Help with Problem 10 in Chapter 7 of Lang's Algebra I have no idea how to proceed with the following problem from Lang's Algebra:
Let $n$ be a positive integer and let $\zeta,\zeta'$ be primitive $n$-th roots of unity.
a) Show that $\frac{1-\zeta}{1-\zeta'}$ is an algebraic integer.
b) If $n\geq 6$ is divisible by at least two primes, show that $1-\zeta$ is a unit in the ring $\mathbb{Z}[\zeta]$.
This has been asked twice before, here and here.
However, I don't fully understand the approach given in either. In the first, I don't understand why $1-\zeta^{p^k}$ is a unit mod $p$ if $\zeta^{p^k}$ is an $m$-th root of unity. In the second, I don't understand how the substitution advised in the comments helps to solve the problem. This is probably supposed to be obvious but please help, I don't get it.
Thanks.
 A: For a): Fix some primitive $n$-th root of unity $\zeta$. Then all other primitive $n$-th roots of unity are of the form $\zeta^k$ with $(k,n)=1$. If this is the case, then there is $k'$ such that $kk'\equiv1\mod n$. Now write $\zeta'=\zeta^{k}$ and $\zeta=(\zeta^k)^{k'}$ and note the following:
$$\frac{1-\zeta}{1-\zeta'}=\frac{1-(\zeta^k)^{k'}}{1-\zeta^k}=\sum_{i=0}^{k'-1}(\zeta^k)^i\in\mathbb Z[\zeta]$$
This proves a) as $\zeta$ is an algebraic integers and hence all its powers and sums of these.
For b): This is best done using a norm argument (as mentioned in one of the linked posts). Let $G=\operatorname{Gal}(\mathbb Q(\zeta)/\mathbb Q)$. Define the norm
$$\mathbf{N}_{\mathbb Q}^{\mathbb Q(\zeta)}\colon\mathbb Q(\zeta)\to\mathbb Q,\,\alpha\mapsto\prod_{\sigma\in G}\sigma(\alpha)\, .$$
Restricted to $\mathbb Z[\zeta]$ the image will be contained in $\mathbb Z$ and $\alpha\in\mathbb Z[\zeta]$ is a unit iff $\mathbf{N}_{\mathbb Q}^{\mathbb Q(\zeta)}(\alpha)=\{\pm1\}$. So computing the norm of $1-\zeta$ in the given cases will suffices for showing that $1-\zeta$ is a unit. Note that $\sigma(\zeta)$ just ranges over the different primitive roots of unity and so
$$\mathbf{N}_{\mathbb Q}^{\mathbb Q(\zeta)}(1-\zeta)=\prod_{\sigma\in G}\sigma(\alpha)=\prod_{(k,n)=1}(1-\zeta^k)$$
Claim. $\displaystyle\prod_{(k,n)=1}(1-\zeta^k)=1$
Proof. Let $\Phi(X)=\frac{X^n-1}{X-1}=X^{n-1}+\cdots+1$ and denote by $\Phi_d(X)$ the $d$-th cyclotomic polynomial. Then
$$\Phi(X)=\prod_{\substack{d\mid n\\d>1}}\Phi_d(X)$$
Note that our polynomial is $\Phi_n(X)$. Now, $\Phi(1)=n$ and for prime $p$ also $\Phi_p(1)=p$. Moreover, $\Phi_{p^r}(1)=p$ for all $r\ge1$. To see this, consider
$$\frac{X^{p^r}-1}{X^{p^{r-1}}-1}=X^{(p-1)p^{r-1}}+\cdots+1$$
and note that this polynomial divides $X^{p^r}-1$ and is of degree $(p-1)p^{r-1}$. But $\varphi(p^r)=(p-1)p^{r-1}$ is the exact number of primitive $p^r$-th roots of unity. Hence $\Phi_{p^r}(X)=X^{(p-1)p^{r-1}}+\cdots+1$ and $\Phi_{p^r}(1)=p$.
We return to $\Phi(1)=n$ written as
$$n=\Phi(1)=\prod_{\substack{d\mid n\\d>1}}\Phi_d(1)\ .$$
Now, $n$ factors $p_1^{r_1}\cdots p_s^{r_s}$ for some distinct primes $p_1,\dots,p_s$. But these factors already occur only considering the prime power factors. Hence the remaining factors, that is $\Phi_d(1)$ for $d>1$ and $d\ne p_i^{r_i}$ have to be $\pm1$. Inductively we see that $\Phi_d(1)=1$ for all $d$ as described, and so in particular $\Phi_n(1)=1$. $\square$
(This argument can be found in Lang's Algebraic Number Theory who credits Bass for it.)
The assumption that $n$ is divisible by at least two distinct primes is used subtly: it ensures that $n$ is not a prime power (or prime) in which case the above argument would collapse as would have no factors like $d$ in the end. As shown the product evaluates to $p$ in case of primes powers.
