Does the following statement hold true for any finite field?
$$a^p\equiv a \qquad(\mathbb{Z_p})$$
I have tought at it this way: all numbers in $\mathbb{Z_p}$ are $\in \{0,\mathbb{Z_p}\}$ and $p*a< a^p< p^p=p\iff a< a^p< p$
I still miss something
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2What does $\leq$ mean in a finite field?? – Ben Grossmann Jul 27 '15 at 13:26
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And do you specifically mean fields of prime order? – Ben Grossmann Jul 27 '15 at 13:27
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Also, what is $z$ supposed to be here? – Ben Grossmann Jul 27 '15 at 13:30
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@Omnomnomnom I mean finite field over the primes. Edited $z$ was meant to be $p$ – gbox Jul 27 '15 at 13:34
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Many symptomatic (also common and related) syntactic confusions visible here. You can order (using $<$) integers, but it does not make sense to say that one element of the field $\Bbb{Z}_p$ would be bigger than another. This is because those elements are really residue classes modulo $p$. Often we conveniently abuse notation and denote a residue class with a representative. This is confusing in the beginning, but becomes second nature after a while. So $$\Bbb{Z}_p={\overline{0},\overline{1},\overline{2},\ldots,\overline{p-1}}.$$ Integers can be congruent to each other modulo $p$, but... – Jyrki Lahtonen Jul 28 '15 at 06:23
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(cont'd) elements of $\Bbb{Z}_p$ are equal. So $$2^5=32\equiv2\pmod{5}$$ and $$7\equiv2\pmod5,$$ but $$\overline{2}^5=\overline{2}$$ and $$\overline{7}=\overline{2}$$ in the field $\Bbb{Z}_5$. For all the elements $a\in\Bbb{Z}_p$ we have $a^p=a$. For all the integers $n$ we have $\overline{n}^p=\overline{n}$ in $\Bbb{Z}_p$ and also $n^p\equiv n\pmod p$. Yeah, those overlines become a pain after a while, but I don't think you have reached that state yet :-) – Jyrki Lahtonen Jul 28 '15 at 06:25
2 Answers
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Each element $a$ of a field $F$ of cardinality $q$ is a root of $$X^q -X,$$ so $a^q = a$ in $F$.
Note though that you need the cardinality of the field (not its characteristic), so the $q$ might not be a prime number.
quid
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2Yes, $a$ is a root of $X^q-X$ iff $a^q-a = 0$. (But, $X^q-X$ is not the $0$ polynomial.) – quid Jul 27 '15 at 14:20
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The fact that this is true for any field of prime order is known as Fermat's little theorem. For other finite fields, however, this need not hold.
Ben Grossmann
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