Lang Algebra Chapter 6 Theorem 9.4

Question

In Section 9 of Chapter 6, Lang constructrs for the first time a non-abelian extension as a splitting field $K$ of $X^n - a$ when the ground field $k$ does not contain primitive $n$-th roots of unity $\zeta_n$. He constructs an injective homomorphism of the Galois group $G(K/k)$ into the matrices of type

$$ M=\begin{pmatrix} 1 & 0 \\ b & d \end{pmatrix} \quad \hbox{with}\ b\in \mathbb{Z}/n\mathbb{Z} \quad \hbox{and} \quad d\in (\mathbb{Z}/n\mathbb{Z})^*,$$

where $(\mathbb{Z}/n\mathbb{Z})^*$ the multiplicative group of integers modulo n.

Afterwards, he states in Theorem 9.4 that if $n$ is an odd positive integer prime to the characterstic and $[k(\zeta_n):k]=\varphi(n)$ (i.e, the primitive roots of unity are not in the ground field), then the above homomorphism is, in fact, an isomorphism. Besides, the commutator group is $G(K/k(\zeta_n))$, so $k(\zeta_n)$ is the maximal abelian subextension of $K$.

Lang proves the theorem inductively over the number of prime factors in the decomposition of $n$ (i.e. if $n=9= 3\cdot 3$, then the proof "would take two steps" to complete). The induction start with $n$ prime is written clearly and I agree with it.

Now come the questions:

1. I do not understand Lang's argument to prove that the commutator subgroup is the subgroup $\begin{pmatrix}1 & 0 \\ b & 1\end{pmatrix}$, $b\in \mathbb{Z}/n\mathbb{Z}$. What is this projection on the diagonal and how does one see that the factor group of the commutator group is isomorphic to $(\mathbb{Z}/n\mathbb{Z})^*$?

By the way, direct computation of $x\, y\, x^{-1}\, y^{-1}$ for two arbitrary matrices $x=\begin{pmatrix} 1 & 0 \\ b & d \end{pmatrix}$, $y=\begin{pmatrix} 1 & 0 \\ c & e \end{pmatrix}$ yields a matrix $\begin{pmatrix} 1 & 0 \\ b(1-e)+c(d-1) & 1 \end{pmatrix}$. As $n$ is odd, $2 \in (\mathbb{Z}/n\mathbb{Z})^*$, so taking $e=1$ and $d=2$, we indeed reconstruct the above-mentioned group.

2. The second (inductive) part of the proof is a complete mystery to me with virtually every sentence being a Theorem which needs an extra proof. Can someone give me some hints?

2.a. Lang writes $n=p\,m$ and takes $\alpha$ to be the root of $X^n-a$. It is clear that $\beta=\alpha^p$ is a root of $X^m -a$. Now comes the argument "By induction we can apply the theorem to $X^m-a$". I agree that we can apply it here but the result is that $K=k(\beta,\zeta_m)$ is the splitting field of $X^m-a$. In the diagram below, there appears $k(\beta,\zeta_n)$ out of nowhere.

2.b. Now Lang proceeds to apply the theorem on $X^p-\beta$ over $k(\beta)$. Bergman's "Companion" reminds to check that $[k(\beta,\zeta_p):k(\beta)]=\varphi(p)$ before applying the theorem and gives a hint "$[k(\zeta_p):k]$ and $[k(\beta):k]$ are relatively prime". I disagree with this hint. Take $n=21= m p$ where $m=3$ and $p=7$. Then $[k(\zeta_p):k]=\varphi(p)=p-1=6$ but $[k(\beta):k]=m=3$ and both are not co-prime. Can we apply the theorem here at all? Or if so what is the argument? If Bergman's argument is valid, where is the fault in my logic?

2.c. Even assuming that we can apply the theorem in 2.b., then we can say something about $k(\alpha, \zeta_p)$ and not $k(\alpha,\zeta_n)$. How do we proceed to checking everything Lang states in the last section of the proof (basically, everything after the diagram)?

EDIT: I have found the essentially same question here

Answer 1

I have finally got all the details of the proof which I post as a separate answer to leave the original questions (and remarks about Bergman's companion book) for the posteriority. The notation is as in the question except that I also write $k(\mu_n)$ for $k(\zeta_n)$.

Theorem 9.4 Let $k$ be a field and $n$ an odd positive integer prime to the charactersitc. Assume that $[k(\mu_n):k]=\varphi(n)$. Let $a\in k$ and suppose that, for each prime $p|n$, $a$ is not a $p$-th power in $k$. Let $K=k(\alpha, \mu_n)$ be the splitting field of $X^n-a$ over $k$. Then the above homomorphism $\sigma\mapsto M_\sigma$ is an isomorphism of $G(K/k)$ and $G(n)$.

The commutator group is $G(K/k(\mu_n))$, so $k(\mu_n)$ is the maximal abelian subextension of $K$. Furthermore, $k(\alpha)\cap k(\mu_n) = k$.

Proof The proof will be carried out inductively over the number of prime numbers (not necessarily distinct) in the decomposition of $n$.

