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Let $F_n$ be the field with $n=2^k$ elements. Let $K_{16}=K_2(\alpha)$.

  1. Which multiplicative orders are possible for $\alpha$?
  2. How many such $\alpha$ do exist?

Here are my thoughts:

  1. The multiplicative groups of $K_{16}$ has $15$ elements, so only $3,5,15$ are candidates for the order of $\alpha$. Since the degree of the field extension must be $4$ I can rule out $3$ as a candidate.
  2. Since the degree of the field extension must be $4$ I looked at the irreducible polynomials of degree 4 over $K_2$ of which I found 4, so there are at most 16 such $\alpha$.

I did not manage to get any further. I would be really grateful for input. Only related question I found is Order of element in field extension

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Assuming $\ K_{2^n}\ $ is the finite field $\ F_n\ $ of order $\ 2^n\ $, then $\ K_{16}\ $ is the splitting field of the polynomial $\ x^{16}-x\ $ over $\ K_2\ $. According to Wolfram alpha, the factorisation of $\ x^{16}-x\ $ into irreducible factors over $\ K_2\ $ is \begin{align} x(x+1)(x^2+x+1)&(x^4+x+1)\\ &(x^4+x^3+1)(x^4+x^3+ x^2+x+1)\ , \end{align} so there are only $3$, not $4$, irreducible polynomials of degree $4$ over $\ K_2\ $. Since $\ K_{16} = K_2(\alpha)\ $ if and only if $\ \alpha\ $ is a root of any of those $3$ irreducible polynomials, the total number of such $\ \alpha\ $ is $12$. The polynomials, $\ x^4+x+1\ $ and $\ x^4+x^3+1\ $ are both primitive, so all their roots have multiplicative order $15$, and since $\ (x+1)(x^4+x^3+ x^2+x+1)=x^5+1\ $, all the roots of $\ x^4+x^3+ x^2+x+1\ $ have multiplicative order $5$.

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  • $\begingroup$ First of all thanks very much, I have a question though: How can I be sure that there are really 12 distinct $\alpha$? $\endgroup$
    – GEO
    Commented Oct 21, 2019 at 8:16
  • $\begingroup$ To make my question more precise: How can I be sure the two primitive polynomials do not share roots? What about the non primitive polynomial? Is there some trick to quickly observe that those two polynomials are primitive? $\endgroup$
    – GEO
    Commented Oct 21, 2019 at 8:27
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    $\begingroup$ The most direct (if tedious) way of showing that no two of the polynomials share a root is to compute their gcd over $\ K_{16}\ $. If $\ \alpha $ were a common root, then the gcd would have to be a multiple of $\ x-\alpha\ $, but every pair of the listed factors of $\ x^{16}-x\ $ has gcd $1$. A more civilised proof is to observe that every element $\ \alpha\ $ of $\ K_{16}\ $ satisfies the equation $\ \alpha^{16}=\alpha\ $, so all sixteen distinct elements of $\ K_{16}\ $ are roots of the polynomial $\ x^{16} -x\ $, and since this polynomial has degree $16$, none of the roots can be repeated. $\endgroup$ Commented Oct 21, 2019 at 9:11
  • $\begingroup$ How would you prove the two mentioned polynomials are primitive? $\endgroup$
    – GEO
    Commented Oct 21, 2019 at 9:15
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    $\begingroup$ An irreducible polynomial of degree $\ n\ $ over $\ K_2\ $ is primitive if $\ 2^n-1\ $ is the smallest value of $\ m\ $ such that the polynomial divides one of the form $\ x^m-1\ $. Here, $\ m\ $ can only be $3$, $5$ or $15$. Clearly, none of the degree $4$ polynomials can divide $\ x^3-1\ $, and \begin{align} x^5-1&= x(x^4+x+1)+x^2+x+1\\ &=x( x^4+x^3+1)+ x^3+x\ , \end{align} so neither of the polynomials $\ x^4+x+1\ $ or $\ x^4+x^3+1\ $ divides $\ x^5-1\ $. $\endgroup$ Commented Oct 21, 2019 at 9:37

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