computation of the zeros of p-adic ¿functions. iizeros of the kubota-leopoldt p-adic l-functions...

MATHEMATICS OF COMPUTATIONVOLUME 62, NUMBER 205JANUARY 1994, PAGES 391-406

COMPUTATION OF THE ZEROS OF p-ADIC ¿FUNCTIONS. II

R. ERNVALL AND T. METSÄNKYLÄ

Abstract. The authors have carried out a computational study of the zeros of

Kubota-Leopoldt p-adic L-functions. Results of this study have appeared re-

cently in a previous article. The present paper is a sequel to that article, dealing

with the computation of the zeros under certain conditions that complicate the

original situation.

1. Introduction

This paper continues the authors' computational study, begun in [2], of the

zeros of the Kubota-Leopoldt p-adic L-functions Lp(s, x) ■ We will discuss

four themes, all of which came up but were left aside in [2].

In the following, the article [2] is referred to as Part I. When referring to an

individual section or proposition ofthat paper we write, say, §1.3 or Proposition

1.5. We retain the basic notation of Part I; in particular, p is an odd prime,

fe(T) = J2%oajTj denotes the Iwasawa power series determined by the first

factor 6 of x , and X = Xg stands for the A-invariant of this power series.Our first question, considered in §2, concerns the computation of the zeros of

Lp(s, x) when the second factor of x is nonprincipal. The remaining sections

deal with the computation of the zeros T\, ... , T\ of fg(T) under certain

conditions that complicate the original situation discussed in Part I. Thus, in

§3 there are two zeros Tx and Ti close to each other, §4 introduces types of

the Newton polygon of fe(T) different from our two basic types, and the final§5 shows how to compute zeros 7^ lying in wildly ramified extensions of Qp .

A sample of numerical results related to each of these sections is given in

Tables VI-X at the end of the paper.We recall that a crucial step in the computation of a zero To of fe(T) is the

determination of the extension E = QP(T0) and of an initial approximation to

of To satisfying the condition

fg(t0) = 0 (modn2^1),

where n is a prime element of &e and y = vn(fg(T0)). To accomplish this,

one first reduces the possible candidates for E to a family of relatively few fields

and expands T0 7t-adically with unknown coefficients; then E and io will t>e

Received by the editor April 30, 1992.1991 Mathematics Subject Classification. Primary 11S40, 11Y05, 11Y40, 11R20, 11R23.Key words and phrases, p-adic L-functions, computation of zeros, factorization of polynomials,

Newton's tangent method, Abelian fields, Iwasawa theory.

©1994 American Mathematical Society

0025-5718/94 $1.00+$.25 per page

391

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392 R. ERNVALL AND T. METSÄNK.YLÄ

obtained by solving successively a sequence of approximation congruences

fe(T0) = 0 (mod O, m= 1,2,... , m0,

denoted by W(nm) below.

When describing the computation of To, we usually assume, to fix ideas,

that ûo = /e(0) does not vanish. If a0 = 0, the procedure is essentially the

same; one just excludes the zero To — 0, which is known to be simple, and then

replaces fe(T) by fg(T)/T. In this case we say that fg(T) is of type Z.

2. L-FUNCTIONS INVOLVING CHARACTERS OF THE SECOND KIND

In this section we are concerned with the zeros so of Lp(s, 6 y/n ), where y/„

is a nonprincipal character of the second kind. Recall that y/n is of order pn

and conductor p"+x (n > 1). We will discuss the questions of how to find

examples of Lp(s, 6tp„) having a zero So , and how to compute this So .

Suppose we have found a zero 7b of some Iwasawa power series fg(T).

By §1.3, if there is a corresponding zero So of some Lp(s, dy/n), then s0 =

log(l-r-7b)/log(l+rf/7) and vp(T0) = \/(p-\)pn~x. In the range of our compu-

tations, the only /7-ordinals of L0 of this form are v3(T0) - 1/2, v$(Tq) = 1/4,

and Vi(To) = 1/6, and the respective values of X are 2, 4, and 6 (or 3, 5, and

7 when f$(T) is of type Z). Since, moreover, the coefficients of fe(T) in all

examples are from %p , Proposition I.5(i) shows that To must belong to QP(ÇP")

in order that So lie in Ds, the domain of definition of Lp(s, x) ■

The case v^(To) = 1/6, which leads to a wildly ramified extension Qi(T0),

will be studied in §5. We found no example of this kind with L0 e (MCç).As for the other two cases, it follows from Proposition I.5(ii) that the con-

dition To € Qp(C/>) is even sufficient for So to be a zero of a Lp(s, dy/x). We

have the following practical criterion.

Proposition 1. Let X = p - 1 and vp(ao) = 1, so that vp(To) = l/(p-l). Then

To e Qp(Cp) if and only if aox = axo ■

Proof. The Newton polygon of fe(T) is of the "first type" discussed in §1.9.

From that discussion it is seen that 7b G Qp( p~yrp), where r is determined by

the conditions

To = xn (mod n2), 0 < x < p,

aox + aXorx"~x = 0 (mod p).

