As in the title, I'm trying to understand the implications of Fermat's Little Theorem in $\Bbb Z/p\Bbb Z [x] $. Fermat states that if $p $ is prime then for all integer $x$, $x^p-x $ is a multiple of $p$. As far as I can see, this should mean that $x^p-x=0$ in $\Bbb Z/p\Bbb Z [x] $, but I'm not sure. Is that correct?
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No, it does not hold for the variable in the polynomial ring, since the statement is for an integer. – Tobias Kildetoft Jul 18 '17 at 19:19
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Polynomials and the functions they determine by evaluation should not be confused, especially over finite rings. The former family is infinite, the latter finite. – Arkady Jul 18 '17 at 19:20
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@DougM Yeah, I removed my comment and made a new one when I saw you had removed it :) – Tobias Kildetoft Jul 18 '17 at 19:21
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@Tobias I see, thank you – Richard Jul 18 '17 at 19:22
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@Richard Note that by the fundamental theorem you might expect this polynomial to have $p$ roots in $\mathbb{Z}/p\mathbb{Z}$ (since it's of degree $p$), and in fact it does - so it can be written as $\prod_{i=0}^{p-1}(x-i)$ in $\mathbb{Z}/p\mathbb{Z}[x]$. Now, from this, can you see how to prove (half of) Wilson's Theorem? – Steven Stadnicki Jul 18 '17 at 19:27
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1See also this answer for some background on formal polynomials vs. polynomial functions. – Bill Dubuque Jul 18 '17 at 21:43
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@StevenStadnicki Just saw your comment, not really... – Richard Jun 16 '18 at 20:37
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The polynomial $x^p-x$ is not the zero polynomial in $\Bbb Z/p\Bbb Z [x]$.
The polynomial function $x^p-x$ is the zero function $\Bbb Z/p\Bbb Z \to \Bbb Z/p\Bbb Z$. That's what Fermat says.
Polynomials and polynomial functions are different objects but they can be identified when the base field is infinite.
lhf
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