Tables of Values of Dedekind Zeta Functions


Explanations about how the values are calculated.
 

Riemann's Zeta Function.

The totally real subfield of a cyclotomic field.
Of roots of unity of order  5
Of roots of unity of order  7
Of roots of unity of order  8
Of roots of unity of order  11
Of roots of unity of order  12
Of roots of unity of order  13

Quadratic Fields.
discriminant  5
discriminant  8
discriminant  12
discriminant  13
discriminant  17
discriminant  21
discriminant  24
discriminant  28
discriminant  29
discriminant  33
discriminant  37
discriminant  40
discriminant  41
discriminant  44

Cyclic Cubic Fields
Of Q(exp(2*Pi*i/7))
Of Q(exp(2*Pi*i/13))
Of Q(exp(2*Pi*i/19))
 

Non-Galois Cubic Fields.
x^3 - 9*x - 6
x^3 - 9*x - 3
x^3 - x^2 - 7*x + 2
x^3 - 11*x - 11
 
 
 
 
 

The  calculations of values of zeta functions were done using PARI (Version 2.0.17 (beta), Windows NT ix86 (ix86 kernel), 32-bit version, running on Windows NT '98). A problem is that the size of the values grows very fast. It was found, experimentally, that setting the precision by  \p 180  works best (for a larger precision the answer PARI supplies is usually wrong!). Even at this precision at some point PARI announces "precision loss in truncation" and very often produces wrong answers before this annoucement is made. One easily recognizes that the result is wrong by a sharp increase in its size and by the denominator being way too big (either in size or having wrong primes dividing it).
 

Calculating Riemann's zeta function.
(09:16) \p2000         (No problem with precision in this function)
realprecision = 2003 significant digits (2000 digits displayed)
(09:16) for(i=1,100,a=round(zeta(1-2*i)*(6*i)!)/(6*i)!;print("zeta(1-",2*i,") = ",a," ",factor(denominator(a))))

Calculating the minimal polynomial.
Here is an example for roots of unity of order 8. (No need arose for a general function).
? ? a=exp(2*Pi*I/8);
? ? round((x-2*real(a))*(x-2*real(a^3)))
%4 = x^2 - 2

Initializing a number field and obtaining some invariants.
? k = nfinit(x^2-5);
? k[3]        (The discriminant)
? 5

Calculating the factorization of primes in the totally real field.
? ? k=nfinit(x^2-5);
? ? forprime(i=2,32,a=idealprincipal(k,i);print(i,"  ",idealfactor(k,a)))
2  Mat([[2, [2, 0]~, 1, 2, [1, 0]~], 1])
3  Mat([[3, [3, 0]~, 1, 2, [1, 0]~], 1])
5  Mat([[5, [-1, 2]~, 2, 1, [-1, 2]~], 2])
7  Mat([[7, [7, 0]~, 1, 2, [1, 0]~], 1])
11  [[11, [-5, 2]~, 1, 1, [3, 2]~], 1; [11, [3, 2]~, 1, 1, [-5, 2]~], 1]
13  Mat([[13, [13, 0]~, 1, 2, [1, 0]~], 1])
17  Mat([[17, [17, 0]~, 1, 2, [1, 0]~], 1])
19  [[19, [-10, 2]~, 1, 1, [8, 2]~], 1; [19, [8, 2]~, 1, 1, [9, 2]~], 1]
23  Mat([[23, [23, 0]~, 1, 2, [1, 0]~], 1])
29  [[29, [-12, 2]~, 1, 1, [10, 2]~], 1; [29, [10, 2]~, 1, 1, [-12, 2]~], 1]
31  [[31, [-7, 2]~, 1, 1, [5, 2]~], 1; [31, [5, 2]~, 1, 1, [-7, 2]~], 1]

Calculating the value. The script used is very simple. Here is an example
? ? \p180
   realprecision = 183 significant digits (180 digits displayed)
? ? k=zetakinit(x^5 + x^4 - 4*x^3 - 3*x^2 + 3*x + 1);for(i=1,16,t=round(zetak(k,1-2*i)*32!)/32!;print("z(",(1-2*i),") =",t,"=",factor(denominator(t))))
z(-1)=-20/33  [3, 1; 11, 1]
z(-3)=1695622/165  [3, 1; 5, 1; 11, 1]
z(-5)=-50936925341420/693  [3, 2; 7, 1; 11, 1]
z(-7) =3543010400763352360091/165=[3, 1; 5, 1; 11, 1]
z(-9) =-3064839249175370545023293582020/33=[3, 1; 11, 1]
z(-11) =150607491678457644611241613057523401954646722/45045=[3, 2; 5, 1; 7, 1; 11, 1; 13, 1]
z(-13) =-22786609701564620802831984165768429157292386800423020/33=[3, 1; 11, 1]
z(-15) =3536453128145555872124660832571626062092079545106341991126179692667/5610=[2, 1; 3, 1; 5, 1; 11, 1; 17, 1]
z(-17) =-82870594275907635610187436559268476911703227525756627620509217091783293008291940/39501=[3, 3; 7, 1; 11, 1; 19, 1]
z(-19) =18101711521606517304423113131909702437120441255887075037380681034586375053865444166752853062/825=[3, 1; 5, 2; 11, 1]
z(-21) =-5351670936024492939166666065913000091973169152733384440393242319773760895634658032449754173452447868870060/8349=[3, 1; 11, 2; 23, 1]
z(-23)=3933397627801024785773628245144706556399758829607091351873360917366068054009441081200629620752513951663521629804029045786187829996892580542856105913948216758885700075519999999513/82759437283573817805748593649155833590143242364518400000000000=[2,48; 3, 21; 5, 11; 7, 6; 11, 4; 13, 3; 17, 2; 19, 2; 23, 2; 29, 1; 31, 1; 37, 1; 41, 1; 43, 1; 47, 1] (Note this result is clearly wrong).
  ***   precision loss in truncation

Non-cyclic cubics. I wanted fields with small discriminant. To find examples I used John Johnes data (http://math.la.asu.edu/~jj/numberfields/), that I assembled into one big list L and ran
for(i=1,100,if(poldegree(L[i])==2,,k=nfinit(L[i]);print(L[i],"   ",polgalois(L[i]), "   ", k[2],"   ",k[3])))