# Kummer generator for the Ribet extension

Let $p$ be an odd prime and let $k\in[2,p-3]$ be an even integer such that $p$ divides (the numerator of) the Bernoulli number $B_k$ (the coefficient of $T^k/k!$ in the $T$-expansion of $T/(e^T-1)$). This happens for example for $p=691$ and $k=12$.

Ribet (Inventiones, 1976) then provides an everywhere-unramified degree-$p$ cyclic extension $E$ of the cyclotomic field $K={\bf Q}(\zeta)$ (where $\zeta^p=1$, $\zeta\neq1$) which is galoisian over $\bf Q$ and such that the resulting conjugation action of $\Delta={\rm Gal}(K|{\bf Q})$ on the ${\bf F}_p$-line $H={\rm Gal}(E|K)$ is given by the character $\chi^{1-k}$, where $\chi:\Delta\to{\bf F}_p^\times$ is the mod-$p$ cyclotomic character''.

Kummer theory then tells us that there are units $u\in{\bf Z}[\zeta]^\times$ such that $E=K(\root p\of u)$. Which units ?

More precisely, there is an ${\bf F}_p$-line $D\subset K^\times/K^{\times p}$ such that $E=K(\root p\of D)$. Which line ?

Let me first add that Herbrand wasn't the first to publish his result; it was obtained (but with a less clear exposition) by Pollaczek (Über die irregulären Kreiskörper der $\ell$-ten und $\ell^2$-ten Einheitswurzeln, Math. Z. 21 (1924), 1--38).

Next the claim that the class field is generated by a unit is true if $p$ does not divide the class number of the real subfield, that is, if Vandiver's conjecture holds for the prime $p$.

Proof. (Takagi) Let $K = {\mathbb Q}(\zeta_p)$, and assume that the class number of its maximal real subfield $K^+$ is not divisible by $p$. Then any unramified cyclic extension $L/K$ of degree $p$ can be written in the form $L = K(\sqrt[p]{u})$ for some unit $u$ in $O_K^\times$.

In fact, we have $L = K(\sqrt[p]{\alpha})$ for some element $\alpha \in O_K$. By a result of Madden and Velez, $L/K^+$ is normal (this can easily be seen directly). If it were abelian, the subextension $F/K^+$ of degree $p$ inside $L/K^+$ would be an unramified cyclic extension of $K^+$, which contradicts our assumption that its class number $h^+$ is not divisible by $p$.

Thus $L/K^+$ is dihedral. Kummer theory demands that $\alpha /\alpha' = \beta^p$ for some $\beta \in K^+$, where
$\alpha'$ denotes the complex conjugate of $\alpha$.

Since $L/K$ is unramified, we must have $(\alpha) = {\mathfrak A}^p$. Thus $(\alpha \alpha') = {\mathfrak a}^p$, and since $p$ does not divide $h^+$, we must have ${\mathfrak a} = (\gamma)$, hence $\alpha \alpha' = u\gamma^p$ for some real unit $u$.

Putting everything together we get $\alpha^2 = u(\beta\gamma)^p$, which implies $L = K(\sqrt[p]{u})$.

If $p$ divides the plus class number $h^+$, I cannot exclude the possibility that the Kummer generator is an element that is a $p$-th ideal power, and I cannot see how this should follow from Kummer theory, with or without Herbrand-Ribet.

If $p$ satisfies the Vandiver conjecture, the unit in question can be given explicitly, and was given explicitly already by Kummer for $p = 37$ and by Herbrand for general irregular primes satisfying Vandiver: let $g$ denote a primitive root modulo $p$, and let $\sigma_a: \zeta \to \zeta^a$. Then $$u = \eta_\nu = \prod_{a=1}^{p-1} \bigg(\zeta^\frac{1-g}{2}\ \frac{1-\zeta^g}{1-\zeta}\bigg)^{a^\nu \sigma_a^{-1}},$$ where $\nu$ is determined by $p \mid B_{p-\nu}$.

Here is a survey on class field towers based on my (unpublished) thesis on the explicit construction of Hilbert class fields that I have not really updated for quite some time. Section 2.6 contains the answer to your question for primes satisfying Vandiver.

• Thanks for the reference to Pollaczek and for the remarks. Ribet's major contribution in this story is not so much the result as the method, which has been exremely fruitful. See for example the notes of Mazur talk at the Ribet conference (math.harvard.edu/~mazur). Waiting anxiously for your lost construction to be found and put online. – Chandan Singh Dalawat Sep 7 '10 at 8:50
• [T]here will be more than one if the irregularity index is $>1$''. Yes, but we are fixing the pair $p$, $k$ (such that $p|B_k$), and then one expects (Iwasawa's conjecture) there to be a unique line $D\subset K^\times/K^{\times p}$ such that $E=K(\root p\of D)$ is everywhere unramified over $K$ and such that $\Delta$ acts on ${\rm Gal}(E|K)$ via $\chi^{1−k}$. – Chandan Singh Dalawat Sep 7 '10 at 9:10
• Thank you very much for putting your excellent notes (rzuser.uni-heidelberg.de/~hb3/publ/pcft.pdf) online. Future generations will owe you a debt of gratitude... (By the way, it is $l^2$, not $l^n$, in the title of Pollaczek's paper). – Chandan Singh Dalawat Sep 8 '10 at 10:05
• @Franz : Great answer ! But Cam (and Washington 15.8) seem to be constructing $E$ by adjoining the $p$-th roots of some (not very explicit) unit of $K$, irrespective of whether $p$ satisfies Vandiver or not. You seem to be skeptical that this can be done. What is going on ? – Chandan Singh Dalawat Sep 9 '10 at 3:16
• There are two cases in Washington's proof; in the second case he constructs a unit that gives a Kummer generator of a suitable unramified construction, but in the first case he does not. Without having gone through the details I bet that if Vandiver holds, we are always in the second case, and that the problematic case is the first case in Washington's proof. – Franz Lemmermeyer Sep 21 '10 at 18:41

