Let $F_n$ be the field with $n=2^k$ elements. Let $K_{16}=K_2(\alpha)$.
- Which multiplicative orders are possible for $\alpha$?
- How many such $\alpha$ do exist?
Here are my thoughts:
- 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.
- 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
Best Answer
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 $5$.