This is an exercise in the book "Local fields an their extensions" by I.B. Fesenko and S.V. Vostokov (Ch. IV, sect. 6, Exercise 10, page 161).

Let $F$ be a local field with finite residue field and let $M$ be the maximal abelian extension of the maximal abelian extension of the maximal abelian extension of $F$. Denote by $B(M/F)$ the union of all upper ramification jumps of finite Galois subextensions of $M/F$. Show that $B(M/F)$ is dense in $[0, \infty)$ and deduce that every nonnegative real number is an upper ramification jump of $M/F$.

Local class field theory describes the maximal abelian extension $F^{ab}$ of $F$, and the upper ramification jumps of finite subextensions of $F^{ab}/F$ are integers by the Hasse-Arf theorem. But I don't know what to say about the maximal abelian extension of $F^{ab}$ and about the maximal abelian extension of this extension.

It would be great if someone could help me.


For concreteness, I’ll prove the fact for $F=\Bbb Q_p$, and I’m sure you’ll have no difficulty modifying those few things that are necessary to come through to a general proof.
Consider the set $\Lambda$ of all positive rational numbers of form $\lambda=r/n$, where $n$ is prime to $p$. I’ll show that such a $\lambda$ is the upper number for a finite extension contained in the maximal abelian extension $M$ of the max ab ext of the max ab ext of $F$. You already know the following:

Lemma: Let $F\subset M_0\subset M$. Then if $\lambda$ is an upper number for the extension $M_0\supset F$, $\lambda$ remains an upper number for $M\supset F$.

So for $\lambda\in\Lambda$, it’s enough to find a finite extension $M_0$ of $F$, contained in $M$ with $\lambda$ among the upper numbers for $M_0\supset F$. Following Fesenko-Vostokov, I’ll use here the covariant Hasse-Herbrand function, which I’ll call $\Psi^K_k$, for an extension $K\supset k$. It’s covariant in the sense that if $k\subset K\subset L$, then $\Psi^L_k=\Psi^L_K\circ\Psi^K_k$. And I’ll presume that you know that if $k$ is a local field and $\mathscr K$ is an extension of $k$ that’s maximal with respect being totally ramified and abelian, then the vertices of $\Psi^{\mathscr K}_k$ are at $(0,0)$, $(1,q-1)$, $(2,q^2-1)$, $(3,q^3-1)$, etc., where $q$ is the cardinality of the residue field for $k$. As always, the upper numbers are the first ($x$)-coordinates of the vertices of the HH function.

With all this out of the way, we can proceed. Let $\lambda=r/n$ be given in $\Lambda$. I’m going to construct an extension $M_0\supset F$ that’s contained in $M$, by finding a tower of finite extensions, $F\subset K_0\subset L_0\subset M_0$, each layer being abelian. Then clearly $M_0$ is finite over $F$ and contained in $M$. And you’ll see that $r/n$ is an upper number for the extension $M_0\supset F$, in other words, the $x$-coordinate of a vertex of $\Psi^{M_0}_F$.

The first layer is unramified, $K_0=F(\zeta)$, where $\zeta$ is a primitive $n$-th root of unity; you see that this extension is of degree $N$, where $p^N-1$ is the smallest number of this form divisible by $n$. And this extension is cyclic. The HH-function here has no vertices, it’s just given by $y=x$ all the way out.

The second layer is gotten by adjoining an $n$-th root of the prime element $p\in F$: $L_0=K_0(p^{1/n})$; this extension is totally ramified of degree $n$, and tamely ramified, because $p$ doesn’t divide $n$. Thus the HH-function has only the one vertex, at the origin, and slope $n$ beyond that: given by $y=nx$. Notice that since the $n$-th roots of unity are in $K_0$, this is a Kummer extension, has cyclic Galois group of order $n$.

For the last step, take a finite extension $M_0\supset L_0$ contained in a maximal totally ramified abelian extension of $L_0$, big enough to have a vertex at $(r,p^{rN}-1)$.

Finally, evaluate $\Psi^{M_0}_F$ at $\lambda=r/n$: $$ \Psi^{M_0}_F(r/n)=\Psi^{M_0}_{L_0}\circ\Psi^{L_0}_{K_0}\circ\Psi^{K_0}_F(r/n)=\Psi^{M_0}_{L_0}(r)\,, $$ but this last-acting function has a vertex at $r$, i.e. the slope jumps there. So the derivative of $\Psi^{M_0}_F$ jups at $r/n$, and we are done.


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