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 ?
 A: 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. 
A: 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.
A: 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. 
