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