Number of forms of a finite field

The number of elements in a finite field must be a prime power, and for every prime power there is only one finite field up to isomorphism.

The finite field with 256 elements, GF(28), is important in applications. From one perspective, there is only one such field. But there are a lot of different isomorphic representations of that field, and some are more efficient to work with that others.

Just how many ways are there to represent GF(28)? Someone with the handle joriki gave a clear answer on Stack Exchange:

There are 28−1 different non-zero vectors that you can map the first basis vector to. Then there are 28−2 different vectors that are linearly independent of that vector that you can map the second basis vector to, and so on. In step k, 2k-1 vectors are linear combinations of the basis vectors you already have, so 28−2k−1 aren’t, so the total number of different automorphisms is

\prod_{k=1}^8\left(2^8 - 2^{k-1} \right ) = 5348063769211699200

This argument can be extended to count the number of automorphism of any finite field.

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