@KConrad: By the same token, Vorhauer's argument sketched in the exercises section of Montgomery and Vaughan's book applies to $\zeta_K(s)$ to yield the weaker inequality $\gamma_K > - (2/3)\log{\sqrt{|D_{K/\mathbb{Q}}|}}$, in lieu of Badzyan's coefficient $1-1/\sqrt{5}$ that still appears to be the best unconditional one available. It does seem like a kind of threshold exponent in the present day theory, although it does improve under GRH to $0.459$, as Badzyan also showed in that paper. For bounded degree fields, or Dirichlet $L$-function, it should improve to any $\varepsilon > 0$...

So my comment was mostly methodological, about the specific saving coefficient of $1/2\sqrt{5}$ coming out of this method. In a way this goes back all the way to Stechkin's work [3] on $\zeta(s)$ that he cited (cf. the estimate (19) there). BTW: I was wondering if perhaps one manifestation of the analytic difficulties inherent in Vinogradov's conjecture is that it is closely related to the lower bound $(L'/L)(1,\chi_q) > -\varepsilon\log(q)$, (once $q \gg_{\varepsilon} 1$), but for sufficiently general L-functions including $\zeta_K(s)$, Badzyan's result does become sharp up to the constant.

@KConrad: I looked at it more carefully, and I think it works actually directly to the function $L(s,\chi)$, replacing his $\zeta_K(s)$. (No zeta factor or positivity property is relevant in Badzyan's argument.) On page 34, he estimates his $Z_K(s) = - \zeta_K'(s) / \zeta_K(s)$ from above by $-n\zeta'(s)/\zeta(s)$, for all real $s > 1$, and applies it for his choice $s = (\sqrt{5}+1)/2$. In our case we simply have to note $-\mathrm{Re}(L'(s,\chi) / L(s,\chi)) \leq -\zeta'(s) / \zeta(s)$, for all real $s > 1$, and the argument goes through verbatim (choice $s$ = the Golden Mean, for Lemma 1).

@KConrad: Thanks for pointing this out! Yes, for quadratic characters it follows straightaway from Badzyan's bound for the Euler-Kronecker constant of the quadratic field. (But even in that case it seems that his coefficient would be the best available, as far as I was aware? Anyway I was viewing the quadratic characters as the most interesting case.) While Badzyan in that paper focuses on the Euler-Kronecker constants of global fields, but impression was that his argument based on Lemma 1 applied implicitly to arbitrary Dirichlet characters, based on the positivity of $\zeta(s)L(s,\chi)$.

If $f(z)$ has just finitely many zeros, there would be a polynomial $P(z)$ such that $f(z) / P(z)$ is entire and non-vanishing, hence equal to $\exp(g(z))$ for some $g(z)$ entire. For $f(z)$ to have order $1$, the only possibility is that $g(z)$ be a linear function. But then $f(z)$ cannot be even. The contradiction shows that your $f(z)$ has indeed infinitely many zeros (and that can be made quantitative: $\sum 1 / |\rho| = +\infty$ is divergent over these zeros taken with multiplicities).

Yes, there is are infinitely many such special parameters $P$ at which $\sigma(P)$ becomes torsion in its fibre. Moreover, all large enough orders are realizable for the torsion point. This comes by a version of the implicit function theorem. When the data is over the algebraic numbers, one further knows that these special parameters $P$ are a set of bounded height (so they are rather sparse). See Masser and Zannier's paper Torsion anomalous points and families of elliptic curves and, for a generalization, Prop. 3.1 in Habegger's paper Special points on fibered powers of elliptic surfaces.

In (2.27) of Iwaniec and Kowalski's book, it is explained how the divergence of $\sum 1/p$ over just the one arithmetic progression $1 \mod{N}$ yields at once the non-vanishing of all $L(1,\chi)$ to the conductor $N$.

@MarkSapir: But the $p_i(n)$ need not be constant. What you write applies to the $p_i \equiv \mathrm{const}$ case, i.e. when $f(X) \in \mathbb{Q}(X)$ is rational. Example: if $a(n) = 1/n$, or $f(X) = -\log(1-X)$, we have $D_n = [1,\ldots,n] = e^{n + o(n)}$, by the prime number theorem. I have no idea how to prove the $k=2$ case; rather trying to get a sense if the statement looks plausible in general.

@MarkSapir: For $k=1$ it shouldn't be too hard to prove that this dichotomy is indeed true. It is also easy when $\sum_{i=0}^{k-1} \deg{p_i} \in \{0,1\}$.

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