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1. • CommentRowNumber1.
   • CommentAuthorColin Tan
   • CommentTimeApr 17th 2014
   • PermaLink
   Author: Colin Tan
   Format: MarkdownItexStated [Fermat's little theorem](http://ncatlab.org/nlab/show/Fermat%27s+little+theorem).

   Stated Fermat’s little theorem.

2. • CommentRowNumber2.
   • CommentAuthorTodd_Trimble
   • CommentTimeApr 18th 2014
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   Author: Todd_Trimble
   Format: MarkdownItexI expanded this a bit.

   I expanded this a bit.

3. • CommentRowNumber3.
   • CommentAuthoradeelkh
   • CommentTimeApr 18th 2014
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   Author: adeelkh
   Format: MarkdownItex"a bit" ;) I added a redirect from Frobenius automorphism to [[Frobenius morphism]].

   “a bit” ;)

   I added a redirect from Frobenius automorphism to Frobenius morphism.

4. • CommentRowNumber4.
   • CommentAuthorColin Tan
   • CommentTimeApr 20th 2014
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   Author: Colin Tan
   Format: MarkdownItexI 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.

   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.

5. • CommentRowNumber5.
   • CommentAuthorTodd_Trimble
   • CommentTimeApr 20th 2014
   • (edited Apr 20th 2014)
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   Author: Todd_Trimble
   Format: MarkdownItexNo, 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.

   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.

6. • CommentRowNumber6.
   • CommentAuthorColin Tan
   • CommentTimeApr 21st 2014
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   Author: Colin Tan
   Format: TextAh, every nonzero element of a field is invertible. I made a mistake.

   Ah, every nonzero element of a field is invertible. I made a mistake.