Let $q$ be an odd prime and let $\zeta$ be a primitive $q^{th}$ root of unity. Let $I$ be the ideal in $\mathbb{Z}[\zeta]$ generated by $1-\zeta$. It is known that we can give $I$ the structure of an integral lattice by using the trace pairing $<x,y> = tr(x\bar{y})/q$, and that this lattice is isomorphic to the root lattice $A_{q-1}$. In particular, the minimal norm vectors in this lattice have norm 2.
My question is: what are the minimal norms in the sublattices $I_n: = <(1-\zeta)^n>$ for $1\le n \le \frac{q-1}{2}$?
My (low confidence) guess is that the minimal norm in $I_n$ is equal to $2n$ for $1\le n \le \frac{q-1}{2}$. This is certainly correct for $n=1$. One can verify that in the case $n=(q-1)/2$ the element $\sqrt{\pm q} \in I_n$ and has norm $q-1$ as predicted.
Additionally, for $n=1, 2, 3, 4, 5, 6$ the elements $(1-\zeta)$, $(1-\zeta)(1-\zeta^2)$, $(1-\zeta)(1-\zeta^2)(1-\zeta^3)$, $(1-\zeta)(1-\zeta^2)(1-\zeta^3)(1-\zeta^5)$, $(1-\zeta)(1-\zeta^2)(1-\zeta^3)(1-\zeta^5)(1-\zeta^7)$, $(1-\zeta)(1-\zeta^2)(1-\zeta^3)(1-\zeta^5)(1-\zeta^7)(1-\zeta^8)$ respectively have norm $2n$ for sufficiently large $q$. (For example, for the last term to work, $q=47$ suffices.)
Unfortunately, as best I can tell, the nice pattern of elements for $n=1, 2, ..., 6$ does not continue. Any help or insight, even if only partial, would be greatly appreciated!
