TITLE: Norm of an element in cyclotomic extension (Exercises VI.19 Lang's Algebra)
QUESTION [5 upvotes]: Let $\zeta$ be a primitive $n^{\rm{th}}$ root of unity. Let $K=\mathbb{Q}(\zeta)$.

If $n=p^r (r\geq 1)$ is a prime power, show that $N_{K/F}(1-\zeta)=p$
If $n$ is divisible by at least two distinct primes then show that $N_{K/F}(1-\zeta)=1$


I tried for $r=1$
Suppose $n=p$ then we have $Gal(\mathbb{Q}(\zeta)/\mathbb{Q})\cong (\mathbb{Z}/p\mathbb{Z})^*=\{\sigma_1,\sigma_2,\sigma_3,\cdots,\sigma_{p-1}\}$ where $\sigma_i(\zeta)=\zeta^i$
Now, 
$N_{K/F}(1-\zeta)=\sigma_1(1-\zeta)\sigma_2(1-\zeta)\cdots\sigma_{p-1}(1-\zeta)=(1-\zeta)(1-\zeta^2)\cdots(1-\zeta^{p-1})$
I could prove that this is $p$  by just expanding.. (as of now i do not have a general proof..)  
When $r=2$ i tried to do similarly but ended up nowhere..
For $r=2$ we want to see what are all elements that are coprime to $p$.. (This is what gives me the galois group)
Coprimes of  $p^2$ are $\{1,2,\cdots, p^2\}-\{p,2p,3p,\cdots,p^2\}$.
Then, 
 $N_{K/\mathbb{Q}}(1-\zeta)=(1-\zeta)(1-\zeta^2)\cdots(1-\zeta^{p-1})(1-\zeta^{p+1})(1-\zeta^{p+2})\cdots(1-\zeta^{p^2-1})$
This is equal to $$\frac{\Pi_{i=1}^{p^2-1}(1-\zeta^i)}{\Pi_{i=1}^{p-1}(1-\zeta^{ip})}$$
I do now know how to make this end up being $p$..
Please help me to clear  this... 
For $n=6$ we have $Gal(\mathbb{Q}(\zeta_6)/\mathbb{Q}=\{ \sigma_1,\sigma_5\}$
$N_{K/F}(1-\zeta)=(1-\zeta)(1-\zeta^5)=1-(\zeta+\zeta^5)+1=1$ (Credits to David Holden)
As i have succeeded in case of $n=6$ i tried to see this for $n=pq$ with $p,q$ are distinct primes..
co primes to $pq$ are $\{1,2,3,\cdots,pq-1\}-\{p,2p,3p,\cdots,(q-1)p, q,2q,3q,\cdots,(p-1)q\}$
Then $$N_{K/F}(1-\zeta)=\frac{\Pi_{i=1}^{pq-1}(1-\zeta^i)}
{\Pi_{i=1}^{q-1}(1-\zeta^{ip})\Pi_{i=1}^{p-1}(1-\zeta^{iq})}$$
I do not know what should be done after this...
Please let me know how to fix with this...

REPLY [2 votes]: We consider 
$$\Phi_n(x)=\prod (x^{\frac{n}{d}}-1)^{\mu(d)}$$
If $n=p^2q^2$ then $$\Phi_{p^2q^2}(x)=(x^{p^2q^2}-1)(x^{p^2q}-1)^{-1}(x^{pq^2}-1)^{-1}(x^{pq}-1)=\frac{(x^{p^2q^2}-1)(x^{pq}-1)}{(x^{p^2q}-1)(x^{pq^2}-1)}=\frac{x^{p^2q-1}+x^{p^2q-2}+\cdots +x+1)(x^{pq-1}+x^{pq-2}+\cdots+x+1)}{(x^{p^2q-1}+x^{p^2q-2}+\cdots+x+1)(x^{pq^2-1}+x^{pq^2-2}+\cdots+x+1)}$$
Evaluating at $1$ we have :
$$\Phi_{p^2q^2}(1)=\frac{p^2q\cdot pq}{p^2q\cdot pq^2}=1$$
More generally, if $n=p_1^{i_1}p_2^{i_2}\cdots p_r^{i_r}$ (all $i_j$ are even, $r$ is also even) 
we have 
$$\Phi_n(x)=\frac{(x^n-1)(x^{p_1p_2}-1)(x^{p_2p_3}-1)\cdots ??}{???}$$
I am not able to write general expression of $\Phi_n(x)$ neatly but $\Phi_n(1)$ is $$\frac{p_1^{i_1}p_2^{i_2}\cdots p_r^{i_r}\cdot(p_1p_2)(p_2p_3)\cdots (p_1p_2p_3p_4)\cdots}{:( ~:(}$$

We consider another formula:
If $p$ is a prime number that does not divide $n$ then we have 
$$\Phi_{pn}(x)=\frac{\Phi_n(x^p)}{\Phi_n(x)}$$
If $p$ divides $n$ then we have $\Phi_{pn}(x)=\phi_n(x^p)$
For $n=pq$ we have $$\Phi_{pq}(x)=\frac{\Phi_q(x^p)}{\Phi_q(x)}\Rightarrow \Phi_{pq}(1)=\frac{\Phi_q(1)}{\Phi_q(1)}=\frac{q}{q}=1$$
For $n=pqr$ we have $$\Phi_{pqr}(x)=\frac{\Phi_{pq}(x^r)}{\Phi_{pq}(x)}\Rightarrow \Phi_{pqr}(1)=\frac{\Phi_{pq}(1)}{\Phi_{pq}(1)}=\frac{1}{1}=1$$
More generally, for $n=p_1p_2p_\cdots p_r $ we have $\Phi_n(1)=1$
Even more generally, for $n=p_1^{i_1}p_2^{i_2}\cdots p_r^{i_r}$ we have 
$$\Phi_{p_1^{i_1}p_2^{i_2}\cdots p_r^{i_r}}(x)=\cdots = \Phi_{p_1p_2\cdots p_r} (x^{p_1^{i_1-1}p_2^{i_2-1}\cdots p_r^{i_r-1}})$$
So, $$\Phi_n(1)=\Phi_{p_1^{i_1}p_2^{i_2}\cdots p_r^{i_r}}(1)=\cdots = \Phi_{p_1p_2\cdots p_r} (1)=1$$
Assuming $N_{K/\mathbb{Q}}(1-\zeta)=\Phi_n(1)$ we conclude that , if $n$ has atleast two distinct primes in its factorization then $$N_{K/\mathbb{Q}}(1-\zeta)=\Phi_n(1)=1$$ 
If $n=p$ we have $$\Phi_p(x)=x^{p-1}+x^{p-2}+\cdots +x+1\Rightarrow N_{K/\mathbb{Q}}(1-\zeta)=\Phi_p(1)=p$$
If $n=p^r$ then,
$$\Phi_n(x)=\Phi_p(x^{p^{r-1}})\Rightarrow \Phi_n(1)=\Phi_p(1)\Rightarrow N_{K/\mathbb{Q}}(1-\zeta)=\Phi_n(1)=\Phi_p(1)=p$$
So, we have 

$N_{K/\mathbb{Q}}(1-\zeta)=p$ if $n=p^r$
$N_{K/\mathbb{Q}}(1-\zeta)=1$  if $n$ is divisible by at least two distinct primes