The exercise 5.41 in Algebraic Number Theory, Second Edition by Richard A. Mollin asks:

Prove that any number field $F$ abelian over $\Bbb Q$ with both degree over $\Bbb Q$ and discriminant a power of an odd prime $p$ must be a cyclic extension of $\Bbb Q$.

He gives the proof: Mollin 5.41 The idea is that $p$ is the only ramified prime in $F$, and is actually totally ramified. Then the order of the inertia group $T_p$ is $[F:\Bbb Q]$, so that $T_p=\mathrm{Gal}(F/\Bbb Q)$. Moreover, $T_p/V_1$ is cyclic, so it sufficient to show that the first ramification group is trivial : $V_1=1$. This is what he claims at the very end of the proof above.

My question is:

Why would $V_1$ be trivial?

For me, this sounds wrong. If the decomposition group $D_p$ is abelian, then we can show that the order of $T_p/V_1$ divides $q-1$, where $q$ is the norm of $p$... which is $p$ itself. So $|T_p/V_1|$ is a power of $p$ and divides $p-1$, so $T_p = V_1$ is the whole Galois group!

Instead, I think that $V_2 = 1$ holds (which is not obvious, but then we would be done by showing that $V_1/V_2$ is cyclic). In any case, do you agree that the claim $V_1=1$ (and then Mollin's final argument) is wrong?

By the way, this exercise is used in the proof of the Lemma 5.14 which is part of his proof of Kronecker–Weber theorem...

Thank you!

  • $\begingroup$ Have you tried computing an example? $\endgroup$ – Mathmo123 Jan 9 '17 at 14:18
  • $\begingroup$ @Mathmo123 : I don't think there is need to compute an example : I have an argument showing that $T_p=V_1$ is the whole Galois group. (The statement of 5.41 is correct, but the proof doesn't seem to be correct to me). $\endgroup$ – Watson Jan 9 '17 at 14:21
  • $\begingroup$ Can you explain your notation, what is $V_n$ in general? $\endgroup$ – Adam Hughes Jan 9 '17 at 14:42
  • $\begingroup$ @Adam Hughes : yes sorry… $V_n$ (denoted by $\mathcal V_n$ in Mollin's book) is the $n$-th ramification group (of any prime ideal $P$ above $p$), i.e. the set of $\sigma$ in $Gal(F/\bf Q)$ such that $\sigma(x)-x \in P^{n+1}$ for any $x \in \mathcal O_F$. $\endgroup$ – Watson Jan 9 '17 at 14:46
  • $\begingroup$ "$T_p/V_1$ is cyclic" is shown in theorem 2 here, p. 267. The order is dividing $q-1$, as shown in Prop J (1), p. 269. $\endgroup$ – Watson Jan 9 '17 at 14:54

Just an informative remark. If you accept to use more powerful tools than just ramification theory, more precisely CFT (which, I guess, is above the level of a first course in ANT (*)), then Mollin's exercise becomes easy. The hypotheses mean that $F$ is an belian extension of $Q$ which is $p$-ramified, i.e. unramified outside the prime $p$. But CFT shows that the maximal such extension is precisely the so called cyclotomic $Z_p$-extension $Q_{\infty}/Q$, whose Galois group over $Q$ is isomorphic to the additive group of the $p$-adic integers $Z_p$. See e.g. https://math.stackexchange.com/a/2087797/300700. So not only $F/Q$ is a cyclic $p$-extension, but it is a finite "layer" of $Q_{\infty}/Q$ .

(*) But I notice that the exercise is a preliminary to the K-W. theorem

  • $\begingroup$ Thank you for your answer. However, it doesn't adress the original question about the first ramification group $V_1$. $\endgroup$ – Watson Jan 10 '17 at 9:02
  • 1
    $\begingroup$ @Watson I completely agree, but I stressed right from the start that it was just an informative remark. I alluded to the K-W theorem because even in this framework (CFT over Q), it is interesting, I think, to point out that the maximal pro-p-abelian p-ramified of Q is its cyclotomic Z_p-extension. $\endgroup$ – nguyen quang do Jan 10 '17 at 10:18

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