Dependencies among roots of irreducible polynomial over GR(8,m)

Question

Let $GR\left( 8,m\right) =\left\{ a_{0}+a_{1}\zeta +\cdot \cdot \cdot +a_{m-1}\zeta ^{m-1}:a_{0},a_{1},\ldots,a_{m-1}\in \mathbb{Z}_{8}\right\} $. Let $i,j,k,l=0,1,\ldots,2^{m}-2$, $i,j,k,l$ are distinct and $% \zeta ^{i},\zeta ^{j},\zeta ^{k},\zeta ^{l}$ $\in GR\left( 8,m\right) $ satisfy the following equation \begin{equation} \zeta^i+\zeta^j=\zeta^k+\zeta^l. \end{equation}

Could you help me to prove or disprove that $\zeta^i \zeta^j =\zeta^k \zeta^l$?

Answer 1

[Edit: Adding some definitions] The ring $GR(8,m)$ has several equivalent descriptions. It is a matter of taste, which one you would call the definition.

1. Pick a primitive polynomial $p(x)$ of degree $m$ in $\mathbb{F}_2[x]$, i.e. the minimal polynomial of a generator of the multiplicative group of the finite field $GF(2^m)$. Hensel lift this to a degree $m$ monic polynomial $f(x)\in \mathbb{Z}_8[x]$ such that $f(x)\equiv p(x)\pmod 2$ and $f(x)\mid x^{2^m-1}-1$ in the ring $\mathbb{Z}_8[x]$. Then $$ GR(8,m)=\mathbb{Z}_8[x]/\langle f(x)\rangle $$ is a commutative ring of characteristic $8$ and cardinality $8^m$, $\zeta$ is the coset of $x$.
2. Let $g$ be a generator of the multiplicative group of $F=GF(2^m)$. The ring of Witt vectors of length 3 is what we need $$ GR(8,m)=W_3(F). $$ See e.g. Jacobson, Basic Algebra II for the details. Here $\zeta=(g,0,0)$ as a Witt vector.
3. Let $\zeta$ be a primitive root of unity of order $2^m-1$. Then the field $\mathbb{Q}_2[\zeta]$ is an unramified extension of degree $m$ of the field of $2$-adic numbers. Let $\mathcal{O}$ be the ring of integers of the extension field. Then $$ GR(8,m)=\mathcal{O}/8\mathcal{O}, $$ and we denote the coset of $\zeta$ with $\zeta$ as well.

[/Edit]

So whichever way we look at it, $\zeta$ is a root of unity of order $2^m-1$ = a generator of the non-zero elements of the Teichmüller set. Consequently the element $g=\zeta+2GR(8,m)$ is a generator of the multiplicative group of the finite field $F=GR(8,m)/2GR(8,m)\cong \mathbb{F}_{2^m}.$

I claim that there are no solutions $i,j,k,l$ with all the exponents distinct in that range. Assume contrariwise that both equations $\zeta^i+\zeta^j=\zeta^k+\zeta^\ell$ and $\zeta^i\zeta^j=\zeta^k\zeta^\ell$ hold. Then by projecting to $F$ we get that the equations $$ g^i+g^j=a=g^k+g^\ell $$ as well as $$ g^ig^j=b=g^kg^\ell $$ also hold for some elements $a,b\in F$.

Consider the polynomial $$ p(x)=x^2+ax+b\in F[x]. $$ It has a factorization $$ (x-g^i)(x-g^j)=x^2-(g^i+g^j)x+g^ig^j=x^2+ax+b=p(x) $$ as well as the factorization $$ (x-g^k)(x-g^\ell)=x^2-(g^k+g^\ell)x+g^kg^\ell=x^2+ax+b=p(x). $$ Because $F$ is a field, the polynomial $p(x)$ can have at most two zeros in $F$. This means that the sets $\{g^i,g^j\}$ and $\{g^k,g^\ell\}$ are actually equal. But the restriction of the projection $GR(8,m)\rightarrow F$ to the Teichmüller set $\{0,1,\zeta,\zeta^2,\ldots,\zeta^{2^m-2}\}$ is a bijection, so this implies an equality of sets $\{\zeta^i,\zeta^j\}=\{\zeta^k,\zeta^\ell\}$ and, a fortiori, of the sets $\{i,j\}=\{k,\ell\} (\subseteq\{0,1,\ldots,2^m-2\})$.

This contradicts the assumption that the exponents are distinct elements within the prescribed range.