Norm of an element in cyclotomic extension (Exercises VI.19 Lang's Algebra) 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...
 A: Extended hints for part (1):


*

*The zeros of $f_r(x)=x^{p^r}-1$ are the numbers $\zeta^i, i=0,1,\ldots,p^r-1$.

*Therefore the zeros of $f_r(x+1)=(x+1)^{p^r}-1$ are the numbers $\zeta^i-1, i=0,1,\ldots,p^r-1$. Notice that the constant term of this polynomial is equal to zero as $\zeta^0-1$ is one of the zeros.

*Therefore the zeros of $g_r(x)=f_r(x+1)/x$ are the numbers $\zeta^i-1, i=1,2,\ldots,p^r-1$.

*Therefore the constant term $g_r(0)$ is, up to sign, equal to the product
$$g_r(0)=\pm\prod_{i=1}^{p^r-1}(1-\zeta^i).$$ You figure out the sign ;-)

*Binomial theorem gives you the value of $g_r(0)$.

*Doing this for both $r$ and $r-1$ should give you the answer.



For the second part the same idea should take you the distance. I haven't really thought about it more than that, so I'm only giving you a few suggestions. Let's denote
$$
g_n(x)=\frac{(x+1)^n-1}x=x^{n-1}+\cdots+n.
$$
Above I used $g_r$ instead of $g_{p^r}$. I apologize for changing my notation from part 1. The same calculation shows that
$$
g_n(0)=\pm\prod_{k=1}^{n-1}(1-\zeta^k).
$$
This time we have
$$
N(1-\zeta)=\prod_{k=1, \gcd(k,n)=1}^{n-1}(1-\zeta^k).
$$
So for example for if $n=pq$ is a product of two distinct primes, we get
$$
\begin{aligned}
N(1-\zeta)&=\frac{\prod_{k=1}^{n-1}(1-\zeta^k)}
{\prod_{k=1,\gcd(k,pq)=p}^{n-1}(1-\zeta^k)\prod_{k=1,\gcd(k,pq)=q}^{n-1}(1-\zeta^k)}\\
&=\pm\frac{g_n(0)}{g_q(0)g_p(0)}.
\end{aligned}
$$
Again leaving it to you to figure out the signs.
At this point a bell rings. There is a well known formula for the cyclotomic polynomials $\Phi_n(x)$ as a fraction involving the polynomials $x^d-1$ with $d\mid n$.
The Möbius function of $n/d$ tells you whether the factor $x^d-1$ goes upstairs, downstairs or is totally absent. I hazard a guess that you are expected to use that for the general case. 
For example, if $n=pqr$ is a product of distinct primes $p,q,r$, then
$$
\begin{aligned}
N(1-\zeta)&=\prod_{k=1,\gcd(k,n)=1}^{n-1}(1-\zeta^k)\\
&=\frac{\left(\prod_{k=1}^{n-1}(1-\zeta^k)\right)\left(\prod_{k=1,pq\mid k}^{n-1}(1-\zeta^k)\right)\left(\prod_{k=1,pr\mid k}^{n-1}(1-\zeta^k)\right)\left(\prod_{k=1,rq\mid k}^{n-1}(1-\zeta^k)\right)}
{\left(\prod_{k=1,p\mid k}(1-\zeta^k)\right)\left(\prod_{k=1,q\mid k}(1-\zeta^k)\right)\left(\prod_{k=1,r\mid k}(1-\zeta^k)\right)}\\
&=\frac{g_{pqr}(0)g_r(0)g_q(0)g_p(0)}{g_{rq}(0)g_{pr}(0)g_{pq}(0)}
=\frac{pqr\cdot r\cdot q\cdot p}{rq\cdot pr\cdot pq}=1.
\end{aligned}
$$
The usual inclusion/exclusion business: We take all the powers. Next we cancel those with exponent of $\zeta$ divisible by $p$, $q$ or $r$. Then we notice that the exponents divisible by two prime factors where cancelled twice, so we need to add them there again ...
A: it seems that, in the case of a primitive $p$-th  root, that the statement $N(1-\zeta)=p$ is equivalent to the assertion that:
$$
\prod_{k=1}^{p-1} \sin(\frac{\pi k}{p}) = \frac{p}{2^{p-1}}
$$
A: 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

