Finite coefficient bounds for the scaled cyclotomic grid

The two inequalities $\mathrm{\mid }u\mathrm{\mid },\mathrm{\mid }{u}^{\bullet }\mathrm{\mid }\le 1$, together with their complex conjugates, give $\mathrm{\mid }{\sigma }_j(u)\mathrm{\mid }\le 1$ for all four embeddings $j=1,3,5,7$; these are the identity, complex conjugation, $\bullet$, and their composite by [F2]. If $k=-m<0$, the definition of $k$ gives $x=(\sqrt{2}{)}^{k}u=(\sqrt{2}{)}^{-m}u\in O$, so $u=(\sqrt{2}{)}^{m}x$. Thus $\mathrm{\mid }{\sigma }_j(x)\mathrm{\mid }\le 2^{-m\mathrm{/}2}$ for every $j$. By [F3], $x\bar{x}=P+\sqrt{2}Q$ with $P,Q\in \mathbb{Z}$. Since $\bullet$ negates $\sqrt{2}$ and commutes with conjugation, the product of all four embeddings is ${\prod }_{j=1,3,5,7}{\sigma }_j(x)=(x\bar{x})(x^{\bullet }\overline{x^{\bullet }})=(P+\sqrt{2}Q)(P-\sqrt{2}Q)=P^2-2Q^2$. It is a nonzero integer because $x\mathrm{\ne }0$ and each embedding is injective. Its modulus is at most $2^{-2m}<1$, a contradiction. This proves $k\ge 0$.

For $k\ge 0$, set $a=(\sqrt{2}{)}^{k}u\in O$. Then $\mathrm{\mid }{\sigma }_j(a)\mathrm{\mid }\le 2^{k\mathrm{/}2}$ for every $j$. Fourier inversion in the power basis gives $a_r=\frac{1}{4}{\sum }_{j\in \{1,3,5,7\}}{\sigma }_j(a){\omega }^{-jr}$: expanding the sum, the coefficient of $a_s$ is $\frac{1}{4}{\sum }_{j\text{ odd}}{\omega }^{j(s-r)}$, which equals $1$ for $s=r$ and $0$ for distinct $r,s\in \{0,1,2,3\}$. Therefore $\mathrm{\mid }{a}_r\mathrm{\mid }\le \frac{1}{4}{\sum }_j\mathrm{\mid }{\sigma }_j(a)\mathrm{\mid }\le 2^{k\mathrm{/}2}$. Since $a_r$ is an integer, it lies in the stated finite interval.