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How to find the degree and the basis for $GF(64)$ over $GF(2)$.

Can anyone show me how to do this? I'm am only familiar with problems such as finding the degree and basis of $Q(\sqrt{2},\sqrt{3})$ over $Q$, since their respective bases are just easy to find.

But when I'm given a Galois field, I'm not sure on how to find its basis.

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First of all, I presume you know that the degree is 6 and that the intermediate fields are GF(4) and GF(8) and that the multiplicate group of a finite field is cyclic. Let $\zeta$ be the cyclic generator, i.e., a primitive 63$^{rd}$ root of unity. Then $\zeta^{9}$ is a primitive 7$^{th}$ root and generates GF(8). 1, $\zeta^{9}$ and its square $\zeta^{18}$ must be linearly independent in GF(8) since otherwise 1 and $\zeta^{9}$ would span GF(8), which is impossible. Since GF(8) has degree 3, these form a basis. Since GF(64) has degree 2 over GF(8), 1 and $\zeta$ form a basis for GF(64) over GF(8). The pairwise products then form a basis of GF(64) over GF(2), namely 1, $\zeta$, $\zeta^{9}$, $\zeta^{10}$, $\zeta^{18}$, $\zeta^{19}$.

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  • $\begingroup$ I like this, and have given it a plus 1. But if $\zeta$ is a cyclic generator, then certainly $\Bbb F_{64}=\Bbb F_2(\zeta)$, and therefore $\{1,\zeta,\zeta^2,\zeta^3,\zeta^4,\zeta^5\}$ is a basis. $\endgroup$ – Lubin Apr 15 '18 at 1:18
  • $\begingroup$ True. Sometimes I think it helps to see the structure a bit better if the basis contains a basis of an intermediate field, if there is one. $\endgroup$ – C Monsour Apr 15 '18 at 1:26
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The general result is this:

For any prime power $q=p^n$, there exists, up to an isomorphism, exactly one finite field $\mathbf F_q$ with $q$ elements, which is the splitting field of the polynomial $\;x^{p^n}-x\;$ and is an extension of degree $n$ of $\mathbf F_p$.

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