Let $K_\infty/\mathbb{Q}$ denote the $\hat{\mathbb{Z}}$-extension of $\mathbb{Q}$. Then for each $n\geq1$, $K_\infty$ has a unique subfield $K_n$ of degree $n$ over $\mathbb{Q}$.

The fields $K_n$ are rather nice:

  • $K_n$ is Galois over $\mathbb{Q}$, and $\mathrm{Gal}(K_n/\mathbb{Q})$ is cyclic of order $n$ (so Kronecker-Weber applies),
  • The subfields of $K_n$ are exactly $K_m$ for $m\mid n$,
  • $K_n$ is totally real,
  • It is conjectured that $K_{p^j}$ always has class number 1 (see this answer).

I was looking through these fields in the LMFDB (links at the end of this question), and I noticed something peculiar: $$\begin{array}{l|l} n&\lim_{s\to1^+}(s-1)\zeta_{K_n}(s)\\\hline 1&1\\ 2&0.623225240141\\ 3&0.3774610891760\\ 4&0.431652451862\\ 5&0.411382062671\\ 6&0.26281913742\\ 7&0.29967691809\\ 8&0.3399574212\\ 9&0.231073814089\\ 10&0.59812750863\\ 11&\mathbf{\color{red}{10.9924154253}}\\ 12&0.272503409905\\ 13&0.480494415585\\ 14&0.303247583959919\\ 15&0.126218760817\\ 16&0.271689000529\\ 17&0.149703203263054\\ 18&0.121386452739\\ \end{array}\qquad\qquad\qquad\begin{array}{l|l} n&\lim_{s\to1^+}(s-1)\zeta_{K_n}(s)\\\hline 19&0.127769546503240\\ 20&0.148293571484876\\ 21&0.325798468160489\\ 22&0.481687105754706\\ 23&0.123796267561194\\ 24&0.139575702501298\\ 25&0.167117342071058\\ 26&0.522724794223398\\ 27&0.174105094869797\\ 28&0.448262864859088\\ 29&0.239734187517100\\ 30&0.106243397680971\\ 31&\\ 32&0.278099976864148\\ 33&\mathbf{\color{red}{10.8513191982336}}\\ 34&\\ 35&\\ 36&0.121939637724450 \end{array}$$ For these values of $n$, the limit always lies in the interval $[0.1,1]$, except for $n=11$ and $n=33$ (but not $n=22$), where the limit is over $10$ times larger!

Why is the residue of the pole of $\zeta_{K_n}(s)$ so large for $n=11$? (and also $n=33$, but not $n=22$)

Here are two potential ways of approaching the question:

  1. If $p$ is an odd prime then $\Delta_{K_p}=p^{2(p-1)}$ and conjecturally $h_{K_p}=1$ so we have $$\lim_{s\to1^+}(s-1)\zeta_{K_p}(s)=(2/p)^{p-1}\mathrm{Reg}_{K_p}.$$ Thus, the question could be phrased as:

Why is the regulator of $K_{11}$ so large?

  1. We have $\lim_{s\to1^+}(s-1)\zeta_{K_n}(s)=\prod_{\chi\neq1}L(1,\chi)$. Here are pictures of $L(1,\chi)$ for $n=7,11,13$. Of course, these pictures don't really explain anything by themselves.

$p=7$ (mostly in the unit circle so the product is small, $\approx0.3$)


$p=11$ (almost entirely outside of the unit circle so the product is large, $\approx11$)


$p=13$ (mostly in the unit circle so the product is small, $\approx0.5$)


Here are the LMFDB links:

$K_1$ $K_2$ $K_3$ $K_4$ $K_5$ $K_6$ $K_7$ $K_8$ $K_9$ $K_{10}$ $K_{11}$ $K_{12}$ $K_{13}$ $K_{14}$ $K_{15}$ $K_{16}$ $K_{17}$ $K_{18}$ $K_{19}$ $K_{20}$ $K_{21}$ $K_{22}$ $K_{23}$ $K_{24}$ $K_{25}$ $K_{26}$ $K_{27}$ $K_{28}$ $K_{29}$ $K_{30}$ $K_{31}$ $K_{32}$ $K_{33}$ $K_{34}$ $K_{35}$ $K_{36}$