Ribet's proof shows that the corresponding galois representation occurs as a factor of the p-torsion of the jacobian of a modular curve (of level p?). So in principle you can write the unit as the value of a modular function of the appropriate level on the torsion point. Might not be so easy to do in practice. Also, if it was easy to write this unit as a cyclotomic integer, presumably someone would have done it a long time ago.

• re: the paranthetical question: the level is $\Gamma_1(p)$. I agree with Felipe's pessimism about being able to get anything more explicit in practice. If there was some sort of easy explicit formula then in some sense one feels that the converse would be due to Kummer rather than Ribet. – Kevin Buzzard Sep 6 '10 at 13:46
• I share the pessimism. But I was hoping that people who know Euler systems might shed some light on the question. – Chandan Singh Dalawat Sep 7 '10 at 2:31

Here is an explicit construction$^*$.

Since there exists such an unramified $p$-extension, by class field theory the $p$-part of the class group of $\mathbb{Q}(\mu_p)$ is non-trivial. Further, specifying the $\Delta$-action gives more; namely, that the $\omega$-eigenspace of the $p$-part of the class group is non-trivial (for $\omega=\chi^{1-k}$). By Herbrand-Ribet, the $\omega$-eigenspace of the class group has the same order as the $\omega$-component of the $p$-part of $($units mod cyclotomic units$)$ in $\mathbb{Q}(\mu_p)$., so this quotient too is non-trivial. Now, following the proof of Theorem 15.8 in Washington's Cyclotomic Fields (roughly), we choose a unit $u$ whose $\omega$-projection $\varepsilon_\omega u$ in this quotient group is:

• Congruent to 1 modulo the prime above $p$ in $\mathbb{Z}[\zeta_p]^+$
• Not a $p$-th power of such a unit.
• *Is* a $p$-th power of an element of the topological closure of the group of these units.

Such a thing exists by the converse to Herbrand-Ribet$^{**}$. Then $\mathbb{Q}(\zeta_p,u^{1/p})/\mathbb{Q}(\zeta_p)$ is everywhere unramified and carries the proper action of $\Delta$, so this is the unit you're looking for.

$^*$: "Construction" may be a bit of an exaggeration. Following the proof of Theorem 15.8, however, I'm not immediately clear on what would be difficult to do explicitly. I think SAGE could handle local units well enough to carry out the construction in the proof. Unless someone comes and shoots down this answer, I might see if I can't get SAGE to do this explicitly. Edit: Chris Wuthrich makes a good point below -- even if there's no theoretical obstruction to doing everything explicitly, at a practical level computations would quickly become infeasible.

$^{**}$: Actually, there's one more case to consider, which amounts to doing a similar construction in a different ("reflected") eigenspace, but I think this is good enough to get the gist of the argument.

• I agree that this works in theory. I am not so sure any computer package can compute the units of $\mathbb{Q}(\mu_p)$ for $p\geq 37$. – Chris Wuthrich Sep 6 '10 at 15:20
• Thanks for the reference to Washington 15.8. But he seems to be proving the existence of $E$ by showing that there exists a unit $\delta\in K^\times$ such that... As I'm already admitting the existence of $E$ (Ribet), the existence of $\delta$ is not a problem: such $\delta$ are called $u$ in the question. It is also understood that the units $u$ will have to be quite special for $E|K$ to be everywhere-unramified, and behave under $\Delta$ in a particular way so that the action of $\Delta$ on ${\rm Gal}(E|K)$ is given by $\chi^{1-k}$. Problem : What is $u$ ? – Chandan Singh Dalawat Sep 7 '10 at 3:31
• I agree that the content of the theorem is not what you were looking for. But once you know Herbrand-Ribet is true (by whatever means), the proof of 15.8 seems to describe exactly how to make $u$ . I had originally intended to trace through the construction via an Euler systems approach (in earlier versions of this response, I made comments to this effect, since edited out), but this seemed to do the trick (though I will certainly concede that the explicitness of the construction might be significantly less than what you had in mind). On the other hand, as above, perhaps something more.. – Cam McLeman Sep 7 '10 at 3:55
• ..explicit is unlikely. I gave up trying to do something in terms of the Euler system elements after seeing the somewhat tricky-to-achieve conditions on $\delta$ that Washington gives in 15.8. In particular, becoming a $p$-th power when passing to the topological closure seems hard to encode simply by sums/products/norms/etc. of cyclotomic units. – Cam McLeman Sep 7 '10 at 3:58
• The difference between $u$ and $D$ is pretty negligible, right? Once you know $u$, $D$ is just the set of $u^j$ for $0\leq j\leq p-1$ (mod $K^{\times p}$). Also: Edited answer for u vs. delta. Wanted to comment lest Chandan's comment above looks odd. – Cam McLeman Sep 7 '10 at 12:59