# Class Numbers and 163

This is a bit fluffier of a question than I usually aim for, so apologies in advance if this doesn't pass the smell test for suitability.

Likely my favorite fun fact in all of number theory is the juxtaposition of two "extremal and opposite" properties about the prime 163 in relation to class numbers:

• $p=163$ is the largest value of $p$ for which the quadratic imaginary number field $\mathbb{Q}(\sqrt{-p})$ has class number equal to one. (Baker-Heegner-Stark)
• $p=163$ is the smallest value of $p$ for which the real cyclotomic field $\mathbb{Q}(\zeta_p+\zeta_p^{-1})$ has class number greater than one. (Schoof)

Of the various ways I know of "pushing up" and "pushing down" class numbers (class field theory, Herglotz-type formulas, Scholz-type reflection theorems), none seem to give any indication that these two class numbers should be related, let alone inversely so. Of course, since the smaller Heegner discriminants don't correspond to analogous real cyclotomic fields with positive class number, this is not surprising.

This leads me to wonder if there's an analytic link between these two quantities -- for example, by relating their zeta-functions and looking at the corresponding class number formulas. My initial, admittedly naive, attempts to extract anything from the relationship between the zeta-functions for $\mathbb{Q}(\zeta_p)$ and $\mathbb{Q}(\zeta_p+\zeta_p^{-1})$ have come up empty. So my question is:

Are there fancier analytic (or other) techniques that might shed some light on the "miracle" above?

Of course, I'm also aware that this juxtaposition might be purely coincidental, a mildly large example of the law of small numbers at work. I might even prefer it that way.

Edit to incorporate some computations and comments from below.

For primes congruent to 7 mod 12, the real cyclotomic field of conductor p contains a unique cyclic cubic subfield. By class field theory, there is a surjection of class groups from the real cyclotomic to the cubic. Since for 163, the cyclic cubic has class number 4, the non-triviality of the class number for the real cyclotomic can be said to "come from" the cubic. In a sense, the coincidence thus reduces to the fact that the first cyclic cubic field ("first" with respect to ordering by conductor) of prime conductor 7 mod 12 with non-trivial class group is the one of conductor 163. The fact that 163 is only the 11th prime in this congruence class may modify (in which direction I'm not sure) your opinion of whether or not this is a coincidence.

Barring any insight as to why the class number of this quadratic and cubic would be related, which may be unlikely given Franz Lemmermeyer's answer, it would be interesting to know if one could devise a clever probabilistic test for evaluating how surprised one should be to see ten class-number-one cubics in a row. I imagine that it's not very unlikely -- I just ran a computation, and class number 1 seems to very prevalent for cyclic cubics of small conductor ($p<5000$), and some heuristic (sorry, Andrew) evidence in the literature seems to agree.

• My computer screen smells the same with this displayed as it usually does. – Will Jagy Sep 10 '10 at 19:24
• That is indeed reassuring. – Cam McLeman Sep 10 '10 at 19:25
• Cam, a possibly easier question is whether the Baker-Heegner-Stark answer had to be less than or equal to the Schoof answer, once the former was known to be finite. – Will Jagy Sep 10 '10 at 19:58
• No, I take it back. Stark has a very accurate test for conjectures based on the first result being the most difficult proof in that area of mathematics. As the number 163 was surely known for both problems, an easy answer to my problem, or your original, gives an easy proof of Baker-Heegner-Stark, therefore there is no such easy relationship. The relationship is a coincidence. Quod Erat Demonstarkum. – Will Jagy Sep 10 '10 at 20:10
• See if I can get Stark's test correct. Use apostrophe for negation, S for Baker-Heegner-Stark, C for some conjecture. If we have the implication S' --> C, then Stark believes C pending further investigation. Why, you ask? Worry not, I shall tell you. The contrapositive is C' --> S. If C actually turned out to be false, this would provide an easy proof of S – Will Jagy Sep 10 '10 at 20:32

If I recall it correctly, the class number of the real cyclotomic field comes from its cubic subfield. Thus you might as well ask whether there is a connection between class numbers of quadratic fields ${\mathbb Q}(\sqrt{-p})$ and real cubic fields with prime conductor $p$ for primes $p \equiv 7 \bmod 12$. What little we know about class numbers comes either from analytic objects (zeta functions, class number formulas), class field theory, or elliptic curves. Since we are talking about abelian extensions of the rationals, I would expect an explanation for such a result to come from class field theory. But nothing I know could even be remotely responsible for such a relation. Indeed the only results that might be within reach would be "independence" results claiming that there are infinitely many fields for which one class number is divisible by a certain prime but the other is not.

The problem why there are so few relations between the two objects has to do with the fact that the compositum of the two fields is a cyclic sextic field, which only has two nontrivial subfields. The units of these subfields do not generate a group of finite index in the whole unit group, which for me means that we should not expect any relation between the class numbers of these objects.

On the other hand you never know; the class number $1$ for the quadratic field is connected with continued fraction expansions (see Zagier's book on quadratic fields and forms) - but real cubic cyclic fields seem to have preciously little to do with continued fractions.

