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.


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

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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