Does the everywhere unramified extension of Q(mu_37) of degree 37 grow into a Z_37-extension? Let $p=37$.  Since $p$ divides the numerator of $B_{32}$, by Ribet's proof of the converse of Herbrand's theorem, we know that the class group of ${\bf Q}(\mu_p)$ has size divisible by $p$. More specifically, the $p$-part of the class group decomposes under the action of $\Delta = {\rm Gal}({\bf Q}(\mu_p)/{\bf Q})$, and we know exactly which eigenspace is non-trivial; in this case, the $\omega^5$-eigenspace is non-trivial where $\omega$ is the mod $p$ cyclotomic character.
Now let $X$ denote the $p$-Hilbert class field of ${\bf Q}(\mu_p)$. As above, $X$ decomposes under the action of $\Delta$, and we have that $X^{(\omega^5)}$ is 1-dimensional over ${\bf F}_p$. Thus, there is some abelian everywhere unramified extension $H_5$ of ${\bf Q}(\mu_p)$ 
such that $\Delta$ acts on ${\rm Gal}(H_5/{\bf Q}(\mu_p))$ by $\omega^5$.
Let $Y$ denote the Galois group of the maximal abelian $p$-extension of ${\bf Q}(\mu_p)$ unramified outside of $p$. By Leopoldt, the ${\bf Z}_p$-rank of $Y$ (modulo its torsion) equals $(p+1)/2$.  We can also decomposing $Y$ under the action of $\Delta$; I presume what happens is that eigenspace of $\omega^0$ is rank 1 over ${\bf Z}_p$ (corresponding to the cyclotomic ${\bf Z}_p$-extension), and the eigenspace of $\omega^i$ is rank 1 over ${\bf Z}_p$ for $i$ odd. In particular, there is some ${\bf Z}_p$-extension $M_i$ over ${\bf Q}(\mu_p)$ such that  $\Delta$ acts on ${\rm Gal}(M_i/{\bf Q}(\mu_p))$ by $\omega^i$ for each odd $i$.
My question: is $H_5$ the first layer of the ${\bf Z}_p$-extension $M_5/{\bf Q}(\mu_p)$?
 A: $\newcommand{\L}{\mathcal{L}}$
$\newcommand{\Q}{\mathbf{Q}}$
$\newcommand{\Z}{\mathbf{Z}}$
$\newcommand{\F}{\mathbf{F}}$
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. 
