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$\begingroup$

Ten years old question asks about Computing (on a computer) the first few (non-trivial) zeros of the zeta function of a number field.

Let $K$ be the number field with the degree 24 defining polynomial

 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

and $\zeta_K$ be the zeta function of $K$.

According to computations with Magma online $\zeta_K$ has double zero at $1/2$, but this could be numerical instability with low precision.

Q1 Does $\zeta_K$ really have double zero at $1/2$?

Q2 Does the double zero implies some properties of $K$?

Q3 Is there construction of number fields which have double or higher order zeros of zeta function at $1/2$?

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 linked question (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.

Q4 Are there other, preferably infinitely many, double zeros of $\zeta_K$?

Magma online code in case someone is interested:

 Z1<x>:=PolynomialRing(Integers());
 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;
 Nf<w>:=NumberField(p1);
 L:=LSeries(Nf);
 I:=Sqrt(-1);
 //t0:=0.99014365233;
 //t0:=1.38830360231;
 t0:=2.35103235859;
 A:=1/2+I*t0;
 Evaluate(L,A);
 Evaluate(L,A : Derivative:=1);
$\endgroup$
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  • $\begingroup$ Is this field Galois? $\endgroup$
    – Will Sawin
    Sep 21, 2021 at 12:34
  • $\begingroup$ Kevin Buzzard asked a question about this polynomial earlier mathoverflow.net/questions/63544 . According to him, it is Galois with Galois group $SL_2(\mathbb{F}_3)$. $\endgroup$ Sep 21, 2021 at 12:36
  • 2
    $\begingroup$ Here is a suggestion: This group has a two-dimensional irrep $\chi$ with real character values people.maths.bris.ac.uk/~matyd/GroupNames/1/SL(2,3).html . Then $L(s,\chi)^2$ should divide $\zeta_K$, so it would be enough to show that $L(1/2, \chi)=0$. Moreover, as $\chi$ is real valued, one should (if my guess is right) be able to use the intermediate value theorem to show that $L(s, \chi)$ has an odd number of zeroes in $(0,1)$, after which the functional equation should force $L(1/2, \chi)$ to vanish. $\endgroup$ Sep 21, 2021 at 12:42
  • 2
    $\begingroup$ J.V. Armitage, Zeta functions with a zero at $s=1/2$, Inv. Math. 15, 199–205 (1971). link.springer.com/article/10.1007/BF01404125 The point is that some of the constituent Artin representations have an odd sign of functional equation. There is an example in the Magma Handbook (largely due to Serre in its construction). magma.maths.usyd.edu.au/magma/handbook/text/1585#18264 $\endgroup$
    – user334725
    Sep 21, 2021 at 12:43
  • 2
    $\begingroup$ I'd expect a symplectic representation has a 50% chance of having odd functional equation, though proving it would be nontrivial. So maybe a subpart of Q3 is to show that there are symplectic irreps (of realized Galois groups) of arbitrarily large degree. $\endgroup$
    – user334725
    Sep 21, 2021 at 13:44

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