  • 9
    $\begingroup$ If we view the zeta function as counting primes of the number field, the small split prime $3$ gives a large contribution to $\zeta_{K_{11}}$. This is quite exceptional - the next $p$ for which $3$ is split in $K_p$ is $1006003$, and the first $p$ for which $2$ is split in $K_p$ is $1093$. Maybe this could provide an explanation? $\endgroup$
    – Will Sawin
    Mar 22, 2021 at 20:17
  • 2
    $\begingroup$ I graphed a few of the partial Euler products and this seems right to me. Do you think that this also explains why this doesn't happen for $n=22$ and does for $n=33$? $\endgroup$ Mar 22, 2021 at 20:30
  • 3
    $\begingroup$ Also, the fact that $3$ splits in $K_{11}$ is equivalent to $3^{10}\equiv1\pmod{11^2}$. So maybe this is a "consequence" of the coincidence $3^5=2\cdot11^2+1$. $\endgroup$ Mar 22, 2021 at 20:34
  • 3
    $\begingroup$ I guess when you go from $K_11$ to $K_{33}$, the primes lying over $3$ ramify, so the local contribution ends up being exactly the same, while going from $K_{11}$ to $K_{22}$ renders the primes lying over $3$ inert which drops the local contribution from $(3/(3-1))^11$ to $(9/(9-1))^11$, a loss of a factor of about $24$, which tracks with the explicit values... $\endgroup$
    – Will Sawin
    Mar 22, 2021 at 20:57
  • $\begingroup$ (I made an unsuccessful attempt to use the spoiler mode to hide the links and make them click visible: it replaces the whole thing with a huge grey area so this is essentially useless) $\endgroup$
    – YCor
    Mar 22, 2021 at 21:04

1 Answer 1


A prime $\ell$ splits in $K_p$ for $p$ odd if and only if $$\ell^{p-1} \equiv 1 \mod p^2.$$

We can express $\lim_{s \to 1} (s-1) \zeta_{K_n(s)}$ as the product of $\frac{1}{ 1- |\mathfrak a|^{-1} }$ over primes $\mathfrak a$ of $K_n$, regualarized by dividing by the product of $\frac{1}{1-\ell}$ over primes $\ell$ of $\mathbb Q$.

From this perspective, large values of $\lim_{s \to 1} (s-1) \zeta_{K_p(s)}$ are "explained" by the existence of small primes that split in $K_p$, which give a large contribution to this product.

For fixed $\ell$, the $p$ for which $\ell$ splits in $K_p$ are the generalized Wieferich primes. Examining that table, we see that for $\ell=3$, $11$ is the smallest Wieferich prime. It is exceptionally small. In particular, if we ignore $\ell> p$ as that is the range where we start expecting to have a split prime for random reasons, then the next-smallest Wiefierch prime for any $\ell<11$ is $p=1093$ for $\ell=2$.

The contribution of the primes lying over $3$ to the zeta function is $(3/2)^{11}\approx 86.5$. If we divide by the Euler factors for primes $<11$ of the zeta function of $\mathbb Q$, or $ 2 \cdot (3/2) \cdot (5/4) \cdot (7/6) $, we get $19.8$, which is on the same order of magnitude as your limit. Thus, this exceptional split prime gives an explanation for the unusually large value.

Examining the table linked above, we might suspect that exceptionally large values might occur for $K_p$ when $p=43, 71, 137 863, 1093, 3511, \dots $ owing to the split primes $\ell=19, 11,19, 13, 2,2,\dots$ respectively.

The distinct behavior of $K_{22}$ and $K_{33}$ is probably related to what happens to these small primes in those two extensions. In $K_{22}$ they are inert, taking the local factor of the zeta function lying over $3$ from $(3/2)^{11}$ to $(9/8)^{11}$, a loss of a factor of $\approx 23.7$ (matching the ratio of the residues in your table), while in $K_{33}$ they are ramified, preserving their norms, and preserving the local factor $(3/2)^{11}$ (explaining why the residue barely changes).

  • $\begingroup$ Great! I'm a little worried about your splitting criterion when $n$ is even. In particular, $\ell=5$ does not split in $K_2=\mathbb{Q}(\sqrt2)$. $\endgroup$ Mar 23, 2021 at 19:25
  • $\begingroup$ @ThomasBrowning That's a good point, I'll do a quick-and-dirty fix. $\endgroup$
    – Will Sawin
    Mar 23, 2021 at 19:32
  • 1
    $\begingroup$ @ThomasBrowning Good point, thanks. It indeed needs to be done prime-by-prime, so it's better to ignore the first sentence and just do the prime version. (I put it there because I thought it would help with 22 later, but, in fact, it doesn't). $\endgroup$
    – Will Sawin
    Mar 23, 2021 at 20:43
  • 1
    $\begingroup$ @paulgarrett Thanks! The reason I don't feel it is completely decisive myself is I don't see why the large primes don't produce a contribution that balances or reverses the contribution of the small primes. The normal reason I would give for cancellation in the sum over large primes is the zeta function, but that's a bit circular. I guess maybe a random model could justify it... $\endgroup$
    – Will Sawin
    Mar 23, 2021 at 20:45
  • 1
    $\begingroup$ @ThomasBrowning The log of this is a sum over primes, it suffices to show that this sum converges. Convergence of this sum follows from a good estimate for the prime number theorem / Landau prime ideal theorem in this number field. I think it's a bit easier than that, though. $\endgroup$
    – Will Sawin
    Mar 23, 2021 at 22:44

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