Since Qp(Cp) = Qp( p~\/--p), the assertion follows. D

For p = 5 our numerical material contains five examples with X = 4 and

v$(ao) = 1, and one similar example of type Z . In all these, T0 $ Qs(Cs) •

Four of the examples are exhibited in Tables IV and V of Part I; the remaining

two are (5, 5708,0) and (5, -12712, 1).Our main case is that of p = 3, X = 2, and ^3(00) = 1 • Applying the

criterion of Proposition 1 to the tables computed in Program A (Part I), one finds

numerous examples of this case which give zeros of L^(s, 6y/x). Including some

analogous cases with X = 3 , we settled 13 examples of this kind completely.

Before describing the computation, we observe that there are two characters

y/x mod 9, say ip+ and y/- , identified by the equations y/±(\ + 3d) = — \± j,

where n — \/-3- A zero Tq with 1)3(7b) = 1/2 may be written in the form

Tq = 3b + en , where b, c G Z3, c ^ 0 (mod 3). Then, by Proposition 1.3, the


COMPUTATION OF THE ZEROS OF p-ADIC ¿-FUNCTIONS. II 393

number so = log(l + 7b)/log(l + 3úf) corresponding to c = +\ (mod 3) is a zero

of L3(s, 8ip+), in fact, the only zero of this function (under the assumption

that X = 2). Its conjugate is of course the zero of L$(s, 8\p-).

After computing L3(s, 8), fe(T), T0 , and So in the usual way, we proceed

with the computation of the functionoo

L3(s, 0^,) = ]Tm-s'.(=0

As in §1.5, this means the approximation of u\ by rational integers congruent to

u\ modulo a prescribed power pM ; this makes sense since u\■ —> 0 as / —» oo .

Here we make use of Washington's formula with <3> = 9d in the way described

in §§1.5,1.11. A lower bound for vp(u'¡) needed in the computation is provided

by Proposition 2 below.

Recall that

(1) Lp(s,dy) = fe(p(\+dp)s-\),

where p = i//(l +dp)~x , \p = ipn . To check the correctness of our L3(s, 0i//x),

we evaluated both sides of this formula for s = 1 . As a final check it was

verified that L3(so , 6y/x) vanishes mod nM> , where (see the end of §1.11)

Mx = min ( 2M, M - 3 + min vK(u'¡) J.

By the remark at the end of the present section, Mx> M.

Example. Let 6 be the character defined by (3, -1147, 1). The followingtable presents L3(s, 8) (1st column) and fe(T) (3rd column) according to

the same principles as the Tables I-V in Part I; see §1.10 for a description.The middle column lists the coefficients u'0, ... , u'6 of L^(s, 8y/±) ; here, n =

\/^3. The coefficients u\ with i > 6 are zero to the accuracy displayed in this

column.

0.110000.011220.010200.000110.001020.000000.00011

0.0200 ±0.0122710.0111 ±0.1102ti0.0110±0.0002tc0.0001 ±0.0220tt0.0011 ±0.0000^0.0000 ±0.000l7T0.0001 ±0.0000;r

0.110000.11021.1000.202.00.

The zeros of fe(T) are TXt2 = 0.10 ± 2.071 (approximately), and the corre-

sponding numbers sx ,2 = 0.0 ± (2.)tt are the zeros of L3(s, 9y/^). (The upper

and lower signs correspond to each other.)

Table VI contains further examples of this kind. In Table Via there are two

similar examples with an additional pair of zeros of fg(T) and L$(s, 8).

We conclude this section by proving the following estimate (cf. Proposition

1.6).

Proposition 2. For any y/ - y/„ (n > 1), let Lp(s, dip) = Ylltou'is' • Then

vp(u\) > ip-\

(1 = 0, 1,


394 R. ERNVALL AND T. METSÄNKYLÄ

Proof. Considering the right-hand side of (1), we write

oo

p( 1 + dp)5 - 1 = p exp (slog( 1 + dp)) - 1 = ^2 d¡s',

where do = p - 1 and ú?, = p(\og(\ + dp))1 //! for / > 1 . This gives theexpansions

oo

(p(i+dPy-iy = J2dusl a = o,i,...)i=0

with i/oo = 1 > djo = 0 (/ > 1), and

du= E dh-d'i (*>o,y>i).

We show that

(2) v,(dy) > ./^ - 1) + (l - j^j\ i (i > 0, j > 0).

This is trivial for j = 0 ; so let j > 1. Denote by / the number of positive

indices in a set {tx, ... , tj} with tx H-1-1¡ = i. For simplicity of notation,

assume that the positive indices are just tx, ... , ty , so that

j fEUÄ) = U - J>p(p - i) + EMdt.)-

Since vp(p - 1) < \/(p - 1) and vp(dh) = tv - vp(tv\) >tv- (tv - \)/(p - 1)

(v = 1,...,/), we obtain the estimate

> / j \ /

EV<U>./*v(/>-i) + (i--¿—¡-IE'"-i/=i \ p j „_,This proves (2).

Equation ( 1 ) now allows us to write

oo J oo oo / J \

E «**' = J™ E °> E rfyj' = Ji™ E E aA si>i=0 7=0 i=0 i=0 \j=0 /

whenever s e L\. Restrict 5 for a moment to the subset of Ds defined by

vp(s) > 0. Then it follows, in view of (2), that

oo / oo

$>'/*' = £ E«AK1=0 1=0 \;=0 /

(see Lemma in [3, p. 53]). Hence,

CO

(3) «Î = EM« (1 = 0,1,...),,=0

and this together with (2) implies the proposition. D


COMPUTATION OF THE ZEROS OF p-ADIC ¿-FUNCTIONS. II 395

Remark. Let, in particular, X = p - 1 and vp(ao) = 1. Then the coefficients of

the series for Lp(s, 8ipx) satisfy the stronger inequality

vP(u'i) > i - j—¡ + 1 (i = 0,l,...).

