No, Colin. A finite field with $p^k$ elements does *not* refer to the integers modulo $p^k$ (which isn’t a field at all). It refers to a degree $k$ algebraic extension of the field with $p$ elements. Up to (non-unique) isomorphism, there is a unique field with $p^k$ elements; it is a splitting field for $x^{p^k} - x \in \mathbb{F}_p[x]$.

There is no “difficulty” in proving Fermat’s little theorem.

]]>I think the correct statement should be: For p a prime, If a finite field has $p^k$ elements, then its group of units has $p^k - p^{k-1}$ elements.

I guess essentially the difficulty of proving Fermat’s little theorem is the difficulty of proving that each of $1,2,\ldots, p-1$ has an inverse modulo p.

]]>“a bit” ;)

I added a redirect from Frobenius automorphism to Frobenius morphism.

]]>I expanded this a bit.

]]>Stated Fermat’s little theorem.

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