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

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 $$\zeta^i+\zeta^j=\zeta^k+\zeta^l.$$

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

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$\mathbb Z_8$ is the $8$-adics or the integers $\pmod 8$? I've seen both, but I've never seen $GR(8,m)$, so I was wondering.. –  Patrick Da Silva Jul 21 '12 at 4:41
I don't know what $\zeta$ stands for. An indeterminate? a root of unity of some order? a generator of some ring? –  Gerry Myerson Jul 21 '12 at 4:46
The notation GR($q$,$m$), for a prime power $q$ and a positive integer $m$, means the length $m$ Witt vectors over the finite field of size $q$. For example, when $m = 1$ it is the field of size $q$ and when $q = p$ is a prime it is (isom. to) the integers mod $p^m$. There is an elementary way to describe GR($q$,$m$) in general without mentioning Witt vectors, but I don't remember that more concrete way right now. (The way GR(8,$m$) is described in the question is hardly a complete definition.) –  KCd Jul 21 '12 at 7:12
KCd has it right. The rings $GR(4,m)$ (and to a lesser extent $GR(8,m)$) were a hot topic in coding theory in the 90s. They can be realized either as length 2 (resp. 3) Witt vectors over a finite field of characteristic two, or quotient rings of the ring of integers of an unramified extension of the 2-adics. Google $\mathbf{Z}_4$-linear codes for more information. Yours truly had the pleasure of playing a small part in that flurry of activity. –  Jyrki Lahtonen Jul 27 '12 at 20:04
$\TeX$ tip: Writing "..." is not the same as writing "\ldots". The result looks different. And the former gives results conventionally considered incorrect by those who care about spelling, punctuation, etc., in that it lacks spacing between the dots. (This may vary with web sites and browsers, but when TeX or LaTeX is used in the normal way, as opposed to the way it's used on the web, that's what happens.) –  Michael Hardy Aug 4 '12 at 18:34

[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.

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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.

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Very illuminating. Thanks. –  Lubin Sep 28 '13 at 23:31