This is seen from (3) and (2) upon observing that vp(Oj) > 1 for j = 0, ... ,

p - 2 and jvp(p - 1) > 1 forj>p-\. In the range of our computations,

vp(u\) attains this lower bound for most values of i.

3. TWO ZEROS CLOSE TO EACH OTHER

This section continues our study of the case X = 2, vp(ao) = 2 in which the

Newton polygon of fe(T) is of the "second type" considered in §1.9. This case

was settled under the assumption that the two zeros Tx and T2 (denoted by

7q' and T¿' in §1.9) satisfy vp(Tx - T2) < 2. We now describe how to deal with

examples in which vp(Tx - T2) is bigger. As before, it is assumed throughout

that TX^T2.

We begin with the following proposition that will also be needed in §4.

Proposition 3. If all the zeros Tx,..., Tx of fg(T) are pairwise distinct, then

x

vP(fe(Tk)) = Y, MTk -Tj) (k=l,..., X).;=im

Proof. By the /7-adic Weierstrass preparation theorem (see §1.2),

fe(T) = ue(T)we(T),

where ug(T) is an invertible power series and wg(T) a monic polynomial of

degree X having Tx, ... , Tx as its zeros. Differentiate this equation to obtain

f¡>(Tk) = ug(Tk)w'e(Tk) and, hence, vp(f'6(Tk)) = vp(w'g(Tk)). D

Going back to the above case, suppose that the rational integers x , y , and

r fulfil the conditions

0 < x < p, 0 < y < p, r = 1 or g (a primitive root of p),

TXt2 = xp±ypy/rp (mod/72).

Such numbers are found in the way presented in §1.9. In particular, if y = 0,

then vp(Tx - T2) > 2, and we may put

Tk=xp + zkp2 (mod/75/2) (k = l,2),

where zx and z2 are either in Z or in 1[^/g]. Looking at the coefficient of

p4 in the approximation congruence mod p5, we infer a condition

d0 + dxzk +a20z2k = 0 (mod/7),

where do and dx are rational integers determined by x and by the first digits

of a0, ... , a4.

If the solutions zx and z2 of the last congruence are distinct, then

vp(Tx - T2) = 2 and Krasner's lemma implies that Tk - xp G <Qp(zk). Thus,

E/Qp is unramified, and we have y = vp(fg(Tk)) = 2 by Proposition 3. In this

case the algorithm may be started with tk = xp + zkp2.



We found two examples of this kind, for the characters (3, 13564, 0) and

(3,-11188, 1), both with E = Q3(\/2). They are presented in detail in TableVII.

Secondly, suppose that zx = z2 = z. Then z G Z, and one proceeds with

the congruence

TXt2 = xp + zp2±up2yfrp (mod/73),

where w G Z, 0 <u < p . Put n = ^frp . To find u and r, one has to solve a

congruence of the form ru2 = I (mod p), where / G Z ; this is in fact obtained

from ^(tt11). If u ¿ 0, we see that E = Qp(n) and vp(Tx - T2) = 5/2,

and Proposition 3 gives y = vK(fg(Tk)) - 5. Hence, we may take ii,2 =

xp + zp2 ± up2n .

In Table VII, the example for the character (3, 13784, 0) belongs to thiscategory. Table Vila shows two similar examples of type Z .

If u = 0, i.e., vp(Tx - T2) > 3, we are in the analogous situation we started

from. There is no example of this in our material.

4. Various types of the Newton polygon

We now show how to deal with some classes of the Newton polygon of fg(T)

which differ from our two basic types. They are called Types 3-6 below; each of

them was met in the computations and will be treated in a generality sufficient

to settle our particular examples.

The examples mentioned in this section appear in Table VIII unless otherwise

stated.Type 3. vp(a0) > 2, vp(ax) = 1 , X = 2.The two zeros Tx and T2 satisfy vp(Tx) = vp(a0) - 1 > 1 and vp(T2) = 1 .

Hence they lie in Qp . This is a slight variant of our "second type", and from

§1.9 it is seen, on putting a02 = 0 , that one may start with tk = xkp (k = 1, 2),

where xx = 0 and x2 is the root of the congruence axx + a2ox = 0 (mod p).

Examples of this kind are obtained for the characters (5,3101, 0), (5,-4371,1 ), ( 11, -723, 1 ), the last two being of type Z . The example (3, -1399, 1 )in Table II (Part I) also belongs to this family.

Type 4. vp(ao) = c > 1 , vp(ak) > c(l - k/X) for k = 1, ... , X - 1 , X (> 1)prime to c and p .

The Newton polygon has one nonzero slope, and this equals -c/X. As in

the case of the "first type", the zeros To (= Tx, ... , Tx) are in fully ramified

extensions E = Qp(tfrp). Because c and p are prime to X, one easily verifies

that y = vn(XaxTQ~x) = (X - l)c. If X = 2, one has to solve for x and r the

congruence aoc + a2orcx2 = 0 (mod p), obtained from W(n2c+X), and then take

to = xnc. For X > 2 the procedure is similar but more complicated.

