuj =

r(i)+r(j)

k=max{r(i),r(j)}

ck(r(i), r(j))u(q(i)+q(j))m+k, where ck(s, t) = umin{s,t};k−max{s,t}+1

xmax{s,t};xmax{s,t}+1;. . .;xk

(the di- vided difference ofumin{s,t} with respect toxmax{s,t}+1, . . . , xk)and xk=xr(k) ifk≥m.

Proof. If r(j) = 0 then j = q(j)m and, by (2), uiuj =ui+j. Hence (5) follows withct(t,0) = 1. Now, suppose that r(j) = 0.Then

(6) uiuj =u(q(i)+q(j))mur(i)ur(j).

(3)

Assume r(i) r(j). Then, by Newton interpolation formula with respect to xr(i);xr(i)+1;. . .;xr(i)+r(j),we have

(7) ur(j)=

r(j)

v=0

ur(j);v+1

xr(i);xr(i)+1;. . .;xr(i)+v v

w=1

X−xr(i)+w−1 . Now, (6) and (7) imply (5).

LetRbe a commutativeK-algebra andx0, x1, . . . , xm−1distinct elements of K. If ν = (ν1, ν2, . . . , νn) Nn, set N(ν) = ν1 +ν2 +· · · +νn, X = (X1, . . . , Xn), Uν =

n

j=1uνj K[X], where uνj K[Xj] are given by (1).

Putq(ν) = (q1), q2), . . . , qn)), r(ν) = (r(ν1), r2), . . . , rn)) and if τ = (τ1, τ2, . . . , τn) Nn, j∈N,defineν+τ = (ν1+τ1, ν2+τ2, . . . , νn+τn) and = (jν1, jν2, . . . , jνn). Order Nn in the following manner: ν < τ if either N(ν) < N(τ) or N(ν) = N(τ) and ν is less than τ with respect to the lexicographical order. Denote by n a symbol such that ν < n for every ν Nn and by R(x0,...,xm−1) [[X]] the set of formal series of the form f =

n

ν=0aνUν, aν R. An element of R(x0,...,xm−1) [[X]] is called a Newton interpolating seriesat x0, x1, . . . , xm−1.Iff, g=

n

ν=0bνUν ∈R(x0,...,xm−1)[[X]], then define addition and multiplication off and g by

(8) f +g=

n

ν=0

(aν +bν)Uν,

(9) f g=

n

ν=0

pνUν, where

(10) pν =

max{α+q(β)m,β+q(α)m}≤να+β

Cν−(q(α)+q(β))m(r(α), r(β))aαbβ, withCν(α, β) =cν11, β1). . . cνnn, βn) andci(s, t) defined in Lemma 2.1.

Consider a non-zero series f =

n

ν=0aνUν R(x0,...,xm−1)[[X]]. If τ is the smallest subscript ν for which aν is different from zero, then τ will be calledthe order of f and will be denoted o(f).We agree to attach the order

n to the element 0 of R(x0,...,xm−1)[[X]]. It follows in the usual way that o(f+g) min{o(f), o(g)},but o(f g)≥o(f) +o(g) does not hold for every f, g∈R(x0,...,xm−1)[[X]].

(4)

Lemma2.2. There exists an injective map ϕ:R[X]→R(x0,...,xm−1)[[X]]

such that

ϕ(P +Q) =ϕ(P) +ϕ(Q) for allP, Q∈R[X],and

ϕ(P Q) = ϕ(P) ϕ(Q),

where the addition and the multiplication in R(x0,...,xm−1)[[X]] are defined by (8)and(9).

Proof. For P ∈R[Xj], j ∈ {1,2, . . . , n} with d= deg(P) m−1 , by means of Newton interpolation formula, put

(11) ϕ(P) =

m−1

k=0

Pk+1(x0;x1;. . .;xk)(Xj−x0). . .(Xj−xk−1).

