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.