A: I want to compute $$\frac{\prod_{i=1}^{p^2-1} (1-\zeta^i)}{\prod _{i=1}^{p-1}(1-\zeta^{ip})}$$
Consider $f(x)=x^{p^2}-1$. 
Roots of $f(x)$ are $\zeta^{i}:0\leq i\leq p^2-1$  ($\zeta$ is primitive $p$th root of unity) 
Roots of $f(x+1)=(x+1)^{p^2}$ are $\zeta^{i}-1:0\leq i\leq p^2-1$
Roots of $g(x)=\frac{f(x+1)}{x}$ are $\zeta^{i}-1:1\leq i\leq p^2-1$
Product of roots of $g(x)$ is $(-1)^{p^2-1}\prod_{i=1}^{p^2-1} (\zeta^i-1)=\prod_{i=1}^{p^2-1} (1-\zeta^i)$
By binomial expansion we see that constant term of $g(x)$ is $\binom{p^2}{p^2-1}=p^2$
Thus, $\prod_{i=1}^{p^2-1} (1-\zeta^i)=p^2$

Consider $h(x)=x^p-1$
Roots of $h(x)$ are $\eta^{i}:0\leq i\leq p-1$ ($\eta$ is primitive $p$th root of unity) 
Roots of $h(x+1)=(x+1)^p$ are $\eta^{i}-1:0\leq i\leq p-1$
Roots of $p(x)=\frac{h(x+1)}{x}$ are $\zeta^{i}-1:1\leq i\leq p-1$
Product of roots of $p(x)$ is $(-1)^{p-1}\prod_{i=1}^{p^2-1} (\eta^i-1)=\prod_{i=1}^{p-1} (1-\eta^i)=\prod_{i=1}^{p-1} (1-\zeta^{ip})$
By binomial expansion we see that constant term of $g(x)$ is $\binom{p}{p-1}=p$
Thus, $\prod_{i=1}^{p^2-1} (1-\zeta^{ip})=p$.

So,$$\frac{\prod_{i=1}^{p^2-1} (1-\zeta^i)}{\prod _{i=1}^{p-1}(1-\zeta^{ip})}=p$$

More generally, 
$$\frac{\prod_{i=1}^{p^r-1} (1-\zeta^i)}{\prod _{i=1}^{p^{r-1}-1}(1-\zeta^{ip})}=p$$

For $n=pq$ we have 
$$N_{K/F}(1-\zeta)=\frac{\prod_{i=1}^{pq-1}(1-\zeta^i)}
{\prod_{i=1}^{q-1}(1-\zeta^{ip})\prod_{i=1}^{p-1}(1-\zeta^{iq})}$$
For similar reasons as above we have $\prod_{i=1}^{pq-1}(1-\zeta^i)=pq$ ; $\prod_{i=1}^{q-1}(1-\zeta^{ip}=q$ ; $\prod_{i=1}^{p-1}(1-\zeta^{iq}=p$
Thus, 
$$N_{K/F}(1-\zeta)=\frac{\prod_{i=1}^{pq-1}(1-\zeta^i)}
{\prod_{i=1}^{q-1}(1-\zeta^{ip})\prod_{i=1}^{p-1}(1-\zeta^{iq})}=\frac{pq}{pq}=1$$
For $n=p^2q$ we have 
$$N_{K/F}(1-\zeta)=\frac{\prod_{i=1}^{p^2q-1}(1-\zeta^i)\prod_{i=1}^{p-1}(1-\zeta^{iq})}
{\prod_{i=1}^{pq-1}(1-\zeta^{ip})\prod_{i=1}^{p^2-1}(1-\zeta^{iq})}=\frac{p^2q\cdot p}{pq\cdot p^2}=1$$
Similar ideas for any $n$ having more than one prime in its prime factorization...
