Let $p$ be an odd regular prime and $F$ be a $p$-rational number field containing $\mu\_p$. Equivalently, there is a unique prime $\mathfrak{p}$ above $p$ in $F$ and the $p$-class group is generated by $\mathfrak{p}$ [Gras' book on Class Field Theory IV.3.5]. Let $F\_\mathfrak{p}$ be the maximal pro-$p$ unramified outside $\mathfrak{p}$ extension of $F$. Is it guaranteed that the inertia group of $\mathfrak{p}$ has finite index in $Gal(F\_\mathfrak{p}/F)$? Somehow is it true that in a cyclic $p$-extension of a $p$-rational field, the unique prime above $p$ must ramify and can not remain inert?



A question on p-rationality of number fields
I asked your question to my advisor Professor Movahhedi who introduced and studied the notion of $p$-rationality in his thesis.
Here I share his answer with you.


The two questions are somehow of different nature.


Denote by $F\_{S\_p}$ the maximal $p$-extension of the $p$-rational filed $F$ unramified outside the prime(s) above $p$. Suppose that $F\_{S\_p}$ has only one prime above $p$; for instance, when our $p$-rational field $F$ contains the $p^{th}$-roots of unity $\mu\_p$ as is assumed. Denote by ${\mathfrak p}$ and ${\cal P}$ the primes above $p$ in $F$ and $F\_{S\_p}$ respectively.
Then the corresponding decomposition group $D({\cal P}/{\mathfrak p})$ is the whole Galois group $G\_{S\_p}:=Gal(F\_{S\_p}/F)$ which can also be identified to the local Galois group after completion at the primes above $p$.
The corresponding inertia group $I({\cal P}/{\mathfrak p})$ is normal in the decomposition group $D({\cal P}/{\mathfrak p})$ and we have a short exact sequence  

$$0 \rightarrow I({\cal P}/{\mathfrak p}) \rightarrow G\_{S\_p}=D({\cal P}/{\mathfrak p}) \rightarrow Gal(E/F) \rightarrow 0, $$
where the inertia field $E$ is the maximal unramified extension of $F$ contained in $F\_{S\_p}$ since ${\cal P}$ is the only ramified prime in $F\_S/F$.
Therefore $E/F$ is a finite (cyclic) extension as was to be shown: the inertia group $I({\cal P}/{\mathfrak p})$ is a normal subgroup of finite index in $G\_{S\_p}=D({\cal P}/{\mathfrak p})$. Moreover the quotient is cyclic.
We also notice that, by maximality of $F\_{S\_p}$, it is clear that $E$ is in fact the maximal unramified $p$-extension of $F$.


As far as the second question is concerned, the answer is negative. For this, you first need to be familiar with the notion of primitive sets of primes which play an important role in the going-up properties :


**Definition** (Movahhedi's thesis, Definition 1, page 42).
A set $S$ of primes of $F$ containing $S\_p$ is called primitive for ($F, \; p$) if
the Frobenius $\sigma\_v({{\tilde{F}}}\_1/F)$ "attached" to the primes $v$ in $S-S\_p$ generate an ${\bf F}\_p$-subspace of ${\rm Gal}({{\tilde{F}}}\_1/F)$ of dimension the
cardinality of $S-S\_p$, where ${{\tilde{F}}}\_1$ is the compositum of
the first layers of the ${\bf Z}\_p$-extensions of $F$.


Now, for each odd regular prime $p$, there exist infinitely many $p$-rational fields $F \supset \mu\_p$ admitting a cyclic extension of degree $p$ (contained in $F\_{S\_p}$) in which the $p$-adic prime of $F$ is inert. Namely, start with the cyclotomic field $k:=Q(\mu\_p)$ which is $p$-rational. Consider a maximal primitive set $T:=\{{\mathfrak p, \mathfrak L\_1, \mathfrak L\_2, \cdots, \mathfrak L\_r}\}$ for $(k,p)$, where $r:=(p+1)/2$ is the number of independent $Z\_p$-extensions of $k$ and ${\mathfrak p}=(\zeta\_p-1)$ is the $p$-adic prime of $k$. By the Čebotarev density theorem, infinitely many such primitive sets exist. Since $k$ is $p$-principal (i.e. $p$ does not divide the class number of $k$), for each $i= 1, 2, \cdots, r$, there exists an integer $\alpha\_i$ in $k$ such that ${\mathfrak L\_i^{t\_i} = (\alpha\_i)}$ for some prime-to-$p$ integer $t\_i$. Now consider the field $F$ obtained by adding to $k$ a $p$-th root of $a:=(\zeta\_p-1)\alpha\_1 \alpha\_2 \cdots \alpha\_r$, where $\zeta\_p \neq 1$ is a $p$-th root of unity : $F=Q(\zeta\_p, \sqrt[p]{a}).$
Then the cyclic Kummer extension $F/k$ is ramified exactly at the primes in the primitive set $T$ (notice that the binomial $X^p-a$ is of Eisenstein type with respect to each prime in $T$) and unramified elsewhere.
Therefore, $F$ is $p$-rational according to the following theorem giving the necessary and sufficient condition for the $p$-rationality to go up along a finite $p$-extension.


