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 ?



Kummer generator for the Ribet extension
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](http://www.rzuser.uni-heidelberg.de/~hb3/publ/pcft.pdf) 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. 