We have two examples with c = 3 and X = 2, namely for (3, 1901, 0) and

(5, 12056, 2), and one similar example of type Z , for (5,-7816, 1). TableIII in Part I gives an example with (5, 1317,0) in which c — 2 and X = 3 .

Type 5. vp(ao) = 2 , vp(ax) > 2, X = 4.As in Type 4, there is but one nonzero slope, this time equal to -1/2. The

zeros 7|, ... , 7*4 lie in fully ramified quartic extensions E = Qp(J/rp) or, if

this is not the case, either in quadratic or biquadratic extensions of Qp (here

"biquadratic" means "a composite of two quadratic").

An example of the former kind is given by (3, 1541, 0). The computation


COMPUTATION OF THE ZEROS OF p-ADIC L-FUNCTIONS. II 397

is analogous to that in Type 4; note that Proposition 3 yields y — 7 (see Table

VIII). On the other hand, for the character (3, 4204, 0) one rules out the

fields Q3(v^±3) by using ^(ti9) , and it turns out that the zeros are contained

in Q3(v/3, \/2). In this case there is no zero of any L}(s, dy/„), n > 0,

corresponding to 70 ; indeed, u3(log(l + 7b)) = 1/2, and so v3(sq) - -1/2 .In these two examples, the approximate values of the zeros given in Table

VIII are easily computed by hand. A third example of the same kind, not

included in the tables, is (3, 7244, 0). In this case we have E = Q3(-v/-3).

Type 6. vp(ao) > 2 , vp(ax) = 1 , X > 3 , X - 1 prime to p .Among the zeros To - Tk , k = I, ... , X, there is one, say Tx , belonging to

Qp and satisfying vp(Tx) > vp(ao) - 1 > 1. The other zeros satisfy vp(Tk) =

\/(X - 1) (k = 2, ... , X). For T0 = Tx , Proposition 3 yields

^ = ¿^-^ = ¿^ = 1.

Thus, it is sufficient to look at &(p3), and this gives t0 = xp with x = -aoi/ax x

(mod p).The computation of the remaining zeros is analogous to that in the "first

type". In particular, E/Qp is a fully and tamely ramified extension of degree

X - 1. We find that vp(Tk - 7)) = 1/(A- 1) for j = 1, ... , X, j ¿k, and so

y = X- 1.To the first zero Tx there always corresponds a zero sx of Lp(s, 8). The

same holds true for T2, ... , Tx in the case X < p, while the situation varies

for X > p .Our examples with X = 3 < p are the characters (5, 4924, 2) and (7, -4072,

5). The character (3, -25528, 1) has X = 5 (computed first by Kobayashi),and s2, ... , s$ lie outside of Ds. So also do 52, s3 in the example (3,3512,0),where X = 3 . Compare this example with (3, 7804, 0) from Table Via: here,

too, X = 3 but 52 and s3 are zeros of Li,(s, 8y/±). Finally, the character

(3, -5051, 1) in Table V (Part I), for which fe(T) is of type Z , also fits thiscategory.

5. Wildly ramified extensions

All examples of wildly ramified extensions E/Qp were found for p - 3.

The degree of the extensions is 3 or 6.

Consider first the case of cubic extensions. Let X = 3 and v3(a0) = 1,

so that Vi(T0) = 1/3 and E/Q3 is indeed wildly ramified of degree 3. Onemay identify E (up to conjugates) by providing a cubic Eisenstein polynomial

r(X) G Z3[X] such that E = Q3(tt) with r(n) = 0. In the table below, the listof 11 such polynomials r(X) corresponds to a complete system of nonconjugate

fields E . This list is extracted from results by Amano [1], who characterizes in

this way all ramified extensions of degree p over any finite extension of Qp .

We will give a further characterization of the fields E, better suited for their

computational identification. Note that every r(X) is of the form

r(X) = X3 - 3aX2 - 3bX - 3c,

where a, b,c G Z with c = 1 (mod 3). Let 71 and E be as above. We

have 3 = e?r3, with e a unit of &e . and from r(n) = 0 it follows that


398 R. ERNVALL AND T. METSANKYLA

e = l/(c + bn + an2). In particular, e = 1 (mod n). Write

CO

e = \ + Y,Wv, e„G{0,±l}.i/=i

A straightforward calculation shows that our fields E can be identified by the

triple (ex, e2, e3) as indicated in the table.

AxA2

BxB2

B3

r(X)

X3 - 3X - 3

X* + 3X- 3A-3 - 3X2 - 12

X3 - 3X2 - 3

X3 - 3X2 + 6

(£i, e2.f3)

(-1.+1.-1)(+1,-1-1,+1)(0,-1,-1)(0,-1,0)

(0,-1,+1)

c,

c3

^2

^3

r(X)

X* + 1X2- 12A-3 + 3X2 - 3

X3 + 3X2 + 6

A-3- 12

A"3-3

A-3+ 6

(ei,£2>í3)

(0,-1-1,-1)(0,+l,0)(0,+l, +1)

(0,0,-1)(0,0,0)(0,0,+1)

Here the fields are labeled with Ax, A2, Bx , etc. The extension E/Q3 is

cyclic for E — Cx, C2, C3 and pure for E — Px, P2, P3.

