Let $M$ be the splitting field of

x^8 + 3*x^7 + 13*x^6 + 17*x^5 + 45*x^4 + 37*x^3 + 11*x^2 + 112*x + 108

over the rationals. If I've understood some tables correctly, the splitting field is (of course) Galois over the rationals, with Galois group isomorphic to $SL(2,\mathbf{Z}/3\mathbf{Z})$.

How might I go about computing (on a computer) the first few (say, ten or so) zeros of the zeta function of this field on the half-line $1/2+it$, $t\geq0$?

That's the question, here's the obligatory extra blurb.

I asked this question on math.stackexchange but got no answer (yet). Here's the link: https://math.stackexchange.com/questions/35941/computing-on-a-computer-the-first-few-non-trivial-zeros-of-the-zeta-function

I tried to compute with the zeta function of $M$ in PARI-GP but zetakinit failed before it even got going. I tried on magma and I could (slowly, and to not much accuracy) compute some values of $\zeta_M(1/2+it)$---but each computation took a while and I didn't really know how to go from "I am computing values of this function" to "I am finding zeros of this function"---it was the latter that I wanted to do.

I ask for the following quite stupid/naive reason. $SL(2,3)$ has a 2-dimensional complex representation which is not induced from a character (it in fact has three such reps). So $Gal(M/\mathbf{Q})$ has a 2-dimensional representation which isn't induced from a character, and hence the analyticity of the Artin $L$-function associated to this representation is not immediately obvious from Hecke/Tate: one instead needs Langlands' result about $A_4$ Galois representations. I unravelled what this said explicitly yesterday, and, unsurprisingly, it boils down to statements vaguely of the form "all the zeros of the zeta function of this number field are also going to be zeros of the zeta function of either that field or this field", where all the fields in question are subfields of the field $M$ above. I just wanted to "really see this happening" so I could look on in wonderment at a "concrete" application of cyclic base change.

More details, for anyone interested: $M$ is obtained by adjoining one root of

x^24 + 3*x^23 - 2*x^22 - 43*x^21 + 81*x^20 + 1579*x^19 + 2434*x^18 - 5192*x^17 + 4678*x^16 - 41425*x^15 + 423527*x^14 + 1352722*x^13 + 5199537*x^12 - 13364304*x^11 - 138065100*x^10 + 228783352*x^9 + 1254448448*x^8 - 3179566016*x^7 + 4205123840*x^6 + 139822208*x^5 - 31439415040*x^4 + 28607489536*x^3 + 330701977600*x^2 - 807251576832*x + 635017424896

to the rationals. If $N$, $R$, $L$ and $K$ are subfields of $M$ of degree $8,12,6,4$ over $\mathbf{Q}$ respectively, and if I got the combinatorics right, then Langlands implies $\zeta_M\zeta_K^2/(\zeta_N^2\zeta_R)$ is entire, hence any zero in the denominator is magically cancelled by a zero in the numerator. if I've got the combinatorics right then this statement should be not be a consequence of the Hecke/Tate theory (that 1-dimensional $L$-functions are holomorphic) but should lie truly deeper. Furthermore, some analogous question where $SL(2,3)$ is replaced by $SL(2,5)$ should be actually inaccessible (at least if $M$ is totally real) because Artin's conjecture is open in this setting. Can one even compute far enough to see Artin's conjecture "looking true" in a non-solvable case? Not sure.