• For the first part: Agreed, this is one of the "pushing down" theorems I meant. If p is 7 mod 12, then the real cyclotomic field contains a real cyclic cubic. For p=163, the cubic is the spliting field of $x^3 + x^2 - 54*x - 169$, and has class number 4. – Cam McLeman Sep 11 '10 at 11:25

After examining the tables of class numbers in Washington and Borevich/Shafarevich, I can't rule out a connection, but if anyone is willing to bet on this I will offer 3:1 odds that it is just a coincidence.

The first few primes where the real part of the cyclotomic class number h is greater than 1 are (h'=imaginary quadratic class number)
163 (h=4, h'=1), 191 (h=11, h'=13) 229 (h=3, h'=10) 257 (h=3, h'=16), ...
which does not look promising. The published tables of class numbers of quadratic/cyclotomic fields contain hundreds of numbers of about this size, so a few meaningless coincidences are to be expected.

• Indeed. I've spent many hours poring over such tables (which have been greatly extended since Washington and Borevich/Shafarevich!), and I certainly agree that meaningless coincidences are sure to abound, and this may be one of them. On the other hand, this is a very specific coincidence, and because of the significance of the two events in question, I think it's worth wondering if there's an explanation. – Cam McLeman Sep 11 '10 at 0:18
• Since it's impossible to prove that something is just a coincidence, I don't see the downside of betting against you. Maybe you could run the bet like insurance; somebody pays you x, and if anybody figures out why it's not a coincidence, you pay 4x back. – Peter Shor Sep 11 '10 at 18:54
• Of course. I expect your check. – Richard Borcherds Sep 11 '10 at 23:22
• As soon as I figure out why it's not a coincidence, I'll send you a check. – Peter Shor Sep 12 '10 at 19:21

Maybe I can help with the outline of the class group of the 163rd cyclotomic ring of integer. You find the document at