Ifd≥mthen there existQs ∈R[Xj], s= 0, . . . ,d

m

, with deg(Qs)< m, such

thatP = [md]

s=0Qsusm.Then defineϕ(P) = [md]

s=0ϕ(Qs)usm. Now, ifP =

s

i=1aiXti ∈R[X1, . . . , Xt], t≤n,whereai ∈R[X1, . . . , Xt−1], define

ϕ(P) = s i=1

ϕ(ai)ϕ(Xt)i

by induction on t. Because a system of polynomials which have different degrees is linearly independent over K, the proof is complete.

Theorem2.3. If R is a commutative K-algebra which is a noetherian ring, then under the addition and the multiplication defined by (8) and (9), the setR(x0,...,xm−1)[[X]] becomes a commutativeK-algebra which is a noethe- rian ring.

Proof. Since for every nonnegative integerssandtthe system of polyno- mialsu(q(s)+q(t))m(X), u(q(s)+q(t))m+1(X), . . . , us+t(X)∈R[X] is linearly inde- pendent overK, by (5), for each i, pi(s, t) =pi(t, s).If we also denoteϕ(R[X]) byR[X],where ϕis given in Lemma 2, we obtain that R[X] becomes a com- mutative K-algebra when the addition and the multiplication are defined by (8) and (9). Consider f =

n

ν=0aνUν ∈R(x0,...,xm−1)[[X]], and take the partial sumfs=

s

ν=0aνUν.Then each coefficient off+g, f gfrom (8) and (9) can be obtained as a sum or a multiplication of polynomials of the form fs and gs.

(5)

Hence the statement that R(x0,...,xm−1)[[X]] is a commutative K-algebra can be reduced toR[X], whenceR(x0,...,xm−1)[[X]] is a commutativeK-algebra.

Since theK-algebra R(x0,...,xm−1)[[X1, . . . , Xn−1]](x0,...,xm−1)[[Xn]] is iso- morphic to R(x0,...,xm−1)[[X1, . . . , Xn]], we prove by induction on n that the K-algebra R(x0,...,xm−1)[[X]] is a noetherian ring. It is enough to show that R(x0,...,xm−1)[[X]] is a noetherian ring. Let I be an ideal in R(x0,...,xm−1)[[X]].

In a similar way to the particular casem= 1, x0 = 0 (see [7], p. 138), for any integeri≥0 denote byEi(I) the set of elements ofR consisting of 0 and the coefficients ofui in the elements of I of orderi. Then Ei(I) is an ideal of R.

Since uium =ui+m,forr ∈ {0,1, . . . , m1}, the non-zero ideals Er+jm(I), j∈N,constitute an ascending sequence. Thus there existsjrNsuch that

Er+jrm(I) =

j∈N

Er+jm(I), r∈ {0,1, . . . , m1}.

Hence there existb1,r, . . . , bnr,r ∈Er+jrm(I) such thatEr+jrm(I) is generated by these elements. Considerg1,r, . . . , gnr,r∈ I such thato(gi,r) =r+jrm for r∈ {0,1, . . . , m1},and

(12) gi,r=bi,ruo(gi,r)+ j>o(gi,r)

di,r,juj, i∈ {0,1, . . . , nr}.

Denote bys the greatest integer among the order of the series gi,r.Now, for every t < s consider c1,t, . . . , cmt,t Et(I) such that Et(I) is generated by these elements and chooseh1,t, . . . , hmt,t∈ I such that, for t < s,

(13) hi,t=ci,tuo(hi,t)+ j>o(hi,t)

ei,t,juj, i∈ {0,1, . . . , mt}.

We shall prove that the ideal I is generated by the elements gi,r, hi,t

given by (12) and (13). Let f =

i=o(f)

aiui be an element of I. If o(f) < s thenao(f) Eo(f)(I) and we can write

(14) ao(f)=

mo(f)

j=1

αjcj,o(f),

where αj R. Hence the order of f mo(f)

j=1 αjhj,o(f) is greater than o(f). It follows by successive applications of this result that we can considero(f)≥s.

