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This question is an exact duplicate of:

I am having trouble understand theorem $9.4$ of Chapter $6$ of Lang's Algebra (pg. 300-301).

The setup is a we have a field $k$ of characteristic not dividing $n$. We know that the splitting field of $f=X^n-a$ is $k(\zeta_n,\alpha)$ where $\alpha$ is a root of $f$ and $\zeta$ a primitive $n^{th}$ root of unity. Any automorphism $\sigma$ of the Galois group of $f$ over $k$ maps $\alpha \mapsto \alpha\zeta^b$ where $b$ is unique modulo $n$, and $\sigma$ induces an automorphism of the cyclic group $\mathbf{\mu_n}=\left\langle \zeta \right \rangle $ via $\zeta \mapsto \zeta^d$ where $(d,n)=1$ and $d$ is uniquely determined by $\sigma$.

We then verify that the map $\sigma \mapsto \begin{pmatrix} \ \ 1 & 0\\ b_\sigma & d_\sigma \end{pmatrix}$ where $b$, $d$ are the integers determined by $\sigma$ in the previous paragraph is an injective homomorphism into the group $G(n)$ of all matrices $\begin{pmatrix} \ 1 & 0\\ a & c \end{pmatrix}$ such that $a \in \mathbf{Z}/n\mathbf{Z}$, $c \in (\mathbf{Z}/n\mathbf{Z})^{*}.$

The question that the theorem addresses is when the above the map is an isomorphism of the Galois group of $f$ and $G(n)$. With $\phi$ being the Euler function, the theorem states:

Suppose $[k(\zeta_n):k]=\phi(n)$ and let $a \in k$. Suppose for each prime $p|n$ that $a$ is not a $p^{th}$ power. Let $K$ be the splitting field of $X^n-a$ over $k$ and $G$ the Galois group. Then the above map is an isomorphism $G \cong G(n)$ with commutator subgroup Gal$(K/k(\zeta_n))$, so $k(\zeta_n)$ is the maximal abelian subextension of $K$.

The proof begins with the case $n=p$ where $p$ is a prime, which I follow. However, following that case Lang writes (bold is what I don't understand):

A direct computation of commutator of elements in $G(n)$ for arbitrary n shows that the commutator subgroup $C$ is contained in the group of matrices $ \begin{pmatrix} \ \ 1 & 0\\ b & 1 \end{pmatrix}$, $b \in \mathbf{Z}/n\mathbf{Z}$ and so must be that subgroup because its factor group is isomorphic to $(\mathbf{Z}/n\mathbf{Z})^{*}$ under the projection on the diagonal.

When $n=p$ is prime I already happened to know that $G \cong \mathbf{Z}/p\mathbf{Z} \rtimes_\varphi (\mathbf{Z}/p\mathbf{Z})^{*}$ from which it is clear (I think?) that the quotient by the image of $\mathbf{Z}/p\mathbf{Z}$ is the maximal abelian quotient, and the fact that the commutator subgroup is nontrivial inside a subgroup of order $p$ means it must be the whole group. However, when $n$ is arbitrary it's not obvious to me why the quotient by the commutator $C$ in $G(n)$ is ismorphic to $(\mathbf{Z}/n\mathbf{Z})^{*}$ nor why $C$ has to be isomorphic to $\mathbf{Z}/n\mathbf{Z}$. If someone could explain what is missing here that would be much appreciated.

I also have a couple of questions on the rest of the argument which I will just link here

rest of proof

1.) On the $3^{rd}-4^{th}$ lines: $\beta$ is a root of $X^m-a$ and by induction we can apply the theorem to $g=X^m-a$.

OK, fine, but what is Lang using the induction for? The splitting field for $g$ is $k(\beta,\zeta_m)$ and has Galois group isomorphic to $G(m)$ and it's maximal abelian extension is $k(\zeta_m)$. I've been staring at this and I don't see what the conclusion is.

2.) lines $3-5$ after the diagram: apply the $1^{st}$ part of proof (case of $n=p$ is prime) to $X^p-\beta$ over $k(\beta)$...shows that $k(\beta,\zeta_n)\cap k(\alpha)=k(\beta)$.

Again I don't know what exactly is being said. The splitting field of $X^p-\beta$ over $k(\beta)$ is $k(\alpha,\zeta_p)$, and it's maximal abelian subextension is $k(\beta,\zeta_p)$. How does Lang's conclusion follow from that?

Thanks to everyone who took the time to read my question, any help is much appreciated.

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marked as duplicate by Yanior Weg, mrtaurho, José Carlos Santos, Paul Frost, Leucippus May 23 at 4:11

This question was marked as an exact duplicate of an existing question.

  • $\begingroup$ As far as I know, it's so-called generalized Kummer theory, which is related to Galois representations (I haven't dug into this area, sorry). I don't know whether this material is clearer. $\endgroup$ – Yai0Phah Nov 25 '15 at 20:09
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I have given a complete proof of the Theorem in my own post here. The OP will probably not profit from it but I hope some desperate algebra students will.

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