We computed 17 examples in which the zeros belong to the fields of this table;

eight of them were selected to be exhibited in Table IX. In the entire sample of

17 fields, Ax and A2 occur five times each, supporting the natural hypothesis

that the values of ex be randomly distributed.

The functions L^(s, 8) and fg(T) were computed with the parameter n =

16. This implies, by Propositions 1.12 and 1.7, that the coefficients a¡ are

obtained mod 38 J (j —

us to determine the field

0, ... , 7). It is seen below that this accuracy allows

E in all examples but one, and to find the zeros

T0 = Y,xvnv (xve{0,±l},xx¿0)

to 1-4 Xy-places. We computed 7b by hand, thus avoiding the rather tedious

implementation of the arithmetic of the fields in question. The results were

checked by computing fe(To). From §1.3 it follows that the corresponding

numbers Sq are not zeros of any L3(s, 8y/n); hence they are ignored.

To find E and T0, start with the approximation congruence ©

reduces to the congruences

«01+030^1=0, aoxex +axxxx + a40 = 0 (mod 3),

(rc5). This

which give us xx and ex . If ex = -1 or +1 , the field is Ax or A2, respectively,

and the coefficients e2, e3, ... are determined by the equation e = 1/(1 ± n).

Better approximations for T0 then follow easily from W(nm) for bigger expo-

nents m.

In Table IX, the characters (3, -3592, 1) and (3, 1781, 0) are examplesgiving the field Ax . The latter requires slightly different techniques since v3(ao)

= 2. In both examples, the zeros Tx , T2, T3 lie in conjugate extensions

generated by the roots n = nk of r(X) = 0 (k = \ ,2,3).Similarly, the characters (3, 281, 0), (3, -311, l),and (3, -2132, 1) yield

examples in which E — A2. The first of these is also mentioned in §1.10, with

another normalization of r(X). The second example is of type Z , while the

third shows a function fg(T) with two zeros (one trivial) in Q3.

Here are the other examples of Ax and A2 computed by us.



Example 1. For nk a zero of X3 - 3X - 3 (/c = 1,2,3),

(3, 1397,0)(3, 5368,0)(3,6712,0)

Tk = nk-n2k- n\ (mod n5k),

Tk = -nk -n2k-n3k + n4k (mod n5k),

Tk = -n\ + n\ (mod n\).

Example 2. (3,-2920, 1): nk zero of X3 + 3X - 3 (k= 1,2,3),

Tk = nk + n¡ + n¡ (mod n¡).

Example 3. (3,-4184, 1): Tx=0,sx=0,

nk zero of X3 + 3X - 3 (k = 2, 3, 4),

Tk = nk-n\ + n\ (mod n\).

If ex = 0, we compute e2 from the congruence

-a0xe2 = a50xx + a2x (mod 3),

which is a consequence of W(n6) .

Let first e2 = 1 , so that £/Q3 is cyclic. Then ^(n1) gives e3 and so fixes

E. When computing further places for To, one should observe that E now

contains all the three zeros of fg(T).

We have two examples of cyclic fields, in fact, of E — Cx . One, included in

Table IX, is for (3, 2504, 0), the other is given below.

Example 4. (3, 5624, 0) : tt zero of X3 + 3X2 - 12,

Tx = -n - n3, T2 = -n + n2 - n3, T$ = -it - n2 (mod 7i4).

Secondly, suppose that e2 = 0. Then E is a pure extension of Q3 and one

finds e3 and x2 by solving simultaneously a pair of congruences produced byg?V).

In Table IX, this case is represented by the characters (3, 4172, 0) and

(3, -1144, 1), which lead to the fields Q^v'Tl!) and (Qbt-v^ó), respectively.The next two examples give our other characters of this kind. Example 6 in-

troduces a case with X = 4 ; here the ultimate identification of E must be left

open.

Example 5. For nk a zero of X3 + 6 (k = 1, 2, 3),

(3,401,0): Tk =-nk + n2 (mod n3),

(3, 4472, 0) : Tk = nk + n\ (mod tt3).

Example 6. (3,6856,0): 7*, = 0.2201, sx = 2.000,

nk zero of X3 - a (k = 2, 3, 4) with a = 4, 1,

or - 2,

Tk = nk (modn2k).

After this, it is also clear how to deal with e2 = -1 corresponding to the

fields Bx, B2, B?,. We have no example of these fields.

Now let us turn to the case of sextic extensions. The basic case is the one

with X = 6 and v^(ao) = 1 . As usual, put E = Q3(70) = Q3(jt) .



Proposition 4. For a fully ramified sextic extension E = Qi(n),let t be the unit

of the ring &E defined by the equation 3 — en6. The field E contains the subfield

Q3(\/3) or <Q>3(\/^3) according as t = 1 ore = -1 (mod n), respectively.

Proof. The polynomial q(X) — X2 ■+ 3 satisfies vn(q'(n3)) = 3 and q(n3) =

n6(l +e) = 0 (mod tt7) provided that e = ±1 (mod n). D

It follows that E is a ramified cubic extension of Q3(a/3) or Q3(\/^3).

Hence, it can be classified in the way shown by Amano in [1]. We do not write

down the complete classification but, rather, show how to handle the specific

examples we have of this case. They are the three examples presented in Table

X, two of them with X = 6 and one with X = 1 and fe(T) of type Z . In theseexamples we chose the truncation parameter r\ = 18 or 20 according as X = 6

or 7, respectively. This enables us to identify E and compute To mod tt3. As

mentioned in §2, we never have E = Q3(Í9) ; hence, we do not compute So.

