Cyclotomic euclidean number fields I´m here because I want to repeat my question about the norm-euclidean algorithm in a particluar cyclotomic integer ring. 
Let $L=\mathbb{Q}(\zeta_{32})$ and $A=\mathbb{Z}[\zeta_{32}]$. On the page 28 of "Cyclotomic euclidean number fields", Reza Akhtar wrote a proof attributed to Hendrik Lenstra Jr. in which he showed that $A$ is not euclidean. However, I do not fully understand this proof.  In the document you can see the following proof:
"We claim in particular, that there are not elements $q, r$ in $A$ such that
$1+(1+\zeta)^5=q(1+6)^6+r$  with $N_{L/\mathbb{Q}}(r)<N_{L/\mathbb{Q}}((1+\zeta)^6)$
Using Proposition A.5 (1), we compute $N_{L/\mathbb{Q}}((1+\zeta)=2$, so
$$N_{L/\mathbb{Q}}((1+\zeta)^6)=64$$
Lemma 7.1: Every element in $A$ which is prime to $1+\zeta$ has norm equivalent with $1$ mod $32$
Proof: ( I do understand this proof, so I will omit).
So if we can find $q,r$ sush that 
$1+(1+\zeta)^5=q(1+6)^6+r$  with $N_{L/\mathbb{Q}}(r)<64$, Lemma 7.1 and Proposition A.5(6) and (9) tell us that $r$ is either a unit or a product of prime powers, each equivalent to $1$ mod $32$. The conditions forces $N_{L/\mathbb{Q}}(r)=1$. It is known that the unit group of $A$ is generated by $(1-\zeta^{i})/(1-\zeta)$, where $i$ is in ${1,2,3....,8}$
We examine the residues of each of these elements in the multiplicative group $M$ of the ring $A/(1+\zeta)^6$
Since          $(2)=((1+\zeta)^{16}$
as ideals, we observe that 
$A/(1+\zeta)^6$=Z/$2$Z[$\zeta$]/$(1+\zeta)^6$
therby greatly simplifying computation. Finally, it can be shown (by a direct computation)
that the subgroup $M$ generated by the residues of this units has order $16$ and hence does not contain the residue of $1+(1+\zeta)^5$, giving a contradiction."
My first two questions are:
1) Why is $N_{L/\mathbb{Q}}(r)=1$?
2) Why the unit group of $A$ is generated by $(1-\zeta^{i})/(1-\zeta)$, where $i$ is in ${1,2,3....,8}$?
Best regards, José
 A: Let $A = \mathbb{Z}[\zeta_{32}]$ be the ring of integers of the cyclotomic field $\mathbb{Q}(\zeta_{32})$.  Jose is referring to Reza Akhtar's senior thesis, Cyclotomic Euclidean Number Fields, where he gives Lenstra's proof that $A$ is not a Euclidean number field.
Jose asks why Akhtar states that the unit group of $A$ is generated by
$$\frac{1-\zeta^i}{1-\zeta}$$
with $1 \leq i \leq 8$.
The above units are known as "cyclotomic units," and it would be more precise to state that the unit group of $A$ is generated by those cyclotomic units and the root of unity $\zeta$.
Generally, given a cyclotomic field $\mathbb{Q}(\zeta_n)$ where $n$ is a prime power, e.g. 32, the group $C$ of cyclotomic units, is generated by $\zeta$ and units of the form $(1-\zeta^i)/(1-\zeta)$, is always a finite index subgroup of the full unit group $E$.  Moreover, we have the remarkable fact that
$$h^+ = [E:C],$$
where $h^+$ is the class number of the maximal real subfield $\mathbb{Q}(\zeta_n + \zeta_n^{-1})$.
In the case of $n=32$, we have that the class number of $\mathbb{Q}(\zeta_{32} + \zeta_{32}^{-1})$ is 1 (in fact, J.P. Cerri proved that it is norm-Eulicdean!).  Thus, $C=E$.
Added later in response to Jose's inquiry:
Perhaps it's best just to do the actual calculation:
Let $X=1-\zeta$. Then we are working over the quotient ring $R = \mathbb{F}_2[X]/(X^6)$.  In $R$, the invertible elements are the elements with nonzero constant term.
The root of unity $\zeta$ corresponds to invertible element $1+X$ in $R$, the cyclotomic unit $1+\zeta+\zeta^2$ corresponds to the invertible element $1+X+X^2$ in $R$, and cyclotomic unit $1+\zeta+\zeta^2+\zeta^3+\zeta^4$ corresponds to the invertible element $1+X^3+X^4$ in $R$.  The other two cyclotomic units, $1+\zeta+\zeta^2+\zeta^3$ and $1+\zeta+\zeta^2+\zeta^3+\zeta^4+\zeta^5$ correspond to elements that are not invertible in $R$, so we can ignore them.
Now consider the subring $M$ of the multiplicative group $R^\times$ of $R$ that is generated by $1+X$, $1+X+X^2$ and $1+X^3+X^4$.  The element $1+(1+\zeta)^5$ corresponds to $1+X^5 \in R^\times$.
To show: $1+X^5 \notin M$.
By brute force calculation, it turns out that $M$ is order 16 (vs $R^\times$ which is order 32), and the elements of $M$ are:
$$\{1, 1+X, 1+X^2, 1+X^3, 1+X^4, 1+X+X^2, 1+X^2+X^4, 1+X^3+X^4, 1+X+X^2+X^3, 1+X+X^3+X^4, 1+X+X^3+X^5, 1+X+X^4+X^5, 1+X^2+X^3+X^5, 1+X+X^2+X^4+X^5, 1+X^2+X^3+X^4+X^5, 1+X+X^2+X^3+X^4+X^5 \}.$$
By inspection, $1+X^5 \notin M$, and we are done.
