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Write out the multiplication table for the field $\mathbb{Z}[\alpha] := \mathbb{Z}_2[x]/(x^3+x+1)$.

Q1. What is the inverse of $\alpha$?

Q2. Determine a generator of the multiplicative cyclic group $\mathbb{Z}[\alpha]^\times$

I've written down the table, but how do I get the inverse of $\alpha$? $\alpha$ can be any number. If $\alpha$ is not an integer, how do I take mod $2$?

And for Q2, I can see that, in my table, all $8$ elements of $\mathbb{Z}_2[x]/(x^3+x+1)$ generate all elements of $\mathbb{Z}_2[x]/(x^3+x+1)$. For example, $x+1$ generates all eight elements. Is this because $x^3 + x + 1$ is irreducible?

jun
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  • Your notation ${\mathbb Z}[\alpha]$ is disturbing! You mean ${\mathbb Z}_2[\alpha]$, correct? (I actually prefer ${\mathbb F}_2$ using to denote the field with $2$ elements, rather than ${\mathbb Z}_2$, which often means something else.) Best wishes - – peter a g Nov 05 '20 at 14:20
  • @peterag it's $\mathbb{Z}[\alpha]$. Yep. it's correct. Should that be $\mathbb{Z}_2[\alpha]$? – jun Nov 05 '20 at 14:21
  • It definitely should be $\Bbb{Z}_2[\alpha]$ (or $\Bbb{F}_2[\alpha]$). Anyway, surely $\alpha$ is shorthand for the coset $x+(x^3+x+1)$. Which means that it is your lucky day, because I once produced the relevant discrete logarthm table for referrals like this :-) Writing all the elements as powers of $\alpha$ makes it easy to produce the multiplication table. It is good for you to work out a number of cases by hand. Only then will you appreciate the benefits of discrete logs. Been there :-) – Jyrki Lahtonen Nov 05 '20 at 14:32
  • Mind you, you won't need the full multiplication table to answer Q1 and Q2 :-) – Jyrki Lahtonen Nov 05 '20 at 14:34

2 Answers2

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The field is $GF(8)=\Bbb Z_2[x]/\langle x^3+x+1\rangle$ and contains a zero of $x^3+x+1$, namely the residue class $\alpha = \bar x = x +\langle x^3+x+1\rangle$. Then $\alpha^3+\alpha+1=0$ in $GF(8)$.

The elements of $GF(8)$ are $0, 1, \alpha,\alpha^2,\alpha^3=\alpha+1,\ldots,\alpha^6, \alpha^7=1$. So the inverse of $\alpha$ is $\alpha^6=(\alpha^3)^2 = (\alpha+1)^2 = \alpha^2+1^2 = \alpha^2+1$.

The multiplicative group $GF(8)^*$ is cyclic and has $\alpha$ as a generator, see above.

Wuestenfux
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For Q1: Here $\alpha$ is not 'any number' - $\alpha$ designates a root of $x^3+x+1$, which has been adjoined to make this field. In effect, $\alpha==x$. So ask: what polynomial over $\mathbb{Z}_2$, when multiplied by x, yields a polynomial in $\mathbb{Z}_2[x]$ with remainder 1 modulo $x^3+x+1$?

And for Q2: Yes it is because $x^3+x+1$ is irreducible over $\mathbb{Z}_2$.

PMar
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