Step 1. Let $n=p$ be a prime number. Let $\alpha$ be a root of $X^p - a$. By Theorem 9.1 in Chapter 6, the polynomial is irreducible, hence, $[k(\alpha):k]=p$. Since $[k(\mu_p):k]=\varphi(p)=p-1$ is prime to $p$, one obtains immediately $k(\mu_n)\cap k(\alpha) = k$. Then $k(\alpha, \mu_n)=k(\alpha) k(\mu_n)$ has degree $p(p-1)$ over $k$ by Theorem 1.14 in Chapter 6, so $G(K/k)\simeq G(n)$, since their orders are equal.

We have shown everything except for the characterisation of the maximal abelian subextension of $k(\alpha, \mu_p)$ which will now show for an arbitrary odd $n$.

Step 1a. I still do not get Lang's argument so I've concocted my own proof. Let $n$ be an arbitrary odd $n$ not dividing $\operatorname{char} k$ and assume that $G(K/k)\simeq G(n)$. Taking $x=\begin{pmatrix} 1&0 \\ b & d \end{pmatrix}$ and $y = \begin{pmatrix} 1&0 \\ c & e \end{pmatrix}$, a direct computation yields $x\, y\, x^{-1} y^{-1} = \begin{pmatrix} 1 & 0 \\ c(d-1)+b(1-e) & 1 \end{pmatrix}$. Since $n$ is odd, $2$ is prime to $n$, hence $2\in (\mathbb{Z}/n\mathbb{Z})^*$. Setting $d=2$ and $b=0$, we see that $\begin{pmatrix} 1 & 0 \\ c & 1 \end{pmatrix}$, $c\in \mathbb{Z}/n\mathbb{Z}$, is in the commutator. It is not difficult to see that the set of such matrices is closed under matrix inversion and matrix multiplication, thus, forms a group, the commutator subgroup $H=[G(K/k),G(K/k]$ of order $n$.

It is easy to check by direct computation that $H$ is precisely the subgroup of $G(K/k)$ which fixes $\zeta$, hence, $k(\mu_n)$. In other words, $H=G(K/k(\mu_n))$. We recall that the commutator subgroup is the smallest normal subgroup such that its factor group $G/H$ is abelian. Hence, by Theorem 1.10 in Chapter 6, $G(k(\mu_n)/k)= G(K/k)/G(K/k(\mu_n)) = G(K/k)/H$ is the largest factor subgroup of $G/k$ which is still abelian, hence, $k(\mu_n)$ is the maximal abelian subextension of $K$.

Step 2. Let us now assume that $n=pm$, $p$ prime (and that the theorem holds for $p$ and $m$). Now, $a$ is not any prime power in $k$ for any primes dividing $n$ and $m$ divides $n$, hence, $X^m - a$ is irreducible over $k$. Let $\beta = \alpha^p$ be its root. By the Theorem's assumption $[k(\mu_n):k]=\varphi(n)$, so it is clear that $[k(\mu_m):k]=\varphi(m)$. Hence, the Theorem holds for $k(\beta,\mu_m)$. In particular, $k(\beta) \cap k(\mu_m) = k$. Now, $k(\beta)\cap k(\mu_n)$ is an abelian subextension of $k(\beta,\mu_m)$ because it is a subfield of an abelian extension $k(\mu_n)$ and a subfield of $k(\beta)\subset k(\beta,\mu_m)$. Hence, $k(\beta)\cap k(\mu_n) \subset k(\beta) \cap k(\mu_m) = k$. In particular, by Theorem 1.14 in Chapter 6, $[k(\beta, \mu_n) : k] = [k(\beta) : k][k(\mu_n) : k] = \varphi(n) m$.

Let us apply the same trick on $X^p-\beta$ over $k(\beta)$. First of all, we know that $[k(\alpha):k]=n=mp$ and $[k(\beta):k]=m$, hence, $[k(\alpha):k(\beta)]=p$ and $X^p-\beta$ is irreducible over $k(\beta)$. As we already know that $k(\mu_n)\cap k(\beta)=k$, so it is clear that $k(\mu_p)\cap k(\beta)=k$, hence, $[k(\beta,\mu_p):k(\beta)]=p$ (again by Theorem 1.14). We may again apply the Theorem on $X^p-\beta$ over $k(\beta)$. Using exactly the same argument, we see that $k(\alpha)\cap k(\beta,\mu_n) = k(\beta)$ since it is an abelian subextension of $k(\alpha,\mu_p)$, hence, $[k(\alpha,\mu_n):k]=[k(\alpha) : k(\beta)] [ k(\beta,\mu_n):k(\beta)][k(\beta):k] = p \varphi(n) m = n \varphi(n)$ and the orders of $G(K/k)$ and $G(n)$ are equal.

We now have by Step 1a that $k(\mu_n)$ is the maximal abelian subextension of $k(\alpha,\mu_n)$ and $k(\alpha)\cap k(\mu_n)=k$. Otherwise $[k(\alpha,\mu_n):k]$ would be strictly less than $n \varphi(n)$.

QED

Note that if $n$ were even, the commutator subgroup's order would merely divide $n$ and $k(\mu_n)$ would possibly be not maximal. This would also break step 2 because we extensively use the fact that the intersection of $k(\alpha)$ with any abelian subextension of $k(\alpha,\mu_n)$ is just the ground field $k$.