# A reduction-step on a proof of Kronecker-Weber theorem

I am trying to follow a proof of the Kronecker-Weber theorem (if $K$ is a number field with abelian Galois group then it is contained in a cyclotomic field) but I cannot understand one step.

I could understand that one can only consider the case where $K$ is an abelian extension of prime power degree (say $p^n$) such that $p$ is the only ramified prime. Specifically, one proves the following:

Every abelian extension of $\mathbb{Q}$ is the compositum of abelian extensions of prime power degree.

Let $G=\text{Gal}\left (K/\mathbb{Q}\right )$. Since $G$ is abelian, it is the direct sum of its Sylow subgroups. If $\left |G\right |=p_1^{n_1}\cdots p_r^{n_r}$ is the prime decomposition of the order of $G$, then we take $G_i$ to be the direct sum of all $p_j$-Sylow subgroups of $G$ with $j\neq i$. The fixed field $K_{G_i}$ has degree $p_i^{n_i}$ for each $i$, therefore $K$ is the composite of all $K_{G_i}$ because that composite has degree $p_1^{n_1}\cdots p_r^{n_r}$ (this is because the degrees are pairwise coprime). $\blacksquare$

Therefore the theorem is reduced to the case where $K$ is an abelian extension of prime power degree. Now, one can prove the following:

Let $K$ be an abelian extension of $\mathbb{Q}$ with $\left [K:\mathbb{Q}\right ]=p^m$. Suppose $q$ is a prime $\neq p$ which is ramified in $K$.

Fix $Q$ a prime of $K$ lying over $q$. Then the ramification index $e=e(Q|q)$ divides $q-1$. Let $L$ be the unique subgroup of $\mathbb{Q}(\xi)$ of degree $e$, where $\xi=e^{\frac{2\pi i}{q}}$.

Let $U$ be a prime of $KL$ lying over $Q$ and $K'$ denote the inertia field $\left (KL\right )_{E\left (U|q\right )}$. Then $E\left (U|q\right )$ is cyclic and $e(U|q)=e$. Also, $q$ is unramified in $K'$ and every prime of $\mathbb{Z}$ which is unramified in $K$ is also unramified in $K'.$

Finally, $KL=K'L$ and $K'$ has prime power degree over $\mathbb{Q}$.

I won't do the details. Essentially, all we did is obtain a new number field which also has prime power degree, where $q$ has no ramification and the primes unramified in $K$ are still unramified in $K'$. Therefore, repeating the procedure we obtain a number field of degree $p^k$ in which the only ramified prime is $p$.

Up to now I understood everything. But by a further reduction we can consider that $K$ is a cyclic extension, but I do not understand the proof, which is the following:

Since any abelian group is a direct sum of cyclic group of prime power order, any abelian extension is a composite of cyclic extensions of prime power degree. The proposition above allows us to assume that $p$ is the only ramiﬁed prime in each of these cyclic extensions. $\blacksquare$

I get that the abelian extension is the composite of cyclic extensions of prime power degree, but I do not understand why the above proposition guarantees that the number field $K'$ constructed above is still a cyclic extension of $\mathbb{Q}$. Any help?

EDIT:

User @reuns asked me to put more detail on the construction of $L$ and $K'$. Specifically, this is a little amount of exercises on Chapter $4$ of Marcus's Number Fields, and the specific exercise is the following:

• I'm probably screwing this up, so I'm chickening out and putting this in a comment. But aren't we in the situation of $E < G < G_1 \times G_2$, where $E$ projects isomorphically onto $G_2$, so that $G/E$ is a subquotient of $G_1$? Namely, $G= {\rm Gal}(LK,{\mathbb Q})$, $E= {\rm Gal} (LK/K')$, $G_1 = {\rm Gal}(K,{\mathbb Q})$, and $G_2= {\rm Gal}(L, {\mathbb Q})$? – peter a g Dec 6 '17 at 6:32
• @reuns Yes, sure. I will add the precise exercises (which are on Marcus's Number Fields, Chapter 4). – solomeo paredes Dec 6 '17 at 17:58
• Ok I remember now why I never studied Marcus. There are many other much better self-contained proofs. – reuns Dec 6 '17 at 18:18
• @peterag I could follow you up to a point. Yes, we are in the case you said, but I think that it is not true that $E$ projects isomorphically onto $G_2$. What is really true is that $E$ projects isomorphically onto $\text{Gal}\left (L/L\cap K'\right )$ (theorem 1.12 on Lang's Algebra, "Galois Theory" section). We would be done if we were able to show that $L\cap K'=\mathbb{Q}$. – solomeo paredes Dec 6 '17 at 18:40
• Maybe this works: it is easy to see that $q$ is totally ramified in $L$ and we know that $q$ is unramified in $K'$. Therefore $K'\cap L=\mathbb{Q}$ (exercise $24$ $(b)$ on Marcus's Number Fields, Chapter $3$). – solomeo paredes Dec 6 '17 at 18:44