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Motivated by this question, I am curious how one can characterize primes that splits completely in the Hilbert class field of $\mathbb{Q}(\zeta_q)$, where $q$ is a prime. Then I realize how much I don't know about this extension over $\mathbb{Q}$! I know this is too broad, but I'll be happy if someone can briefly write down what we can say about this extension in general. At the very least

Is the extension $H/\mathbb{Q}$ abelian, if $H$ is the Hilbert class field of $\mathbb{Q}(\zeta_q)$, where $q$ is a prime? I guess it's generally not. If so, when is it abelian?

and

What is the degree of $H/\mathbb{Q}$? What do we know about the ideal class groups of $\mathbb{Q}(\zeta_q)$?

Actually, I am not even sure what the right question to ask here is. A quick google search yields some relevant things, such as this MO post and Franz Lemmermeyer's papers.. They seem to say that:

  1. We can't pin down the Hilbert class field of cyclotomic field unless it's imaginary quadratic or $\mathbb{Q}$ itself. But can we say anything about whether it's abelian, say?

  2. There is a connection to the main theorem of Iwasawa theory. I know that this is too demanding, but if someone's willing to sketch a quick introduction to Iwasawa theory and how it connects to this problem here, I would be super grateful.

Thanks!

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The extension $H/\mathbf Q$ is never abelian, unless $H=\mathbf Q(\zeta_q)$ (which happens for just a few values of $q$, since the class number of $\mathbf Q(\zeta_n)$ grows very fast). Indeed, according to Kronecker-Weber, every abelian extension of $\mathbf Q$ is contained in a cyclotomic extension. If $H/\mathbf Q$ were abelian, we would have $\mathbf Q(\zeta_q) \subseteq H \subseteq \mathbf Q(\zeta_n)$ for some $n$; then, examining the ramification, we see that $n=q$ and so $H=\mathbf Q(\zeta_q)$.

The degree of $H/\mathbf Q$ is equal to $q-1$ times the class number $h_q$.

In general, the primes which split completely in the Hilbert class field are those which are principal in the base field.

The class group of $\mathbf Q(\zeta_n)$ is a very complicated thing. Whole books are devoted to it. It is a fascinating topic with deep connections to the theory of the Riemann zeta function.

I recommend Washington's book Introduction to Cyclotomic Fields. If you are interested specifically in Iwasawa theory, I would recommend the book Cyclotomic Fields and Zeta Values by Coates and Sujatha, which is a great introduction to this truly unbelievable theory.

