# Special value of Dedekind Zeta Functions

Is there a way to compute the values of the following special kind of Dedekind Zeta Function:

Assume $a,b \in \mathbb N$ and $K = \mathbb Q(\sqrt{5}) \subset L= \mathbb Q(\sqrt{a+b\sqrt{5}})$. Set $d = \sqrt{a+b\sqrt{5}}$ which is also totally negative and furthermore a discriminant of L.

Now my problem is, can I calculate the (exact or numerical) values of the Dedekind Zeta function $\zeta_L(s)$ for s positive odd integer?

Thanks for any help!

edit retag close merge delete

Sort by ยป oldest newest most voted

Let us get an explicit field $L$, i will ignore any restriction on $a,b$.

a, b = 7, 3    # chosen so that we have a nice discriminant below
L.<u> = NumberField(sqrt(a + b*sqrt(5)).minpoly())
print(f"L is <{L}> with discriminant {L.disc()}")


So our test situation is:

L is <Number Field in u with defining polynomial x^4 - 14*x^2 + 4> with discriminant 1600


Numerically i tried then with a precision equal to $300$ (bits):

sage: LZ = L.zeta_function(prec=300)
sage: LZ(3)
1.01897545804910243658865624113355956226224991920025923733956382609692854856490756716927418
sage: LZ(5)
1.00101170969216360222562623040290551783199837820002159868621703090382967126817104154134429
sage: LZ(7)
1.00006145739122202145223364281491843791554787649331378421635463305298859721049129811714303


Note that the code goes to pari per default. So let us see what happens in there exactly:

? lfun(x^4 - 14*x^2 + 4, 3)
%42 = 1.0189754580491024365886562411335595623
? \p 100
realprecision = 115 significant digits (100 digits displayed)
? lfun(x^4 - 14*x^2 + 4, 3)
%43 = 1.018975458049102436588656241133559562262249919200259237339563826096928548564907567169274179589141566


So an alternative is

sage: f = (sqrt(7 + 3*sqrt(5))).minpoly()
sage: pari.lfun(f, 3)
1.01897545804910


(And we can go also with similar code to have a better control of the precision.)

more

I hoped to get at least $\zeta_L(3)$ as a linear combination with rational coefficients of values of the trilogarithm computed in points of $L$... no success so far, but i will come back. The discriminant also shows in the formula through its square root, this being the reason for the choice of the sample field $L$.

( 2024-02-17 22:07:01 +0200 )edit

Hi dan, thanks a lot for your answer! This is very helpful!

( 2024-02-18 09:21:38 +0200 )edit