Marvin J. Greenberg provided an elementary proof of the Kronecker–Weber theorem [here](https://www.jstor.org/stable/2319208) (Amer. Math. Monthly, 81 (1974), no. 6, 601-607).
An argument in the lemma 4 was found to be wrong as noticed in *[Correction to "An Elementary Proof of the Kronecker-Weber Theorem"](http://www.jstor.org/stable/2319794)*  (Amer. Math. Monthly, 82 (1975), no. 8, 803):



> 
> Joe L. Mott informed me that the argument for the case $m>1$ in Lemma 4, Volume 81, (1974) 606, is incorrect. What is correct is that $V\_i$ is the unique subgroup of $G$ of index $\lambda$, where $i$ is the smallest index such that $V\_i \neq G$.
> 
> 
> 


The lemma 4 is:



> 
> Let $K$ be an abelian extension of $\Bbb Q$ of degree $\lambda^m$, $\lambda$ an odd prime, in which $\lambda$ is the only ramified prime. Then $K/\Bbb Q$ is cyclic.
> 
> 
> 


The case $m=1$ is dealt correctly. **However, I wasn't able to see where exactly Greenberg's proof fails for the case $m>1$** (the numbering and square brackets are mine) :



> 
> Returning to the case $m > 1$, we will show that $K/\Bbb Q$ is cyclic by showing $V\_2$ *[i.e. the second ramification group of $\lambda$ in $K$]* is the unique subgroup of the Galois group $G = V\_1$ of index $\lambda$.
>  Let $H$ be any subgroup of index $\lambda$ in $G = \mathrm{Gal}(K/\Bbb Q)$, $K'$ its fixed field, $G' \cong G /H$ the Galois group of $K'$ over $\Bbb Q$, $V\_j'$ the $j$-th ramification group of $K'$.
> 
> 
> (1) By restriction to $K'$, $V\_j$ maps into $V\_j'$. According to the sublemma *[i.e. the case $m=1$]*, $V\_2'$ is trivial. Hence $V\_2 \leq H$.
> 
> 
> (2) Applying this, in particular to the case where $H = V\_j$ is the first ramification group which is not all of $G$, we see that $j = 2$ and $V\_2$ has index $\lambda$. Hence $V\_2$ is the unique subgroup of index $\lambda$.
> 
> 
> 


I see no problem with (1) : let $f : G \to G' \cong G/H$ be the restriction to $K'$. Then $f(V\_j) \subset V\_j'$. Since $V\_2' = 1$, we get $V\_2 \subset \ker(f) = H$.


I see no problem with (2), since $V\_{j-1} / V\_j$ is non trivial, and embeds in $O\_K/(\lambda)$ (fact 3 in Greenberg's paper), which has cardinality $\lambda$ because $\lambda$ is totally ramified in $K$ (see the very beginning of the proof of Lemma 4). So $V\_j$ has indeed index $\lambda$ in $G$.




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I tried to find a counter-example of an abelian extension $K/\Bbb Q$ whose degree and discriminant are both powers of an odd prime $\lambda$, such that the second ramification group $V\_2$ of $\lambda$ (in $K$) is the whole Galois group, but it was without success. Maybe the following [MAGMA](http://magma.maths.usyd.edu.au/calc/) code could help:



```
p := 3; r := 3;
a := RootOfUnity(p^r);
M := MinimalPolynomial(a + 1/a);    #here K=NumberField(M) will be the subfield of Q(a) fixed by a subgroup of order p-1 in Gal(Q(a))
R<x> := PolynomialRing(RationalField());
K<a> := NumberField(M);
OK := RingOfIntegers(K); OK;
print " ";print " ";

Gal, _, Map := AutomorphismGroup(K); Gal;
P3 := Decomposition(OK, 3)[1][1]; P3;
print " ";print " ";

V2 := RamificationGroup(P3, 2); V2;
print "Cardinality of V2 is ", #V2;

```

Thank you very much for your help.