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  • $\begingroup$ Ah, Kronecker-Weber, good point! If that's the case, is there any way to characterize the set of primes $p$ that splits completely in $H$, using only arithmetic data of the ground field? $\endgroup$ – user27126 Nov 29 '13 at 5:09
  • $\begingroup$ Dear @Sanchez: I have edited my answer to answer this. $\endgroup$ – Bruno Joyal Dec 1 '13 at 18:13
  • $\begingroup$ Thanks! I know this (actually that's how I came up with the question). I meant to characterize the set of primes in $\mathbb{Q}$, that splits completely in $H$ (the Hilbert class field of $\mathbb{Q}(\zeta_q)$ in this case, not necessarily abelian like you said) in a way similar to class field theory. I know that congruence conditions won't be possible here, but what can I say in general? $\endgroup$ – user27126 Dec 1 '13 at 18:18
  • $\begingroup$ @Sanchez Ah, I see. Then I don't know. I don't think that there is a simple criterion. $\endgroup$ – Bruno Joyal Dec 1 '13 at 18:21
  • $\begingroup$ Thanks! This should really be a new question, but here it goes anyway: What would non-abelian reciprocity law tell us in this case (characterizing the set of completely splitting primes in $H$)? It's not congruence conditions on $p$, but what kind of condition would it be? $\endgroup$ – user27126 Dec 1 '13 at 20:04
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Take the $37^{\text{th}}$ cyclotomic ring of integers, the homomorphism $\sigma:\zeta\to\zeta^2$ with the primitive root $2$ modulo $37$ and the cylotomic integer $g(\zeta)=-\zeta^{13}+\zeta^{25}+\zeta^{27}+\zeta^{30}+\zeta^{32}$. The ideal $\langle g(\zeta)\rangle$ factorizes to $\langle g(\zeta)\rangle=\langle149, \psi_{149}(\zeta)\rangle^{2}\cdot\langle149, \sigma^{11}\psi_{149}(\zeta)\rangle$. The second generator $\psi_{149}(\zeta)$ of the prime ideal can be taken from the first theorem in chapter $4.12$ of Edwards in Fermat's last theorem. In Dedekind domains it is then easy to prove that the greatest common divisor of the ideals $\langle 149\rangle$ and $\langle\psi_{149}(\zeta)\rangle$ is the prime ideal $\langle149, \psi_{149}(\zeta)\rangle=\langle 149\rangle+\langle\psi_{149}(\zeta)\rangle$. The process of factorizing cyclotomic ideals is described in the chapters $4.11$ff of Edwards. I outlined this process in my answer to this question. The cyclotomic integer $$h(\zeta)=\zeta-\zeta^{4}-\zeta^{7}-\zeta^{8}-\zeta^{9}-\zeta^{11}-\zeta^{12}-\zeta^{13}-\zeta^{14}-\zeta^{15}+\zeta^{16}-\zeta^{18}-\zeta^{19}+\zeta^{21}-\zeta^{22}-\zeta^{23}-\zeta^{24}-\zeta^{25}-\zeta^{26}-\zeta^{28}-\zeta^{29}-\zeta^{30}-\zeta^{33}+\zeta^{36}$$ factorizes to $\langle h(\zeta)\rangle=\langle149, \sigma^{11}\psi_{149}(\zeta)\rangle\cdot\langle149, \sigma^{29}\psi_{149}(\zeta)\rangle$. The $2$ factorizations give the class equivalence $\langle149, \psi_{149}(\zeta)\rangle^{2}\sim\langle149, \sigma^{29}\psi_{149}(\zeta)\rangle$. We square this equivalence and get

$${\langle149, \psi_{149}(\zeta)\rangle^{4}}\sim {\langle149, \sigma^{29}\psi_{149}(\zeta)\rangle^2}\sim { \sigma^{29}\circ \left\{\langle149, \psi_{149}(\zeta)\rangle^{2} \right\}}\sim {\sigma^{29}\circ\langle149, \sigma^{29}\psi_{149}(\zeta)\rangle}\sim {\langle149, \sigma^{2\cdot29}\psi_{149}(\zeta)\rangle}.$$

Consecutively squaring gives $\langle149, \psi_{149}(\zeta)\rangle^{2^k}\sim\langle149, \sigma^{29k}\psi_{149}(\zeta)$. With $\sigma^{36}=\sigma^0=\text{id}$ we get $\langle149, \psi_{149}(\zeta)\rangle^{2^{36}}\sim\langle149, \psi_{149}(\zeta)$. We take the letter I for the principal class and finally get $\langle149, \psi_{149}(\zeta)\rangle^{2^{36}-1}\sim I$. Therefore the class order of the prime ideal $\langle149, \psi_{149}(\zeta)\rangle$ divides the number $2^{36}-1$.

Now we take the cyclotomic integer $s(\zeta)=\zeta-\zeta^{4}-\zeta^{17}+\zeta^{25}-\zeta^{27}-\zeta^{34}$. It factorizes to $\langle s(\zeta)\rangle=\langle149, \sigma^{12}\psi_{149}(\zeta)\rangle\cdot\langle149, \sigma^{26}\psi_{149}(\zeta)\rangle\cdot\langle149, \sigma^{27}\psi_{149}(\zeta)\rangle$. With $24\cdot29\equiv12\mod36$ we first get $\sigma^{24\cdot29}=\sigma^{12}$ and then the calculation above gives the equivalence $\langle149, \psi_{149}(\zeta)\rangle^{2^{24}}\sim\langle149, \sigma^{12}\psi_{149}(\zeta)$. Similarly we get $\langle149, \psi_{149}(\zeta)\rangle^{2^{22}}\sim\langle149, \sigma^{26}\psi_{149}(\zeta)$ and $\langle149, \psi_{149}(\zeta)\rangle^{2^{27}}\sim\langle149, \sigma^{27}\psi_{149}(\zeta)$. This gives