Put p = y/(-l)z3, where z = 0 or 1. By Proposition 4, E = Q3(/?, it),where n is a zero of an Eisenstein polynomial r(X) e Qi(p)[X]. From [1] we

find that it suffices to consider the polynomials

r(X) = X3 - apX2 - bpX - cp,

where a, b, c G Z and c = 1 (mod p). Let k be the unit in Q3(/>) such that

p = K7T3. A similar argument as before yields k = l/(c + bn + an2), and we

may write

CO

K = 1 +YKvn" ' K„G{0,±1}.i/=l

With the usual notation To - YlTLi xv71" (where xx ^ 0) we get from W(n8)

the congruences

(4) (-l)za0i+«60 = 0, -2a0XKX = (axx + (-l)za10)xx (mod 3).

The former determines p, and the latter tells us, first, whether or not kx = 0.

If kx ^ 0—as is the case in our examples—then b ^ 0, and we arrive, by [1],

at the polynomials

r(X) = X3 + pX-p,

which generate six ramified extensions of Q3(/>). The six zeros of f$(T) are

contained in these fields, one in each. It also follows that k = 1/(1 ± tt) and

so, in particular, k2 = 1. Hence the approximation congruence ^(tt9) enables

us to compute x2.In case one is interested just in knowing whether or not E — Q3(Í9), it is

worth noting that (4) yields the following necessary conditions for this equality:

«oi = «60 ; «it = «70 • Indeed, we have z = 1 and kx =0 in this case.In our examples obtained for the characters (3, -7108, 1 ) and (3, -20692,

1), the first congruence in (4) gives the result z = 0 and z = 1, respectively,

and the second gives kx = xx . For the third character (3, -1832, 1) we find

that z = 0 and kx = -xx . See Table X for the full result.



Table VI. X-2, ao^O, sx¡2 zeros of Lp(s. 8y/±)

(3, 653, 0)

0.211000.001020.020010.000220.002200.000020.00020

0.02000.00220.02220.00020.00210.00000.0002

± O.OOOOti±0.111171±0.0011 ti±0.021 In± 0.000271± 0.000171± O.OOOOti

(3, -379, 1)

0.200000.001110.021220.001120.002000.000020.00021

0.00000.00020.02000.00120.00220.00000.0002

±0.110071±0.2212ti± 0.00227t± 0.01017t± 0.000271± 0.00027t± O.OOOOti

0.211000.02022.2121.012.02.

0.200000.01212.2102.211.10.

(3, 1153, 0)

0.110000.002100.010020.000120.001210.000010.00012

0.0110 ± O.OOOOti0.0010 ±0.101l7t0.0111 ±0.0010710.0001 ±0.0212ti0.0010 ±0.0002710.0000 ±0.00017t0.0001 ± O.OOOOti

r,,2 0.12 ±2.27t

:0.0±(l.)7t

(3, -1336, 1)

0.220000.001020.022120.001220.002100.000000.00022

0.01000.00200.02100.00100.00200.00000.0002

±0.210071±0.122Û7t±0.000l7i± 0.02007t± 0.OOO27I±0.0001n± O.OOOOti

71

Tl,2

■Î1 .2

0.20 ± 2.071

1.0±(0.)7t

0.110000.02021.2022.010.12.

0.220000.01002.2002.000.10.

Table Via. X = 3, 52,3 zeros of Lp(s, 8y/±)

00.0122220.0012010.0000210.0001200.0000100.0000110.000001

r, =Tí

Tl.3

*2,3

(3, -827, 1)

0.0200 ±0.01227t0.0112 ±0.02227f0.0010 ± 0.0021710.0000 ±0.0010ti0.0001 ±0.000l7t

0, i, =0

■■ 0.0000 ±2.210tt

: 1.011 ±0.11tc

00.2000110.020112.21000.0222.021.21.

0.011000.012200.000000.000200.000210.000000.00001

(3, 7804, 0)

0.0020 ±0.0212ti0.0120 ±0.0222tiO.OOOO ±0.000 Itc0.0002 ±0.00217!0.0002 ± O.OOOOti

T¡ =0.2100,

Tt

Í2.3

j, = 2.001

= v/=3

= 0.1 ± 2.07t

= (0.)±(1.)TI

0.011000.10100.1121.212.22.



Table VIL X = 2, a0¿0, vp(Tx -T2)>2

(3, 13564, 0)

0.0110000.0101110.0120110.0011020.0012100.0000100.0001010.0000000.0000020.0000020.000002

0.01100000.1122101.211011.11011.2211.021.10.

(

Tl.2

*1.2

V2: 0.10201 ±0.01110í

: 1.0111 ± 0.1120Í

(3, 13784, 0)

0.0220000.0111100.0202210.0012120.0020010.0000100.0002210.0000000.0000010.0000020.000001

0.02200000.2221002.111122.02120.1210.022.22.

n = n/3r,.2 =0.10122 ±0.011271

j,,2 = 2.1221 ± 0.20171

(3, -11188, 1)

0.0110000.0110000.0100000.0021120.0010210.0000100.0001120.0000020.0000020.0000010.000002

0.011000000.12222011.0120212.121011.20002.1000.120.10.