Where exactly is the flaw in this proof of the Kronecker–Weber theorem?
As promised, here are my calculations of the higher ramification groups in certain cyclotomic extensions.


Let $G$ denote the Galois group of the field $L$ of $p^n$-th roots of unity, and consider the Hilbert subgroups for the completely ramified prime $p$. The decomposition group and the inertia both are equal to $G$. For computing the higher ramification groups, observe that if
$\sigma(\zeta) - \zeta \in {\mathfrak p}^j$ for some $j \ge 1$, then 
$\sigma(\alpha) - \alpha \in {\mathfrak p}^j$ for all $\alpha \in {\mathbb Z}[\zeta]$. Thus we only need to look at the action of $\sigma$ on roots of unity.


Let $\sigma\_a: \zeta \to \zeta^a$. Then clearly


$$ (\zeta^{a-1}-1) = \begin{cases} 
 {\mathfrak p} & \text{ if } a-1 \in {\mathbb Z} \setminus p{\mathbb Z}, \\
 {\mathfrak p}^p & \text{ if } a-1 \in p{\mathbb Z} \setminus p^2{\mathbb Z}, \\
 {\mathfrak p}^{p^2} & \text{ if } a-1 \in p^2{\mathbb Z} 
 \setminus p^3{\mathbb Z}, \\
 \cdots & \quad \cdots \\
 {\mathfrak p}^{p^{n-1}} & \text{ if } a-1 \in p^{n-1}{\mathbb Z} 
 \setminus p^n{\mathbb Z}.
 \end{cases} $$


This gives us the higher ramification groups:


$$ \begin{array}{lllll}
 V\_1 & = \ V\_2 & = & \ldots = V\_{p-1} & \simeq {\mathbb Z}/p^{n-1}{\mathbb Z}, \\
 V\_p & = \ V\_{p+1} & = & \ldots = V\_{p^2-1} & \simeq {\mathbb Z}/p^{n-2}{\mathbb Z}, \\
 V\_{p^2} & = \ V\_{p^2+1} & = & \ldots = V\_{p^3-1} & \simeq {\mathbb Z}/p^{n-3}{\mathbb Z}, \\
 & \ \ldots & & & \\
 V\_{p^{n-2}} & = \ V\_{p^{n-2}+1} & = &\ldots = V\_{p^{n-1}-1} & \simeq {\mathbb Z}/p{\mathbb Z}, \\
 V\_{p^{n-1}} & = \ 1 & & & 
 \end{array} $$


Next we use Herbrand's theorem on the determination of ramification groups for subfields. Let $H$ be the subgroup of $G$ fixing the subextension $K$ of degree $p^{n-1}$; then $H$ has order $p-1$, and it only has the neutral element in common with the higher ramification groups. This implies that the higher ramification groups in $L/K$ are all trivial, and that $n\_2 = n\_3 = \ldots = 1$ in the notation of his article over [here](http://visualiseur.bnf.fr/ark:/12148/bpt6k31445) (modern representations of his result invoke the "upper numbering" which I haven't looked at for some time), and this in turn implies that the ramification groups "collapse" by a factor of $n\_1 = p-1$. By this I mean the following: instead of $p-1$ ramification groups $\simeq {\mathbb Z}/p^{n-1}{\mathbb Z}$ there is only 
one such group for $K/{\mathbb Q}$, namely $V\_1$. Instead of $p^2-p$ ramification groups $\simeq {\mathbb Z}/p^{n-2}{\mathbb Z}$ there are now $p$ such groups, namely $V\_2 = \ldots = V\_{p+1} \simeq {\mathbb Z}/p^{n-2}{\mathbb Z}$, and so on. In particular, $V\_2 \ne G$ whenever $K$ is cyclotomic of prime power degree. In particular there cannot exist examples with $V\_2 = G$ by Kronecker-Weber.


As a sanity check I computed the discriminant of $K$ (in the case $n = 3$) using Hilbert's formula for the different and obtained disc$(K) = p^k$ with $k = 3p^2 - p - 2$, which agrees with calculations done by pari for $p =3, 5, 7$.


