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Marvin J. Greenberg provided an elementary proof of the Kronecker–Weber theorem here (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" (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 $G$.


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 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.

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  • $\begingroup$ I'm pretty sure that for $m > 1$, the first ramification group $V_j \ne G$ has index $j > 2$. There are results on jumps in the filtration of ramification groups in Serre's Local Fields from which this follows if it is correct -) $\endgroup$ – Franz Lemmermeyer Feb 13 '17 at 19:11
  • $\begingroup$ @FranzLemmermeyer : thank you for your comment. I don't understand when you're writing "the first ramification group $V_j \neq G$ has index $j>2$", do you mean that $V_i$ has index $>2$ (why would it be $i$?) where i is the smallest index such that $V_i \neq G$? $\endgroup$ – Watson Feb 13 '17 at 19:16
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    $\begingroup$ I meant that for $m > 1$ we have $V_1 = V_2$. $\endgroup$ – Franz Lemmermeyer Feb 13 '17 at 22:38
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    $\begingroup$ If ${\mathfrak p} = {\mathfrak P}^2$ in a quadratic extension $L/K$, then ${\mathfrak P}^2 \cap O_K = {\mathfrak p}$. $\endgroup$ – Franz Lemmermeyer Feb 24 '17 at 21:42
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    $\begingroup$ All such extensions are cyclotomic, and for cyclotomic extensions computing the ramification groups is a simple exercise: do it for the full cyclotomic field and then climb down to the $p$-extension using Hilbert's formula for the different. I think I'll find the time to do the calculation next week if you're willing to wait a little while. $\endgroup$ – Franz Lemmermeyer Feb 26 '17 at 9:39
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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 (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 $p =3, 5, 7$.

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