Take the 53$^{rd}$ cyclotomic ring of integer and the homomorphism $\sigma: \zeta\to\zeta^2$ with the primitive root 2 modulo 53. Kummer's class number formula gives the prime 4889. Because every ideal is invariant under the homomorphism $\sigma$$^{52} and the class group must be cyclic, the homomorphism \sigma can only send every class C to the class C^\xi where \xi must be a (not necessarily primitive) 52^{nd} root modulo 4889 so that \xi^{52}\equiv 1 \bmod{4889}. There are 52 different prime ideals \sigma^k\langle 107, Ψ_{107}(\zeta)\rangle = \langle 107, \sigma^k\Psi_{107}(\zeta)\rangle. The second generator \Psi_{107}(\zeta) of the prime ideal can be taken from the first theorem in chapter 4.12 of Fermat's last theorem by Edwards. In Dedekind rings it is then easy to prove that the greatest common divisor of the ideals \langle 107\rangle and \langle\Psi_{107}(\zeta)\rangle is the prime ideal \langle 107, \Psi_{107}(\zeta)\rangle. A more detailed analysis of this ring then gives the class equivalence \sigma\langle 107, \Psi_{107}(\zeta)\rangle \sim \langle 107, \Psi_{107}(\zeta)\rangle^{3637}. The 'zip code' 3637 is of no further interest. However it is a primitive 52^{nd} root modulo 4889 and we have a clear-cut relationship between the cyclic class group of this ring and the cyclic Galois group \operatorname{Gal}(\mathbb{Q}(\zeta)/\mathbb{Q}) = \{\operatorname{id} = \sigma^0, \sigma^1, \dots, \sigma^{51}\}. Now take the 41^{st} cyclotomic ring of integer, the homomorphism \sigma: \zeta \to \zeta^6 with the primitive root 6 \bmod{41} and the cyclotomic integer g_1(\zeta) = \zeta^6 - \zeta^8 - \zeta^{20}. The ideal \langle g_1(\zeta)\rangle factorizes to \langle g_1(\zeta)\rangle = \langle 83, \Psi_{83}(\zeta)\rangle^2\cdot\langle 83, σ^8\Psi_{83}(\zeta)\rangle. The process of factorizing cyclotomic ideals is described in the chapters 4.11ff of Edwards. I outlined this process in my answer to this question on math.stackexchange. The cyclotomic integer g_2(\zeta) = - \zeta^4 + \zeta^{10} + \zeta^{21} + \zeta^{28} + \zeta^{39} factorizes to \langle g_2(\zeta)\rangle = \langle 83, \sigma^8\Psi_{83}(\zeta)\rangle\cdot\langle 83, \sigma^{28}\Psi_{83}(\zeta)\rangle so that we get the class equivalence \langle 83, \Psi_{83}(\zeta)\rangle^2 \sim \langle 83, \sigma^{28}\Psi_{83}(\zeta)\rangle. Squaring this equivalence gives$$\langle 83, \Psi_{83}(\zeta)\rangle^4 \sim \langle 83, \sigma^{28}\Psi_{83}(\zeta)\rangle^2 \sim \sigma^{28}[\langle 83, \Psi_{83}(\zeta)\rangle^2] \sim \sigma^{28}\langle 83, \sigma^{28}\Psi_{83}(\zeta)\rangle = \langle 83, \sigma^{2·28}\Psi_{83}(\zeta)\rangle.$$Consecutive squaring gives$\langle 83, \Psi_{83}(\zeta)\rangle^{2^k} \sim \langle 83, \sigma^{28k}\Psi_{83}(\zeta)\rangle$and we get the general relationship$\langle 83, \Psi_{83}(\zeta)\rangle^n \sim \langle 83, \sigma^{4k}\Psi_{83}(\zeta)\rangle$for an integer$n$because$\operatorname{gcd}(28, 40) = 4$. For$k=10$we get$\langle 83, \Psi_{83}(\zeta)\rangle^{1024} \sim \langle 83, \Psi_{83}(\zeta)\rangle$or the ideal$\langle 83, \Psi_{83}(\rangle)\rangle^{1023}$is principal. Kummer's class number formula gives$11^2$so that the class order of the ideal$\langle 83, \Psi_{83}(\zeta)\rangle$is$11$with$1023 = 3\cdot 11\cdot 31$. Then we have 4 class subgroups$\langle 83, \sigma^{m+4k}\Psi_{83}(\zeta)\rangle, m \in \{0,1,2,3\}$. But the class number is$11^2$so that the combination of the classes of 2 subgroups gives the classes of the remaining 2 class subgroups. For example the cyclotomic integer$s_2(\zeta) = \zeta^{28} + \zeta^{36} - \zeta^{39}$factorizes to$\langle s_2(\zeta)\rangle = \langle 83, \sigma^{14}\Psi_{83}(\zeta)\rangle\cdot\langle 83, \sigma^{24}\Psi_{83}(\zeta)\rangle\cdot\langle 83, \sigma^{33}\Psi_{83}(\zeta)\rangle$. Therefore the homomorphism$\sigma$sends the class of one class subgroup to the next class subgroup. And we only have a clear relationship between the homomorphism$\sigma^{4k}$of the Galois group of this ring and the classes because the homomorphism$\sigma^4$does not send the classes of one class subgroup to the next. However the class group$\mathbb{Z}/11\mathbb{Z} \oplus \mathbb{Z}/11\mathbb{Z}$has been confirmed. The matter gets worse with the 163$^{rd}$cyclotomic ring of integers. We take the homomorphism$\sigma: \zeta \to \zeta^2$with the primitive root$2 \bmod{163}$. I could only determine the class group of the prime ideals that factorize prime integer to maximal 27 conjugated prime ideals. The analysis gives a class subgroup of$\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}$. Take the four classes$c$,$d$,$e$,$f$, each the generator of one cyclic class subgroup$\mathbb{Z}/2\mathbb{Z}$with class order 2. Then each ideal of order 2 can be assigned to a class$(c^r, d^s, e^t, f^u)$. We take the letter I for the principal divisor. The ideals$\langle q, \Psi_{q}(\zeta)\rangle$with$q = 61, 199, 347all have six conjugates (see chapter 4.9 of Edwards). The analysis similar to that above gives \begin{align*} \langle 61, \sigma^0\Psi_{61}(\zeta)\rangle &\sim (c, I, I, I)\\ \langle 61, \sigma^1\Psi_{61}(\zeta)\rangle &\sim (I, I, e, I)\\ \langle 61, \sigma^2\Psi_{61}(\zeta)\rangle &\sim (I, d, I, I)\\ \langle 61, \sigma^3\Psi_{61}(\zeta)\rangle &\sim (I, I, I, f)\\ \langle 61, \sigma^4\Psi_{61}(\zeta)\rangle &\sim (c, d, I, I)\\ \langle 61, \sigma^5\Psi_{61}(\zeta)\rangle &\sim (I, I, e, f)\\ \langle 199, \sigma^0\Psi_{199}(\zeta)\rangle &\sim (I, d, e, I)\\ \langle 199, \sigma^1\Psi_{199}(\zeta)\rangle &\sim (I, d, I, f)\\ \langle 199, \sigma^2\Psi_{199}(\zeta)\rangle &\sim (c, d, I, f)\\ \langle 199, \sigma^3\Psi_{199}(\zeta)\rangle &\sim (c, d, e, f)\\ \langle 199, \sigma^4\Psi_{199}(\zeta)\rangle &\sim (c, I, e, f)\\ \langle 199, \sigma^5\Psi_{199}(\zeta)\rangle &\sim (c, I, e, I)\\ \langle 347, \sigma^0\Psi_{347}(\zeta)\rangle &\sim (c, I, I, f)\\ \langle 347, \sigma^1\Psi_{347}(\zeta)\rangle &\sim (c, d, e, I)\\ \langle 347, \sigma^2\Psi_{347}(\zeta)\rangle &\sim (I, d, e, f) \end{align*} and we have\langle 347, \sigma^0\Psi_{347}(\zeta)\rangle \sim \langle 347, \sigma^3\Psi_{347}(\zeta)\rangle$. Here the action of the Galois group on the prime ideal$(61, \sigma^0\Psi_{61}(\zeta))$does not exhaust the classes of the class subgroup$\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}$. It does not even form a class subgroup of$\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}\$. Hence we have lost all relationship between the class group and the Galois group.