Suppose that $F$ is a field of order 1024 and $F^*=\langle a\rangle$. List the elements of each subfield of $F$.

Question

This is how far my answer is: $1024 = 2^{10}$ There are four divisors of $10$. Thus, there are four subfields of $F$, each with order $2^1, 2^2, 2^5, 2^{10}$.

Subfield of order $2: {0,1}$

Subfield of order $4: {?,?,?,?}$ Subfield of order $32: {?}$ Subfield of order $1024: {F}$

How do I find the elements of the subfields of order $4$ and of order $32$? Thanks!

Answer 1

If $a^m$ is a generator of a subfield of order $2^k$ where $k$ divides $10$, we have $a^{m(2^k-1)} = 1$.

Since $a^{n} = 1 \iff (2^{10} - 1) | n$, we have $ (2^{10} - 1) | m(2^k-1)$.

We can just take $m = \frac{2^{10}-1}{2^k-1}$ because $\;k|10 \implies (2^k - 1) | (2^{10}-1) \implies \frac{2^{10}-1}{2^k-1} \in \mathbb{Z}$.

The subfield of order $4$ is $\langle a^{1023/3} \rangle = \langle a^{341} \rangle$ and the subfield of order $32$ is $\langle a^{1023/31} \rangle = \langle a^{33} \rangle$.