(6)

Then we can write

(15) ao(f)=

nr

j=1

βjbj,r,

whereβj ∈R and r=r(o(f)).Hence the order of f

no(f)

j=1 βjuqm(o(f))−jrgj,r is greater thano(f).By successive applications of this result, for everyn we get elementsvj,r,n ∈R(x0,...,xm−1)[[X]] such that the order off−m−1

r=0 nr j=1

n

i=1vj,r,igj,r

is greater thann ando(vj,r,i) tends to infinity withifor fixedjand r. Hence vj,r=

i=1vj,r,i∈R(x0,...,xm−1)[[X]] andf =

m−1 r=0

nr

j=1vj,rgj,rbelongs to the ideal generated by the elementsgj,r, hj,t.

Theorem 2.4. If R is a commutative K-algebra and f =

i=0aiui R(x0,...,xm−1)[[X]],then f is a unit in R(x0,...,xm−1)[[X]] if and only if

(16) dj(f) =

j i=0

cj(i, j)ai

are units inR for everyj∈ {0,1, . . . , m1}.

Proof. Since f is a unit in R(x0,...,xm−1)[[X]] if and only if there exists g =

i=0biui R(x0,...,xm−1)[[X]] such that p0 = 1 and pi = 0, for all i with i≥1,wherepi are given by (10), we have

(17)

max{s+q(t)m,t+q(s)m}≤is+t

ci−(q(s)+q(t))m(r(s), r(t))asbt=δi,0,

whereδi,j is the Kronecker delta symbol. But, for a fixedi,the corresponding equation of (17) contains elements bj with j i and the coefficient of bi is dr(i) given by (16). Then, if i1≡i2 (modm),the coefficients ofbi1 and bi2 in the corresponding equations to (17) fori=i1 andi=i2, respectively, are the same. Hence, if dj(f) are units inR for every j∈ {0,1, . . . , m1},thenf is a unit inR(x0,...,xm−1)[[X]].

Now, let f be a unit in R(x0,...,xm−1)[[X]]. Then a0b0 = 1 and d0(f) = c0(0,0)a0 =a0 is a unit inR. We prove by induction onj thatdj(f) is a unit inRfor everyj∈ {0,1, . . . , m1}. Sinced0(f) is a unit, we can suppose that dj(f) anddj(g) =

j

i=0cj(i, j)bi are units in R forj < h≤m−1, and we have to prove thatdh(f) also is a unit inR. LetI be the ideal generated bydh(f)

(7)

inR. Add to equation (17) fori= 1 multiplied by ch(1, h1),the equation fori= 2 multiplied by ch(2, h1) and so on, finally, the equation for i=h multiplied bych(h, h1).We get

ch(1, h1) 1

t=0

1

s=1−t

c1(s, t)as

bt+ch(2, h1) 2 t=0

2

s=2−t

c2(s, t)as

bt+· · ·

(18) +ch(h, h1) h t=0

h

s=ht

ch(s, t)as

bt∈ I.

Since dh(f) =

h

i=0ch(i, h)ai = a0 +

h

s=1ch(s, h)as ∈ I, we replace a0 by

h

s=1ch(s, h)as.Thus, it follows by (18) that ch(1, h1)

h s=1

ch(s, h)asb1+c1(1,0)a1b0+c1(1,1)a1b1

+

+ch(2, h1)

h s=1

ch(s, h)asb2+ 1

t=0

2

s=2−t

c2(s, t)as

bt+ 2 s=1

c2(s,2)asb2

+· · ·

+ch(h1, h1)

h s=1

ch(s, h)asbh−1+

h−2

t=0

h−1

s=h−1−t

ch−1(s, t)as

bt+

+

h−1

s=1

ch−1(s, h1)asbh−1

+ch(h, h1) h−1

t=0

h

s=ht

ch(s, t)as

bt∈ I, or

(19)

h−1

t=0

h

s=1

min{

s+t,h}

k=max{s,t}

ch(k, h1)ck(s, t)−ch(t, h1)ch(s, h)

as

bt∈I, where ci(j, k) = 0 if i < max{j, k} or i > j+k. But, by Lemma 2.1, the coefficientes,t ofasbtin (19) is

