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Let $\xi\in GR(4^{m})$ and $\xi$ is of order $2^{m}-1$. If $\pm\xi^{j}\pm\xi^{k}$ is not invertible for $0\leq j, where $m\geq 2$, then why $\pm\xi^{j}\pm\xi^{k}=2\lambda$, where $\lambda\in GR(4^{m})$?

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    Galois rings with $4^{m}$ elements is denoted by $GR(4^{m})$2012-07-27

2 Answers 2

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I guess $R = \operatorname{GR}(4^m)$ is the Galois ring of characteristic $4$ and order $4^m$. This ring is local with the maximal ideal $2R$. So for any $r\in R$, $r$ is invertible if and only if $r\notin 2R$.

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I just want to provide a more detailed proof, but azimut already gave the correct idea.

Since $R=GR(4^m)=GR(p^s,p^{sm})$ with $p=2,s=2,m\in\mathbb{N}$ is a local ring, it has only one maximal ideal and the elements of this ideal are zero divisors, in fact, the maximal ideal of Galois ring is $(p^{s-1})$ i.e. $(2)$ is the maximal ideal of $GR(4^m)$. Then, the elements of $R\setminus(2)$ are invertible in $R$, thus the elements that you propose $\pm\xi^j$ and $\pm\xi^k$ must be in $(2)$ (because there are not invertible) then $\pm\xi^j=2\lambda_1$ and $\pm\xi^k=2\lambda_2$ with $\lambda_1,\lambda_2\in GR(4^m)$, finally $\pm \xi^j\pm\xi^k=2(\lambda_1\pm\lambda_2)=2\lambda$ with $\lambda\in GR(4^m)$.