$\DeclareMathOperator\Frob{Frob}$Iwaniec and Kowalski, in their famous book Analytic Number Theory states a strong form
of the effective Chebotarev density theorem page 143, and prove it assuming both GRH for Artin's $L$-function and Artin's conjecture
(that the Artin $L$-function have no pole except maybe at $s=1$). But they also add, with no argument nor reference,
that the same theorem is true assuming only GRH, not the Artin's conjecture. I'd like to know how this can be proved.
Let me explicit my question for a reader that would not want to open Iwaniec and Kowalski's book. Let $L/\mathbb Q$ be a finite extension
Galois group $G$, and let $\rho$ be an irreducible complex representation of $G$, that one can as well suppose non-trivial (as Artin's conjecture is known for the trivial representation). For $x>0$ a real, let $\psi (\rho,x) = \sum_{p^n < x} \operatorname{tr}\rho(\Frob_p^n) \log p$, where the sum is on all prime power of the form $p^n$, where the prime $p$ is unramified for $\rho$. Then the crucial point in proving Iwaniec-Kowalski's effective Chebotarev is the following estimate
$$(1) \quad \psi(\rho,x) = O(x^{1/2} \log x \log x^d q(\rho)),$$
where $q(\rho)$ is the Artin's conductor of $\rho$ and the implied constant is absolute. Estimate (1) is proved under GRH and Artin in Iwaniec-Kowalski (Theorem 5.15)
How to prove (1) under GRH alone, as Iwaniec and Kowalski suggest is possible ?
I have already given some thoughts to the question, but I am not able to solve it. Surely one of the ideas involved should be the following:
even assuming only GRH for the Artin's $L$-function, not Artin's conjecture, we know that $L(\rho,s)$ has no pole except
maybe on the critical line, for by the theorem of Brauer, $L(\rho,s)$ is a quotient of products of Hecke $L$-functions, and those functions have no poles on the critical strip except perhaps at $s=1$ by a deep result of Hecke, and no zeros either, under GRH, except maybe on the critical line. Moreover, one also knows using the same argument of Brauer that $L(\rho,s)$ is
meromorphic function of order $1$, that is quotient of two entire functions of order $1$, and therefore that $L(\rho,s)$ has a nice Weierstrass product formula (like (5.23)).
Then the natural way to go is to try to follow the proof given by Iwaniec and Kowalski
under GRH and Artin, using the above remarks to avoid using Artin. I see several issues with that method, the most important being the following:
a crucial step in the method is the estimate of the number of zeros $N(T)$ (with their positive multiplicity) of $L(\rho,s)$ on the critical segment between $s = 1/2 -iT $ and $s=1/2 + iT$ – see Theorem 5.8.
This estimate is obtained by integrating $L'/L$ on a suitable rectangle intersecting the critical line on that segment. Yet in the presence of poles
this method will not count the number $N(T)$ of zeros, but the difference $N(T)-P(T)$ where $P(T)$ is the number of poles (with their multiplicity) on the same segment. So even a good estimate for $L'/L$ hence of $N(T)-P(T)$ will not prevent $N(T)$ and $P(T)$ to be arbitrary large. Then, when one computes $\psi(\rho,x)$ by some explicit formula (such as (5.53)), both zeros and poles contribute and if these are too many, that is if $N(T)+P(T)$ is too large, the precise estimate (1) will be completely ruined.
Edit after Sausage and Frank's comments and Denis's answer : I will answer Frank's question and this will allow me to explain why Sausage and Denis's suggestion are not sufficient (or so I think — perhaps I am missing something). The estimate (1) of Lagarias-Odlyzko leads easily by linear combination and then applying Cauchy-Schwarz and the fact the sum of the square of the dimension of irreducible representation of $G$ is $|G|$ to the following form of Chebotarev, for $C$ a subset of $G$ invariant by conjugation, and $M$ the product of primes
ramified in $L$ :
$$(2) \quad \psi(C,x) = \frac{|C|}{|G|}Li(x) + O \left( \sqrt{x} \sqrt{|C|} \log x (\log x + \log M + \log |G|)\right).$$
Here $\psi(C,s) = \sum_{p^n < x, \Frob_p^n \in C} \log p$.
Formula (2) is stated in a slightly weaker form page 144 of Iwaniec-Kowalski, weaker just because they use a substandard majoration for discriminant. Formula (2) was also stated and proved earlier, by Murty, Murty, and Saradha under GRH and Artin: see Modular Forms and the Chebotarev Density Theorem,
American Journal of Mathematics, Vol. 110, No. 2 (Apr., 1988), pp. 253-281. Curiously, this paper is not mentioned in Iwaniec-Kowalski.
Now let us compare (2) with the form of effective Chebotarev proved by Lagarias and Odlyzko, and slightly improved soon after by Serre (because Lagarias and Oslyzko use the same substandard lower bound on the discriminant than later Iwaniec and Kowalski, as Serre use better bounds).
$$ (3) \quad \psi(C,x) = \frac{|C|}{|G|}\mathrm{Li}(x) + O \left( \sqrt{x} |C| \log x (\log x + \log M + \log |G|)\right)$$
It is clear how (2) is better than (3): we gain a factor $\sqrt{|C|}$.
How comes Lagarias and Odlyzko and Serre didn't see it? Well, essentially this is because
they use, for counting the zero of $L(\rho,s)$, the zero on $\zeta_L(s)$. The problem is that
while $L(\rho,s)$ is an $L$-function of degree $\dim \rho$ (and the various $L(\rho,s)$ therefore have in quadratic average a degree equal to $\sqrt{|G|}$), the function $\zeta_L$
is an $L$-function of degree $|G|$. So to get (1) or (2), one needs to get an optimal bound for each $L(\rho,s)$ individually, as a function of degree $\dim \rho$, and the argument
suggested by Sausage can not be sufficient, nor can the one given by Denis, since it is
already used by Lagarias and Odlyzko.
So the question remains :
How to prove (1), or (2), without assuming Artin ?
Or is it possible that there is a mistake in Iwaniec-Kowalski and proving (2) without Artin is not possible ?
Best Answer
Since no one answered my question, I have asked the author of the book. Emmanuel Kowalski told me that this remark they make (namely that the form (1) or (2) they give of Chebotarev can be proved using GRH alone, without Artin) is mistaken. In the state of our knowledge, Artin is necessary to get such a precise form.