(20) es,t =

min{s+t,h} k=max{s,t}

ch(k, h1)ck(s, t)−ch(t, h1)ch(s, h) =

=

min{s+t,h} k=max{s,t}

umin{k,h−1};h−max{k,h−1}+1

xmax{k,h−1};. . .;xh

−umin{s,t};k−max{s,t}+1

xmax{s,t};. . .;xk

−ut;2(xh−1;xh)us;1(xh).

(8)

Hence, ifs+t=h and s≥t, then (21)

es,t=

h−1

k=s

uk;2(xh−1;xh)ut;ks+1(xs;. . .;xk)+uh−1;1(xh)ut;hs+1(xs;. . .;xh)

−ut;2(xh−1;xh)us;1(xh) = 1

xh−xh−1 ((us(xh)−us(xh−1))ut;1(xs)+

+ (us+1(xh)−us+1(xh−1))ut;2(xs;xs+1) +· · ·+ (uh−1(xh)−uh−1(xh−1))·

·ut;hs(xs;. . .;xh−1)) +uh−1;1(xh)ut;hs+1(xs;. . .;xh)−ut;2(xh−1;xh)·

·us;1(xh) = 1

xh−xh−1 ((us(xh)−us(xh−1))ut;1(xs) + (us(xh)(xh−xs)

−us(xh−1)(xh−1−xs))ut;2(xs;xs+1)+· · ·+(us(xh)(xh−xs). . .(xh−xh−2)

−us(xh−1) (xh−1−xs). . .(xh−1−xh−2))ut;hs(xs;. . .;xh−1) + (xh−x0). . . (xh−xh−2) (xh−xh−1)ut;hs+1(xs;. . .;xh))−ut(xh)−ut(xh−1)

xh−xh−1 us(xh) =

= us(xh)

xh−xh−1(ut;1(xs) + (xh−xs)ut;2(xs;xs+1) +· · ·+ (xh−xs)·

·. . .(xh−xh−2)ut;hs(xs;. . .;xh−1) + (xh−xs). . .(xh−xh−1)·

·ut;hs+1(xs;. . .;xh)−ut(xh) +ut(xh−1)) us(xh−1) xh−xh−1·

·(ut;1(xs) + (xh−1−xs)ut;2(xs;xs+1) +· · ·+ (xh−1−xs). . .(xh−1−xh−2)·

· ut;h−s(xs;. . .;xh−1)) = us(xh)

xh−xh−1 (ut(xh)−ut(xh) + ut(xh−1))

us(xh−1)

xh−xh−1ut(xh−1) =ut(xh−1)us;2(xh−1;xh) =ch−1(t, h1)ch(s, h1), and similarly fors+t=h ands < t.

Now, supposes+t < hand s≥t. Then es,t=

s+t

k=s

uk;2(xh−1;xh)ut;ks+1(xs;. . .;xk)−ut;2(xh−1;xh)us;1(xh) =

= 1

xh−xh−1 ((us(xh)−us(xh−1))ut;1(xs) + (us+1(xh)−us+1(xh−1))·

·ut;2(xs;xs+1) +· · ·+ (us+t(xh)−us+t(xh−1))ut;t+1(xs;. . .;xs+t))

−ut;2(xh−1;xh)us;1(xh) = 1

xh−xh−1((us(xh)−us(xh−1))ut:1(xs)+

+ (us(xh)(xh−xs) us(xh−1) (xh−1−xs))ut;2(xs;xs+1) +· · ·+ (us(xh)·

·(xh−xs). . .(xh−xs+t−1)−us(xh−1) (xh−1−xs). . .(xh−1−xs+t−1))·

(9)

