Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 2 / 12I / c

Since $\alpha$ generates $K^{\times }$, it cannot lie in a proper subfield, so its minimal polynomial over ${\mathbb{F}}_2$ has degree $d$. Write it asThe identity $P(\alpha )=0$ and characteristic two giveMultiplying by ${\alpha }^n$ and applying the linear map $T$ yieldsThis is a binary linear recurrence of order at most $d$, so $(x_n)$ is the output of a linear-feedback shift register of length at most $d$.

Because ${\alpha }^{2^d-1}=1$,so the period divides $2^d-1$.

Conversely, suppose $r$ is an eventual period. Then for every sufficiently large $n$,Any block of $2^d-1$ consecutive powers of $\alpha$ runs through every element of $K^{\times }$. Hencefor every $y\in K$, including $y=0$. The assumed nondegenerate bilinear form forces ${\alpha }^r-1=0$. Since $\alpha$ generates the cyclic group $K^{\times }$ of order $2^d-1$, this means $2^d-1\mid r$. The least positive period is thereforeas in a trace-generated maximal-period linear-feedback sequence.