í7),2

s\.2

■ y/20.102122 ±0.012001í

1.01000 ±0.12121{

Table Vila. X = 3, ao = 0, vp(T2 - T3) > 2

(3, -2564, 1)

o0.002011200.002002010.002111010.000011120.000001100.000012220.000001120.000000100.000001000.000000010.000000000.00000001

o0.01222200.2112101.001011.20002.0121.20

1.21.

7Ï = O , si

J2,3

*2,3

O

= \/2^3= 0.20112 ±0.010l7t

= 1.2022 ± 0.210a

(3, -8804, 1)

O0.001012020.002200210.001110210.000211210.000021210.000020220.000001110.000000100.000002120.000000120.000000010.00000002

O0.02111220.2221102.111020.01000.1200.211.00.

7*1=0,

n

T2,i

s2,i

sx = O= v/3

= 0.10001 ±0.0102tt

= 2.2121 ±0.220tt


COMPUTATION OF THE ZEROS OF p-ADIC /.-FUNCTIONS. II 403

Table VIII. Various types of the Newton polygon

(5, 3101, 0)

0.0040120.0401130.0311100.0044030.0003420.0000130.0000420.000003

0.004010.423303.210121.41141.2341.314.20.

7, = 0.040243

5, =0.42140

7*2 =0.224113

s2 =2.12443

(3, 1901, 0)

0.00200.00120.01200.00100.00110.00000.0001

0.00200.0221.111.00.

= n/3,2 = 0.020 ± 0.1 lTt

2 = 0.12 ± 2.071

(3, 3512, 0)

0.011000.010220.001110.001220.000120.000000.00001

0.011000.21010.0101.120.22.

7',

S\

n

7-2.3

*2,3

= 0.1120

= 2.121

= v/3

= 0.1 ± 1.271

= (1.)±(12.);

(7, -4072, 5)

0.00150400.03343320.00210230.00504450.00060130.00004240.00000140.0000006

0.00150400.2454600.364542.04240.2600.123.02.

r,S\

n7*2.3

*2,3

= 0.0352103= 0.230602

= 0.101 ±3.45ti

= 0.163 ±22.271

-4371, 1)

00.0003030.0032340.0033020.0000030.0000210.0000030.000001

00.0032040.313013.11423.4041.304.10.

7Ï

Si

r,S3

0, s, =0

0.041020.4240

0.443404.2101

(11 , -723, 1)

0.0009aaa0.008a 1090.00629060.00084a30.00006570.00000280.0000006

00.00826790.7457591.473206.76830.1168.701.3

7',

7*2

Î2

7*3

S}

= 0, s, = 0= 0.0246697

= 0.33543a= 0.42a5014

= 6.7852a0

(5, 12056, 2)

0.00424120.00042210.01342440.00213040.00002100.00000020.00003020.0000043

0.00424120.0040011.221401.11100.1121.214.21.

71 = VI7*1,2 = 0.000443 ± 0.104207t

5,,2 = 0.02023 ± 1.222Itt

(3, -25528, 1)

0.022000.022210.000000.001200.000020.00000

0.022000.22100.2100.220.11.

7',

S\

It

37*2.3

32*2,3

n1

37*4,5

32-*4*. 5

(3, 1541, 0)

01100002120022100001000020000100001

0.011000.01210.2100.201.20.

71 = ^=3

7*|,2 =0.1 ±(0.)7t + (l.)Tr ±(l.)7t3

Si ,2 = ±l/7t (mod 7t)

.. ' r . 4/ ^71 = Í4 " V -3

7*3.4 = 0.1 ± (O.)tt' + (I.)tt'2 ± (I.)ti'3

S3.4 = ±1/ti' (mod n')

(5, -7816, 1)

00.00020220.00013110.00302110.00034430.00003200.00000110.00000130.0000003

00.0020440.012323.00301.1321.302.33.

'i 0 , s, = 0

n = \/5

7*2 3 =0.04202 ±0.121l7t

s2,3 = 0.4414 ± 1.23071

= 0.2120= 2.111

= v/3

= 0.0±0.l7t + (0.)7t2 ±(0.)7t3

= 0.0 ± 0.27t + (0.)7t2 ±(0.)7t3

= Í4-v-3

= 0.0±0.l7t' + (0.)7t'2 ±(0.)7t'3

= 0.0±0.2ti' + (0.)7t'2 ±(0.)ti'3

(5, 4924, 2)

0.03133210.03120320.00200040.00241020.00010130.00001340.00000200.00000440.0000001

0.03133210.2212130.120023.04304.0043.133.20.

7*1

S|

n7*2.3

•52,3

0.110124

4.43041

v-5:0.22± 1.2271

: 1.3 ±44.3ti

(3, 4204, 0)

0.011000.001210.000120.000220.000120.000000.00002

0.011000.01010.0220.021.21.

7*1.2

7*3,4

\/3. ( = V20.0+((l.)±(l.){)7t

0.0-((l.)±(l.)f)7t



Table IX. Zeros in wildly ramified cubic extensions

(3, -3592, 1)

0.22000000.02211200.00200120.00022210.00002010.00000000.00001210.00000020.00000010.0000001

0.22000000.2121020.122012.10121.2011.202.00.

