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While looking for alternative proofs for the theorem, I came across the following link

The proof

Let $p$ be a prime number. By Fermat's little theorem, all non-zero elements of the field must be the roots of the polynomial $P(x)=x^{p-1}-1$.

$x^{p-1}-1= \prod_{r = 1}^{p-1}(x-r)$

Now, either $p=2$, in which case $a \equiv -a \pmod 2$ for any integer $a$, or $p-1$ is even. In either case, $(-1)^{p-1} \equiv 1 \pmod{p}$, so that

$x^{p-1}-1=\prod_{ r=1}^{p-1}(x-r)=\prod(r-x)$

If we set $x=0$ then we get the theorem.

My question is, if we have assumed that $x$ is non-zero in the beginning, how can we substitute it back in the end to get the theorem?

Can someone verify if this proof is correct or not and, if it is, explain why doing this is allowed?

LoneStar
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    You've proved that the polynomial $P(x)$ has the $p-1$ roots modulo $p,$ so that $P(x)=(x-1)(x-2)(x-3)\cdots(x-(p-1))$ modulo $p.$ This equality is an equality of polynomials, not just functions on $1,2,\dots,p-1$, so it has to be true for $x=0$ too. – Thomas Andrews Mar 06 '19 at 18:06
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    Do you understand the difference between polynomial functions and formal polynomials, e.g. over $,\Bbb F_p,,$ by $,\mu$Fermat, $, x^p -x ,$ equals the constant function $,0,,$ but it is not equal to $,0,$ as a formal polynomial since, by definition, formal polyomials are equal iff their corresponding coefficients are equal. – Bill Dubuque Mar 06 '19 at 18:17
  • @BillDubuque I did not know the difference. Thanks. So basically, $P(x) = (x−1)(x−2)(x−3)⋯(x−(p−1)) \forall x \in \mathbb{F}_p$ since degree of $P$ = number of roots in $ \mathbb{F}_p$. Meaning that at least in $\mathbb{F}_p$, they are equal.

    Is my understanding correct?

     – LoneStar Mar 06 '19 at 18:28
  • Have you yet studied fields? – Bill Dubuque Mar 06 '19 at 19:07
  • @BillDubuque No. I've studied group theory but not fields. Was my comment incorrect? – LoneStar Mar 06 '19 at 19:21
  • Be more precise and complete about which set of values of x is the set applicable to a given sentence, and incorporate that information right into the sentence.......You may have seen the following way to prove Wilson's Theorem: Multiplication modulo the prime $p,$ on the set ${1,...,p-1}$ is a commutative group. What do you get when you take the product of all the members of a finite commutative group? – DanielWainfleet Mar 07 '19 at 11:48

1 Answers1

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It is a special case of the following

Theorem $ $ Suppose $f(x)$ and $g(x)$ are polynomials of degree $n$ whose coefficients lie in a field $F$ (e.g. $\Bbb Q,\, \Bbb R$ or $\Bbb Z_p =$ integers $\bmod p$). If $f$ and $g$ have equal degree $n$ and equal lead coefs and they have equal values at $n$ distinct points $\in F$ then they have equal corresponding coefs $\,f_i = g_i\,$ for all $\,i\,$ (i.e. $f = g$ as formal polynomials), therefore $f$ and $g$ have equal values at all points $\in F$.

Proof $\ $ By hypothesis $f$ and $g$ have equal lead terms which cancel in $\,h := f-g\,$ so $\,\color{#c00}{\deg h < n}.\, $ Thus $\,h\,$ is zero at the $\color{#c00}{n > \deg h}\,$ distinct points where $f$ and $g$ have equal values. By a well-known theorem, if a polynomial over a field has $\rm\color{#c00}{more\ roots}$ than its degree then it is the zero polynomial, i.e. all its coefs are zero, so $\, h_i = f_i - g_i = 0$.

Bill Dubuque
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