Let $K_n$ be the field $\mathbf Q[\cos(\frac{\pi}{2^{n+1}})]$ (the real subfield of the cyclotomic field $\mathbf Q[e^{\frac{i\pi}{2^{n+1}}}]$).

Is there anything known about the growth of the special values of the Dirichlet $\zeta$ function $$\zeta_{K_n}(2)=\prod_{\mathfrak p}\frac{1}{1-\frac{1}{N(\mathfrak p)^2}}$$ when $n\to \infty$ ?

The best I could do was to use the inequalities $$\left(1-\frac{1}{p^2}\right)^{ab}\leq \left(1-\frac{1}{p^{2a}}\right)^{b}\leq \left(1-\frac{1}{p^{2ab}}\right)$$ (that are verified in our range) to obtain the inequalities $$2\ \left(\frac{\pi^2}{3}\right)^{2^n}\ =\ 2^{1-2^n}\ \zeta(2)^{2^n}\geq \zeta_{K_n}(2)\geq 2\ \left(1-\frac{1}{2^{n+1}}\right)\ \zeta(2^{n+1})$$ that are quite unilluminating since the left hand side tends to infinity exponentially fast, while the right hand side converges to 2. The only thing we use here is that $2$ is totally ramified and that the other primes are not ramified. I guess it is possible to use more of our knowledge of the arithmetic in $K_n$ to obtain a precise idea of the behaviour of the sequence $(\zeta_{K_n}(2))_n$ ... (maybe by a standard use of the Chebotarev theorem - I'm quite ignorant in this kind of questions).

I'm almost certain that $\zeta_{K_n}(2)$ tends to infinity exponentially fast (and that's what I'd like to check).

  • $\begingroup$ I don't understand the lower bound $2 \left( 1 - \frac1{2^{n+1}} \right) \zeta(2^{n+1})$. The local factor at $2$ is $4/3$, not $2$, and my answer indicates that $4/3$ is in fact the limit as $n \to \infty$ of $\zeta_{K_n}(2)$. $\endgroup$ – Noam D. Elkies Mar 3 '16 at 4:27
  • $\begingroup$ @NoamD.Elkies yes, right, I made a mistake there ... $\endgroup$ – few_reps Mar 3 '16 at 7:11

Actually $\zeta_{K_n}(\sigma)$ is bounded for any fixed $\sigma > 1$.

Let $N = 2^n = [K_n : {\bf Q}]$. Then all the local factors of $\zeta(\sigma)$, other than the factor $(1-2^{-\sigma})^{-1}$ for the prime above $2$, are of the form $(1 - q^{-\sigma})^{-g}$, where $q$ is a prime power congruent to $\pm 1 \bmod 4N$, and $g \mid N$. Thus $$ \zeta_{K_n}(\sigma) < \frac1{1-2^{-\sigma}} \prod_{m=1}^\infty \frac1{(1 - (2mN)^{-\sigma})^N} $$ and the product approaches $1$ as $n \to \infty$ because its logarithm behaves like $\sum_{m=1}^\infty N/(2mN)^\sigma = \zeta(\sigma) \, / \, 2^\sigma N^{\sigma-1} \to 0$.

Since each local factor exceeds $1$, it follows that in fact $\zeta_{K_n}(\sigma) \to (1-2^{-\sigma})^{-1}$ as $n \to \infty$.

  • $\begingroup$ @NoamDElkies So my guess (coming from computations of mass formulae) was wrong ... thanks anyway, it's comforting to see that the question has an elementary answer. $\endgroup$ – few_reps Mar 3 '16 at 7:16

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