I read : if χ is principal, then the corresponding Dirichlet L-function has a simple pole at $s = 1$.
Probably I need to know what principal is and that will solve my question.
My confusion was this.
Let $\zeta_R(s)$ be the Riemann zeta function with a simple pole at $1$.
Let $\zeta_1(s)$ be the Dirichlet L-function defined as the analytic continuation of
$$ \prod_{p_1} (1+\frac{1}{p_1^s - 1}) $$
where $p_1$ are the primes $ 1$ mod $4$.
Let $\zeta_2(s)$ be the Dirichlet L-function defined as the analytic continuation of
$$ \prod_{p_2} (1+\frac{1}{p_2^s - 1}) $$
where $p_1$ are the primes $ 3$ mod $4$.
Now we have $\zeta_R(s) = (1+\frac{1}{2^s-1}) \zeta_1(s) \zeta_2(s) $
But $\zeta_R(s)$ has a simple pole at $1$ thus if $\zeta_1(s)$ and $\zeta_2(s)$ both have a pole at $1$ then we have a paradox ; because the product of 2 functions with a simple pole at $s=1$ equals a function with a pole of order $2$ at $s=1$.
So what is going on here ?
Did I misunderstood ?
Which one ( $\zeta_1$ or $\zeta_2$ ) is missing a pole at $1$ ?
And where is the location of the pole for the one that has no pole at $1$ ?
Those aren't quite the Dirichlet $L$-functions you intended, is part of the problem. In general, $$ L(s,\chi)\;=\;\sum_n {\chi(n)\over n^s} \;=\; \prod_p {1\over 1 - {\chi(p)\over p^s}}$$ In fact, it appears that the two Dirichlet series you wrote do blow up as $s\to 1^+$ (by a quantitative form of Dirichlet's theorem on primes in arithmetic progressions). In fact, they cannot have meromorphic continuations, since, as you observe, the combination of their alleged poles at $s=1$ would have to give the simple pole of $\zeta(s)$, which is impossible.