$${\langle149, \sigma^{12}\psi_{149}(\zeta)\rangle\cdot\langle149, \sigma^{26}\psi_{149}(\zeta)\rangle\cdot\langle149, \sigma^{27}\psi_{149}(\zeta)\rangle}\sim {\langle149, \psi_{149}(\zeta)\rangle^{2^{24}}\cdot\langle149, \psi_{149}(\zeta)\rangle^{2^{22}}\cdot\langle149, \psi_{149}(\zeta)\rangle^{2^{27}}}\sim {\langle149, \psi_{149}(\zeta)\rangle^{2^{24}+2^{22}+2^{27}}}\sim I.$$

Hence the class order of the prime ideal $\langle149, \psi_{149}(\zeta)\rangle$ divides the number $2^{24} + 2^{22} + 2^{27}$. But then the class order of the prime ideal $\langle149, \psi_{149}(\zeta)\rangle$ divides the greatest common divisor $\gcd (2^{36}-1, 2^{24} + 2^{22} + 2^{27}) = 37$. Kummer's class number formula just gives the prime $37$! We finally get $\langle149, \psi_{149}(\zeta)\rangle^{2^{5}}\sim\langle149, \psi_{149}(\zeta)\rangle^{32}\sim\langle149, \sigma\circ\psi_{149}(\zeta)\rangle$ and $\xi=2^5=32$ is a primitive root modulo $37$. Therefore there are $36$ prime ideals $\langle149, \sigma^k\psi_{149}(\zeta)\rangle$ and each occupies a different class of the $36$ non-principal classes.

The exponent f of a prime q modulo the prime p is the least positive integer f such that $q^f\equiv1 \mod p$. Kummer already knew that a prime q with exponent f modulo p factorizes to $e=\frac{p-1}{f}$ prime ideals $\langle q, \sigma^k\psi_{q}(\zeta)\rangle$,$0\le k\lt e$, with $\langle q, \sigma^k\psi_{q}(\zeta)\rangle=\langle q, \sigma^{k+e}\psi_{q}(\zeta)\rangle$(see chapter $4.9$ of Edwards). Then all prime ideals that divide a prime q with an exponent $f\gt1$ must be principal because otherwise the homomorphisms $\sigma^0,\sigma^e,\sigma^{2e},\dots,\sigma^{e(f-1)}$ would send the prime ideal

$${\langle q, \sigma^{0e}\psi_{q}(\zeta)\rangle}= {\langle q, \sigma^{e}\psi_{q}(\zeta)\rangle}= {\langle q, \sigma^{2e}\psi_{q}(\zeta)\rangle}=\dots= {\langle q, \sigma^{(f-1)e}\psi_{q}(\zeta)\rangle}$$

to several classes because $\xi$ has order $36$ modulo $37$. Therefore only prime ideals that divide a prime q with order $f=1$ can be non-principal. If a prime ideal is not principal one conjugate $\langle q, \sigma^{k}\circ\psi_{q}(\zeta)\rangle$ must be equivalent to the ideal $\langle 149, \psi_{149}(\zeta)\rangle$ and the class group is well understood.

The procedure of determinating the class group of cyclotomic rings of integers by factorizing cyclotomic integers as demonstrated above has been done for a couple of cyclotomic rings of integers. The result can be taken from here. The data that was used in order to compile this document can be taken from here.

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  • $\begingroup$ While this link may answer the question, it is better to include the essential parts of the answer here and provide the link for reference. Link-only answers can become invalid if the linked page changes. - From Review $\endgroup$ – AugSB Nov 19 '16 at 15:06
  • $\begingroup$ Please consider joining your two accounts, doing so makes housekeeping easier for everybody. $\endgroup$ – Alex M. Nov 25 '16 at 13:00

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