**Theorem** (Movahhedi's thesis, Theorem 2, page 50).
Let $F/k$ be a finite (Galois) $p$-extension of number fields.
Then $F$ is $p$-rational precisely when $k$ is $p$-rational and the set $S\_k$ of those primes in $k$ lying above $p$ or ramified in $F$ is primitive for ($k,p$).
Moreover, if $F$ is $p$-rational, then the extension $S\_F$ of $S\_k$ to $F$ remains primitive for ($F,p$).


Besides, since $p$ does not divide the class number of $k$, applying Chevalley's genus formula to the cyclic extension $F/k$, we have the following equality for the order of the class number of $F$ fixed by the Galois group $G:=Gal(F/k)$


$$|Cl(F)^G|=\frac{\prod\_{v}e\_v(F/k)}{\textstyle [F:k][U\_k:U\_k\cap N\_{F/k}(F^\*)]} $$
with the usual notations :
$e\_v(F/k)$ is the ramification index of $v$, $U\_k$ is the unit group and $N\_{F/k}$ is the norm map corresponding to the extension $F/k$.


An easy calculation (maximality of the chosen primitve set $T$ and Dirichlet's unit theorem) then shows that, in our situation, $p$ divides the class number of $F$. Therefore, there exists a non-trivial cyclic $p$-extension of $F$ (namely the $p$-Hilbert class field of $F$) contained in $F\_{S\_p}$ in which the $p$-adic prime ${\mathfrak p}$ is inert.
Notice that smaller primitive sets may produce a field $F$ with the same properties if the norm index $[U\_k:U\_k\cap N\_{F/k}(F^\*)]$ is small.


Numerically, one can take $p=3$, $k:=Q(\zeta\_3)$ and let ${\mathfrak L\_7}$ and $\mathfrak L\_ {19}$ each be one of the primes above $7$ and above $19$, respectively. Then the set $T=\{{\mathfrak p, \mathfrak L\_7, \mathfrak L\_ {19}}\}$ is a maximal primitive set for $(k,3)$
[see Movahhedi's thesis, bottom of page 49].
It is not hard to see that we can take ${\mathfrak L\_7}=(\zeta\_3-2)$ and ${\mathfrak L\_{19}}=(\zeta\_3-7)$, so that  

$$a:=(\zeta\_3-1) (\zeta\_3-2) (\zeta\_3-7)= 33\zeta\_3-3.$$  

Hence, one corresponding such field is $F=Q(\zeta\_3, \sqrt[3]{33\zeta\_3-3})$, which is both $3$-rational and admits an unramified cyclic extension $E$ of degree $3$ (inside $F\_{S\_p}$) in which $3$ remains inert. This field $F$ is obtained by adding to $Q$ a root of $X^6 + 39X^3 + 1197$ which is the minimal polynomial of $\zeta\_3 \sqrt[3]{33\zeta\_3-3}$.


One could also take for $F$ the "smaller" field $F:=Q(\zeta\_3, \sqrt[3]{\zeta\_3-7})$, in which only the primes ${\mathfrak p}=(\zeta\_3-1)$ and ${\mathfrak L\_{19}}=(\zeta\_3-7)$ are ramified and have the same properties :
$F$ is both $3$-rational and admits an unramified cyclic extension $E$ of degree $3$ (inside $F\_{S\_p}$) in which $3$ remains inert. Indeed, $F$ is $3$-rational according to the above theorem since the set $\{{\mathfrak p, \mathfrak L\_ {19}}\}$ is , a fortiori, primitive for ($k,3$). On the other hand, since $N\_{F/k}(2+\sqrt[3]{\zeta\_3-7})=1+\zeta\_3$, then $N\_{F/k}((2+\sqrt[3]{\zeta\_3-7})^2)=\zeta\_3$ and the norm index in the above genus formula is therefore trivial. Accordingly, the class number of $F$ is divisible by $3$. Notice also that this $3$-rational field $F$ is obtained by adding to $Q$ a root of $X^6 +15X^3 + 57$ to $Q$ which is the minimal polynomial of $\zeta\_3 \sqrt[3]{\zeta\_3-7}$.