·ut;t+1(xs;. . .;xs+t)(ut(xh)−ut(xh−1))us(xh)) = us(xh)

xh−xh−1 (ut;1(xs)+

+(xh−xs)ut;2(xs;xs+1)+. . .+ (xh−xs). . .(xh−xs+t−1)·ut;t+1(xs;. . .;xs+t)

−ut(xh) +ut(xh−1)) us(xh−1)

xh−xh−1(ut;1(xs) + (xh−1−xs)·

·ut;2(xs;xs+1) +· · ·+ (xh−1−xs). . .(xh−1−xs+t−1)·ut;t+1(xs;. . .;xs+t)) =

= us(xh)

xh−xh−1 (ut(xh)−ut(xh) +ut(xh−1)) us(xh−1)

xh−xh−1(ut(xh−1)) =

=ut(xh−1)us(xh)−us(xh−1)

xh−xh−1 =ut;1(xh−1)us;2(xh−1;xh) and

(22) es,t =ch−1(t, h1)ch(s, h1).

Similarly, ifs+t < h andt≥swe also obtain (22).

Now, it follows by (19)–(22) that (23)

h−1

t=0

ch−1(t, h1) h

s=1

ch(s, h1)as

bt∈ I, or h1

t=0

ch−1(t, h1)bt

h

s=1

ch(s, h1)as

=dh−1(g) h

s=1

ch(s,h1)as

∈I. Sincedh−1(g) is a unit, we obtain

(24)

h s=1

ch(s, h1)as∈ I. Because

h s=0

ch(s, h)as, h s=1

ch(s, h1)as ∈ I, we have

ch(h, h1) h s=0

ch(s, h)as−ch(h, h) h s=1

ch(s, h1)as∈ I, or

(25) E =

h−1

s=0

(ch(h, h1)ch(s, h)−ch(h, h)ch(s, h1))as∈ I.

(10)

But, by Lemma 2.1,

(26) E =

h−1

s=0

(uh−1;1(xh)us;1(xh)−uh;1(xh)us:2(xh−1;xh))as=

=

h−1

s=0

(uh−1(xh)us(xh)−uh−1(xh) (us(xh)−us(xh−1)))as=

=uh−1(xh)

h−1

s=0

us(xh−1)as =uh−1(xh)dh−1(f).

Now, since uh−1(xh) and dh−1(f) are units in R, by (25) and (26), we obtain thatI =R and dh(f) is a unit in R.The proof is complete.

Corollary2.5. If R is a commutative K-algebra and f =

n

ν=0aνUν is an element ofR(x0,...,xm−1)[[X]], then f is a unit in this algebra if and only if for every ν = (ν1, ν2, . . . , νn) Nn with νi ∈ {0,1, . . . , m1}, 1 i n, we have

(27) dν(f) =

τiνi

Cν(τ, v)aτ,

where Cν(τ, v) =cν11, v1)cν22, v2). . . cνnn, vn) are units in R.

Proof. Since the K-algebra R(x0,...,xm−1)[[X1, . . . , Xn]] is isomorphic to R(x0,...,xm−1)[[X1, . . . , Xn−1]](x0,...,xm−1)[[Xn]], the result follows by induction from Theorem 2.4.

3. A BANACH ALGEBRA OF NEWTON INTERPOLATING SERIES

Consider a field K and a non-trivial non-archimedean absolute value | | onK.Forx, y∈Kputd(x, y) =|x−y|.Then (K, d) is a metric space and we can introduce the customary topological concepts into such of space in terms of the metric. IfK is a locally compact field, then it is called a local field.

If A is a commutative ring with identity and is a non-archimedean norm on A, consider the sets see ([2], Chapter 1): A= {x∈A;x ≤1}, A= {x∈A;x<1} . Then A is a commutative ring with identity and A is an ideal in A . Denote the residue ring A / A by A. Let A = K be a commutative field with a non-trivial non-archimedean absolute value| |.Then K is a local ring called the valuation ring of | | and K is the maximal ideal of K . If K is a local field, then it is a complete field and the residue field K is a finite field of order m = ps, where p is the characteristic of K ([1],

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