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    $\begingroup$ I though L-calc (Mike Rubinstein's package) now in SAGE was Taylor-made for this. Then again, maybe the conductor is too big. It take sqrt(conductor) time to compute it. There is also some ideas of S. Omar to isolate low zeros, using explicit formula. portal.acm.org/citation.cfm?id=1789756 $\endgroup$
    – Junkie
    Apr 30, 2011 at 18:27
  • $\begingroup$ Magma (Dokchitsers) almost instantly divides up the L-function from the degree 24 polynomial into 7 Artin parts, with conductor at most $163^2$. All CheckFunctionalEquations are operable to 30 digits in less than a second. At this point, just feeding the coefficients to Rubinstein should avail. $\endgroup$
    – Junkie
    Apr 30, 2011 at 18:32
  • $\begingroup$ So the question is: what gives me confidence that these Artin L-functions are analytic? From a philosophy of science, one answer is: the CheckFunctionalEquation of the Dokchitsers says so, and with poles at a low height, this would not be true. This is just as valid as evidence as subsets of zeros cancelling, I would suspect. Another idea is to: for each nonlinear Artin L-function, isolate it as a $\zeta$-quotient. So then the 3-dim constituent is, your notation, $\zeta_K/\zeta$ where $K$ is quartic. Computing zeros of $\zeta_K$ is feasible, see if they contain zeros of $\zeta$. Worthwhile? $\endgroup$
    – Junkie
    May 1, 2011 at 16:48
  • $\begingroup$ Now I just had a new idea (see edit in answer below too). I perceive that $\zeta_M$ splits as a product of 3 degree 8 $L$-functions, from the octic field, upon twists by linear characters. Writing $\sigma$ for the 8-dimensional Galois representation of this octic field $N$, I compute that $\zeta_M=L(\sigma)L(\sigma\omega)L(\sigma\bar\omega)$ where $\omega$ is a nontrivial linear character, and $L(\sigma)=\zeta_N$ of course. Each of these three $L$-functions is known easily(?) to be analytic away from the pole, so this gives aside information, and the zeros for these 3 are probably computable. $\endgroup$
    – Junkie
    May 2, 2011 at 2:27

3 Answers 3


Step I: Put the degree 24 polynomial into Magma, make it a number field, and call LSeries on it. This divides the $L$-function into a product of 7 distinct ones (Dokchitsers code, under an attribute called "prod" on the L-series object), given by Artin representations. So my plan was to compute zeros for each of these $L$-functions, with them being a subset of those of $\zeta_M$ naturally, and recombine. As representations for $SL(2,3)$ this is as $$1\oplus \omega\oplus\bar\omega\oplus 2\tau_2\oplus 2\tau_2\omega\oplus 2\tau_2\bar\omega\oplus 3\kappa_3.$$ Or as an $L$-function product $$\zeta_M=\zeta\cdot L(\omega)\cdot L(\bar\omega)\cdot L(\tau_2)^2\cdot L(\tau_2\omega)^2\cdot L(\tau_2\bar\omega)^2\cdot L(\kappa_3)^3.$$

EDIT: Oh I see now, you wanted it for $M$ w/o assuming analyticity of Artin constituent parts, but Magma automatically dissembles it, and assumes. But now at this point, I see you were trying to avoid this decomposition perhaps, but then I really don't see how you could work with $\zeta_M$ directly, as the conductor is too big, being $163^{16}$. From the standpoint of scientific evidence, it is probably enough that CheckFunctionalEquation in Magma for each of the above constituents gives an answer indistinguishable from zero, though the feel of zeros is also nice.

Another way to test individual analyticity is to use subfields. There is a unique (up to isomorphism) quartic subfield $K_4\subset M$, with $$\zeta_K=\zeta\cdot L(\kappa_3).$$ The conductor of $\zeta_K$ is small enough to compute with it directly (no decomposition), so the analyticity of $L(\kappa_3)=\zeta_K/\zeta$ can be checked (numerically) by seeing if Riemann zeros are a subset of those of $\zeta_K$. EDIT: Of course, $\kappa_3$ is induced from the a nontrivial linear character of the $Q_8$ subgroup of $SL(2,3)$, so analyticity already follows by induction.

There is also a unique (up to isomorphism) sextic subfield $L_6\subset M$ with $$\zeta_L=\zeta\cdot L(\omega)\cdot L(\bar\omega)\cdot L(\kappa_3),$$ so the linear parts can be isolated here by $\zeta_L/\zeta_K$, though maybe superfluous ("easy" theorem)? In that matter, already the cubic subfield $C_3\subset M$ has $\zeta_C=\zeta\cdot L(\omega)\cdot L(\bar\omega)$ if desired.

