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    • CommentRowNumber1.
    • CommentAuthorColin Tan
    • CommentTimeApr 17th 2014
    • CommentRowNumber2.
    • CommentAuthorTodd_Trimble
    • CommentTimeApr 18th 2014

    I expanded this a bit.

    • CommentRowNumber3.
    • CommentAuthoradeelkh
    • CommentTimeApr 18th 2014

    “a bit” ;)

    I added a redirect from Frobenius automorphism to Frobenius morphism.

    • CommentRowNumber4.
    • CommentAuthorColin Tan
    • CommentTimeApr 20th 2014

    I think the correct statement should be: For p a prime, If a finite field has p kp^k elements, then its group of units has p kp k1p^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,,p11,2,\ldots, p-1 has an inverse modulo p.

    • CommentRowNumber5.
    • CommentAuthorTodd_Trimble
    • CommentTimeApr 20th 2014
    • (edited Apr 20th 2014)

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

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

    • CommentRowNumber6.
    • CommentAuthorColin Tan
    • CommentTimeApr 21st 2014
    Ah, every nonzero element of a field is invertible. I made a mistake.