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I need to find all generators of the field $\mathbb{Z}_2[x][x^3+x^2+1]^*$

The star is defined as follows: $ F[x]^*m(x) = \{ a(x) \in F[x]m(x) | gcd(a(x), m(x))=1\}$

So this means we only look at the irreducible polynoms as possible generators. The new set contains this polynoms: $1, x, x+1, x^2+x+1$.

So the set has order 4. The number of generators is equal to the number of co-prime numbers less then 4. So there are two generators (as 1, 3 are co-prime to 4). I know that 1 can't be a generator so I only need to check $x, x+1, x^2+x+1$.

Is this correct up to this point? Is there a clever way to check the remaining polynomials instead of just check all the powers?

Johny Dow
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  • Possible duplicate: https://math.stackexchange.com/questions/3048263/how-to-find-minimal-polynomial-in-gf23/3048314#3048314 – Wuestenfux Jan 07 '19 at 13:09
  • I can't understand your definition of $*$: you seem ot be looking at all polynomials which are multiples of $m(x)$ and are coprime to $m(x)$? – ancient mathematician Jan 07 '19 at 14:31
  • I'm looking at all polynomials which are in the Field mod $x^3+x^2+1$ and co-prime to $x^3+x^2+1$. – Johny Dow Jan 07 '19 at 15:56
  • The group has seven elements. In addition to the four you listed there are the cosets of $x^2+x$, $x^2+1$ and $x^2$. What do you know about groups of order seven? Hint: seven is a prime number. – Jyrki Lahtonen Jan 07 '19 at 19:28
  • I don't understand why you would only look at the cosets of irreducible polynomials. Also, here you can see how the coset of $x$ (denoted $\alpha$ there) is a generator of $\Bbb{Z}_2[x]/\langle x^3+x+1\rangle^*$. You can mimick the calculation there. – Jyrki Lahtonen Jan 07 '19 at 19:30
  • @JyrkiLahtonen so basically every element is a generator? – Johny Dow Jan 08 '19 at 09:25
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    Correct. In a cyclic group of prime order any element other than the neutral element generates the whole group. – Jyrki Lahtonen Jan 08 '19 at 09:56

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