Assume $p$ is an irregular prime for which
Vandiver's conjecture holds, e.g. $p<12'000'000$. This conjecture asserts that $p$ does not divide the $+$-part of the class group.
Then there is no capitulation in the class group from the first layer of the cyclotomic $\mathbb{Z}_p^{\times}$-tower to any other in this tower. See Proposition 1.2.14 in Greenberg's book, which says that the capitulation kernel lies in the $+$-part. See also the discussion on page 102 where it is discussed what happens when Vandiver's conjecture does not hold.
Generally capitulations in Iwasawa theory are well studied. The capitulation is linked to the question of whether there are non-trivial finite sub-$\Lambda$-modules in the Iwasawa module $X$, here the projective limit of the $p$-primary parts of the class groups in the tower, or equivalently the Galois group mentioned in the question.
$\newcommand{\L}{\mathcal{L}}$
$\newcommand{\Q}{\mathbf{Q}}$
$\newcommand{\Z}{\mathbf{Z}}$
$\newcommand{\F}{\mathbf{F}}$
The answer is yes. It suffices (as Brian mentions) to show that the corresponding space
of extensions is one dimensional.
Let $p$ be an odd prime, let $k$ be an integer, and let $\omega$ be the mod-$p$ cyclotomic character. The extensions we are interested in correspond to elements of the Selmer group $H^1_{\L}(\Q,\omega^k)$, where
$\L$ is defined by the following local conditions: unramified away from $p$, and
unrestricted at $p$ (so $\L_p = H^1(\Q_p,\omega^k)$). The dual Selmer group $\L^*$
corresponds to classes which are unramified outside $p$, and totally split at $p$.
Let us further suppose that $k \not\equiv 0,1 \mod p-1$, so neither $\omega^k$ nor
$\omega^{1-k}$ have invariants over $\Q_p$, and so $|\L_p| = p$. A formula of Wiles relating the Selmer group to the dual Selmer group implies (under the condition on $k$) that
$$ \frac{|H^1_{\L}(\Q,\omega^k)|}{|H^1_{\L^*}(\Q,\omega^{1-k})|}
= \frac{|\L_p|}{|H^0(G_{\infty},\omega^k)|} =
\begin{cases} p, & k \equiv 1 \mod \ 2, \\\
1, & k \equiv 0 \mod \ 2. \\ \end{cases}$$
(This result also follows more classicaly from so-called mirror theorems, and, phrased slightly differently, occurs in Washington's book on cyclotomic fields.)
In your situation, $k$ is odd, and so the space
$H^1_{\L}(\Q,\omega^k)$ is one dimensional
if and only if $H^1_{\L^*}(\Q,\omega^{1-k})$ vanishes.
Since $1-k$ is even, however, the vanishing of this latter group is essentially the same as Vandivier's conjecture (or rather, the $\omega^{1-k}$-part of Vandivier's conjecture). Since Vandiver's conjecture is true for $p = 37$, everything is ok in this case.
Update: Brian reconciles this answer with his previous comments in the comments to this answer below.
(If you don't assume Vandivier's conjecture, then it isn't clear what question to ask for general $p$, since
$H^1_{\L^*}(\Q,\omega^k)$ could have dimension $\ge 2$.)
Dear Rob,
For your second question (in the comments below), let $\Gamma$ denote the $\omega^{k}$-part of the maximal
extension of $\Q(\zeta_p)$ unramified outside $p$. Then there
is an exact sequence:
$$0 \rightarrow \Z_p \rightarrow \Gamma \rightarrow C \rightarrow 0,$$
where $C$ is the $\omega^k$-part of the class group tensor $\Z_p$. You ask
whether the $\Z_p$-extension is totally ramified. This
boils
down to the following question: Is the image of $\Z_p$ saturated
in $\Gamma$?
This is equivalent to asking
that the sequence above remains exact after tensoring
with $\F_p$. Equivalently, it is the same as asking that
$$\Gamma/p \Gamma \simeq \F_p \oplus C/pC \simeq
\F_p \oplus H^1_{\L^*}(\Q,\omega^k).$$
Yet $\Gamma/p \Gamma \simeq H^{1}_{\L^*}(\Q,\omega^k)$. Thus, using
Wiles' formula again, this is the same as asking that:
$$p = \frac{|H^1_{\L}(\Q,\omega^k)|}{|H^1_{\L^*}(\Q,\omega^k)|}
= \frac{p |H^1_{\L^*}(\Q,\omega^{1-k})|}{|H^1_{\L}(\Q,\omega^{1-k}|},$$
or equivalently:
The $\Z_p$-extension is totally ramified if and only if
there does not exist a $\omega^{1-k}$ extension of $\Q(\zeta_p)$ which
is unramified outside $p$ but ramified at $p$.
Note that:
If $p$ is regular, then there is no such extension, and so the
$\Z_p$-extension is totally ramified. (This is obvious more directly.)
If $p$ is irregular, then mirror theorems imply that there does exist
a $\omega^{1-k}$-extension unramified outside $p$; if Vandiver's conjecture holds this extension is ramified, and so the
$\Z_p$-extension is not totally ramified (this also
follows from the first answer).
If $p$ is irregular, and the $\omega^k$ part $C$ of the class group is cyclic
(a consequence of Vandiver's conjecture), then
$H^1_{\L}(\Q,\omega^{1-k})$ has order $p$. Thus, in this case,
the $\Z_p$-extension is not totally ramified if and only if Vandiver's
conjecture holds (for the $\omega^{1-k}$ part of the class group).
This example shows that one can't really give a particularly clean
answer, since cyclicity of the class group is seen as a strictly weaker
property than Vandiver. This shows that you can't really expect a cleaner answer than I gave above (since proving Vandiver from cyclicity sounds very hard).
BTW, there is a little check mark button next to this answer, if you press it, it goes green, which lights up the dopamine receptors in my brain.
Best Answer
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.