Ttfc zero of A-3 -

(A: = 1,2,3),

Tk = -T-k - "Í

(mod 7t|)

ÎX

(3, 2504, 0)

0.22000000.01200120.00122010.00100200.00021020.00002200.00000010.00000120.00000020.00000220.0000001

0.22000000.2202020.000201.11121.1121.022.21.

7i zero of X} + 3X2 - 12,

T, = Tt - Tt3,

7*2 = 71 + 7t2 + 7I3 ,

7*3 = 7t - K2 - 7i3 (mod 7i4)

(3, 1781, 0)

0.01100000.00110210.00202210.00202100.00020120.00000110.00002220.00000100.00000010.00000110.0000001

0.01100000.0211200.220011.21000.2011.200.21.

7t,i zero of X1 -

(k = 1, 2, 3),T- — 2 3Tk = -nlk - n'k

(mod nl)

IX

(3, 4172, 0)

0.12100000.02001110.00110200.00222210.00000220.00002120.00002210.00000010.00000020.0000011

0.12100000.1101010.211102.00222.2121.122.02.

nk zero of X1

(k= 1, 2, 3),

Tk = nk + k\

(mod Tt3.)

12

(3, 281, 0)

0.20000000.02001220.00100100.00010110.00010020.00002210.00001210.00000210.00000020.00000010.0000002

0.20000000.1001110.202001.10110.0201.221.02.

Ttt zero of X3 + 3X

(£ = 1,2,3),

Tk = Kk + n¡ + n*k

(mod 7t^)

(3, -1144, 1)

0.22000000.02102000.00020020.00012220.00020100.00000100.00001110.00000010.00000000.00000020.0000001

0.22000000.2010010.212212.11202.1112.111.12.

nk zero of X3 + 6

(it = 1, 2, 3),

Tk = -K* + "k

(mod Tt3.)

(3, -311, 1)

00.02121210.00210220.00122110.00011200.00002020.00000200.00000110.00000010.00000200.0000002

00.1020100.022000.22122.1022.221.00.

r\ = o, s, = o

nk zero of X3 + 3X ■

(/c = 2,3,4),

Tk=nk + n¡ + %\

(mod n*)

(3, -2132, 1)

00.00110010.00222020.00010120.00021120.00002220.00002210.00000210.00000010.00000010.0000001

00.0211010.211020.21210.0011.112.00.

T, = 0, s, = 0

7*2 = 0.211, s2 = 1.12

nk zero of X3 + IX - 3

(jfc = 3,4, 5),

Tk = %k (mod n\)



Table X. Zeros in wildly ramified sextic extensions

(3, -7108, 1)

0.121000000.020100220.001110220.001121100.000122110.000020120.000010020.000001000.000000220.000002110.000000220.000000010.00000001

0.121000000.21220120.0022120.212200.02100.2012.122.11.

kx , Tii, it-i zeros of A"3 - VÏX - %/3

7*! s -Tti + n2 (mod 7C3)

Ti = -ni + n\ (mod n\)

7*j = -7T3 + n2 (mod n\)

iti,, Tt5 , Ttf, zeros of A-3 + \ThX - VÎ

T4 = Tt4 + n\ (mod 7C3)

7*5 = tc5 + it2 (mod tt.3,)

T6 = 7t6 + xj (mod n\)

(3, -1832, 1)

00.0111010220.0011222110.0020002110.0001122110.0000101220.0000120110.0000022220.0000002110.0000012010.0000002020.0000000220.0000000100.000000002

00.221101100.01011010.1101010.210010.11120.2211.001.01

7*, = 0 , sx = 0

it2 . ni. ^4 zeros of A-3 - \fi>X - V3

T2 = 7i2 + n2 (mod n\)

7*3 = 7Î3 + n2 (mod 7t|)

7*4 =: 7t4 + tt^ (mod 7rJ)

7i5, 7t6 , 7i7 zeros of A"3 + v^A" - v^3

T¡ = -n¡ + n2 (mod 7r|)

7*6 = -7t6 + tt| (mod 7T3)

7*7 =i -7t7 + tí 2 (mod nl)

(3, -20692, 1)

0.121000000.000111010.000022200.000002010.000002010.000000110.000001210.000000200.000000010.00000001

0.121000000.00120000.0001020.012010.00200.0101.002.02.

7tx , iti, ni zeros of X3 -3X

Tx = -n 7T, (mod n

T2 = -iti - n2 (mod n\)

7*3 = -tit, - it2 (mod n\)

7t4 , 7r5 , 7t6 zeros of A-3 + v/-3A" -

7*4 = 714- it\ (mod 7T3)

75 s ne, - Tt2 (mod 7i|)

T6 = n6- n\ (mod tc3)

^3



Bibliography

1. S. Amano, Eisenstein equations of degree p in a p-adic field, J. Fac. Sei. Univ. Tokyo Sect.

lAMath. 18(1971), 1-21.

2. R. Ernvall and T. Metsänkylä, Computation of the zeros of p-adic L-functions, Math. Comp.

58 (1992), 815-830; Supplement, S37-S53.

3. L. C. Washington, Introduction to cyclotomic fields, Springer, Berlin and New York, 1982.

Department of Mathematics, University of Turku, SF-20500 Turku, Finland

E-mail address: taumetsQsara.cc.utu.fi


computation of the zeros of p-adic ¿functions. iizeros of the kubota-leopoldt p-adic l-functions...

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