There is a unique (up to isomorphism) octic subfield $N_8\subset M$ with $$\zeta_N=\zeta\cdot L(\tau_2\omega)\cdot L(\tau_2\bar\omega)\cdot L(\kappa_3).$$ The integer ring has discriminant $163^4$ so $\zeta_N$ is likely still computable directly. By quotient $\zeta_N/\zeta_K$, this can tell about the product of conjugate 2-dimensional $L$-functions. EDIT: see below for another idea, using twists of $\zeta_N$.

Finally the unique duodecic subfield $R_{12}\subset M$ has $$\zeta_R=\zeta\cdot L(\omega)\cdot L(\bar\omega)\cdot L(\kappa_3)^3.$$ So this gives nothing new, and the discriminant is too large anyway.

Note that none of the parts has $L(\tau_2)$ directly, only $\zeta_M$ itself that is too hard to compute. EDIT: However, by Rankin-Selberg I think(?) it follows that the analyticity of $L(\tau_2)$ is equivalent to that after twisting by $\omega$ to get $L(\tau_2\omega)$.

Answer?: For that matter, recurring to the octic subfield $N_8$, instead of using just the Dedekind $\zeta$-function of $N_8$, twisting it by $\omega$ could be profitable, achieving $$L(\sigma_8\omega)=L(\omega)\cdot L(\tau_2\bar\omega)\cdot L(\tau_2)\cdot L(\kappa_3),$$ $$L(\sigma_8\bar\omega)=L(\bar\omega)\cdot L(\tau_2)\cdot L(\tau_2\omega)\cdot L(\kappa_3),$$ where $\sigma_8=1\oplus\tau_2\omega\oplus\tau_2\bar\omega\oplus\kappa_3$ is for the Dedekind of $N_8$. Also $\kappa_3\omega=\kappa_3$ nicely. Noting here the twisting gives $$L(\sigma_8)L(\sigma_8\omega)L(\sigma_8\bar\omega)=\zeta_M,$$ this could provide a way to compute $\zeta_M$, as each part of the left is known as analytic I presume. The conductor of $L(\sigma_8)$ is $163^4$, and that for the twists is $163^6$. Or one can avoid $\zeta_M$ alternatively, as my computation is $$L(\tau_2)^2={L(\sigma_8\omega)L(\sigma_8\bar\omega)\zeta^2\over\zeta_{N_8}\zeta_{L_6}},$$ where each factor on the right should be known (easily?) to be holomorphic away from the $\zeta$-pole. Note this exploits the linear characters of $SL(2,3)$, and you have none for your next case of $SL(2,5)$.

In all cases, zeros need to be computed, and the right tool is L-calc. But I don't know if it is really feasible to go too far for $L(\sigma_8\omega)$ of conductor $163^6$, without intensive effort.

Part II: Zeros of $L(\tau_2)$ computation (example): I compute its first few zeros for the 2-dimensional representation of real character. For this representation, with $10^5$ coeffs (taking 6sec in Magma), I exported these to Lcalc (a somewhat hackish tool of M. O. Rubinstein, included in Sage I suppose but I did it direct), which returns after 12 seconds, listing the imaginary parts of the first 10 zeros on the half-line (also proving RH up to that point by A Turing method):


The first zero is at $s=1/2$ as the sign is $-1$. The second is about $s\approx 1/2+0.990i$

For reference, here are the L-calc settings I used, hackish as I say:

0.5 0.0
0.5 0.0
-1.0 0.0

As explained in their help, the first "1" says the coeffs are integers, the second "0" says they are not special, the 3rd "100000" says they are that many, the 4th "0" says not periodic, the 5th "2" says degree 2, the 6th ".5" and "0.5 0.0" say the form of the gamma factor, the "51.884" is $\sqrt{163^2/\pi^2}$ as the analytic conductor, the "-1.0 0.0" is the sign, and the final "0" says no poles. Then the 100000 coeffs are given as integers.

  • $\begingroup$ I think there might be a misunderstanding here. I am asking how to compute the zeros of the (Dedekind) zeta function of $M$ on the critical line. I don't know what lcalc is (although of course I could probably try to find out, when I'm not putting kids to bed) but my guess is that even though there are a lot of numbers in your answer, they're not the numbers I'm interested in. Or have I misunderstood your answer? I am wary of the fact that it starts with a statement which is true but whose relevance I don't understand... $\endgroup$ Apr 30, 2011 at 19:19
  • $\begingroup$ I'm guessing these are not exactly what you want, as these are imaginary parts of zeros of $L$-function of the 2-dim Artin representation. What L-calc does is list the imaginary parts of zeros on the half-line, incidentally proving RH by a Turing condition. I ran the above with "-z 10" to get 10 zeros. The first is at $s=1/2+0i$, via the sign being $-1$. $\endgroup$
    – Junkie
    Apr 30, 2011 at 19:31
  • $\begingroup$ I see now. Google has also told me what Lcalc is. Funnily enough the imaginary parts of the zeros of the L-function of the 2-d Artin rep are precisely the zeros that I do not want :-) I want to see all the zeros that are cancelling! But I see how to do it now from what you've said. Unfortunately, doing the calculation this way will, I think, just take me around in a circle :-( because the cancellation I want to see will probably be assumed in the calculation---it's equivalent to the statement that the real 2-d L-function you were computing with has no poles. $\endgroup$ Apr 30, 2011 at 19:43
  • $\begingroup$ In summary, what I was asking was "how do I compute the zeros of this $L$-function?" and your answer is "write it as a product of seven $L$-functions, one of which has a simple pole at 1 and no other poles, and the other six are analytic but this is a profound theorem in some cases. Now compute the zeros of all seven (using programs that probably assume there are no poles anyway) and take the union". Unfortunately I wanted to do the calculation to give an "independent check" that all six were analytic! :-) $\endgroup$ Apr 30, 2011 at 19:47
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    $\begingroup$ [remark: the answer has now been clarified and in particular my earlier comment about "a statement which is true but whose relevance I don't understand" is no longer valid] $\endgroup$ Apr 30, 2011 at 22:28

You might want to download the package ComputeL developed by Tim Dokchitser, see


It is PARI based, and will compute the Dedekind zeta function of a number field defined by a polynomial. See the example file ex-nf.

EDIT: I see the previous answer already referred to Tim Dokchitser's package, via implementations into Magma. Still, the stand alone package says it does something other than what Magma does, namely compute the full Dedekind zeta function rather than decomposing it into a product of Artin L-functions.

Nonetheless, as pointed out above, the fact that the conductor is $163^{16}$ makes the computation impractical.


First we observed that $\zeta_K$ numerically has double zero at $1/2$ and asked question about it.

In comment, David E Speyer suggested $L(s,\chi)^2$ should divide $\zeta_K$.

We believe that all zeros of $L(s,\chi)$ will be double zeros of $\zeta_K$.

Checking the first three zeros of the other answer (they are with very low precision) give $|\zeta_K(\rho)| \sim 10^{-22}$ and $|\zeta_K'(\rho)| \sim 10^{-11}$. This numerical results appear to suggest that the first three complex zeros are double.

Magma online code in case someone is interested:

 p1:=x^24 + 3*x^23 - 2*x^22 - 43*x^21 + 81*x^20 + 1579*x^19 + 2434*x^18 - 5192*x^17 + 4678*x^16 - 41425*x^15 + 423527*x^14 + 1352722*x^13 + 5199537*x^12 - 13364304*x^11 - 138065100*x^10 + 228783352*x^9 + 1254448448*x^8 - 3179566016*x^7 + 4205123840*x^6 + 139822208*x^5 - 31439415040*x^4 + 28607489536*x^3 + 330701977600*x^2 - 807251576832*x + 635017424896;
 Evaluate(L,A : Derivative:=1);

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