Let us consider the following result:
Note that in the ring of polynomials with coefficients $mod p$, the product $x(x-1)(x-2)...(x-p+1)$ has all possible roots, and equals $x^p-x$ by Fermat's Little Theorem and unique factorization in this ring.
I am trying to prove it but without any suscess. I wante to see a proof or a reference about this.
My second question is about the validity of this result for multivariate polynomials, say three variables.
On
For single variables, the standard approach would be to first prove the following lemma:
If $f(x)$ is a polynomial of degree $n >0$ in $F_p[x]$ and $f(a)=0$ for some $a\in F_p$, then there exists a polynomial $g(x)$ of degree $n-1$ with $$f(x) = (x-a)g(x).$$
You can prove this lemma by proving that polynomial division works in $F_p[x]$, and then showing that if $a$ is a root of $f(x)$, the remainder when dividing $f(x)$ by $x-a$ must be zero.
You can then prove that $x^p-x$ has the factorization you list by inductively pulling out roots.
All of this is covered in any intro text on elementary number theory.
By Fermat's Little Theorem, $x^p - x \in \mathbb{F}_p[x]$ has root $x=a$ for all $a \in \mathbb{F}_p$; therefore, each $x - a \mid x^p - x$. Since $\mathbb{F}_p[x]$ is a UFD (because $\mathbb{F}_p$ is a field) and each $x - a$ is irreducible and distinct, it follows that $x(x-1) \cdots (x-p+1) \mid x^p - x$. The left and right hand sides have equal degree and therefore the quotient must be a constant polynomial; and by looking at the coefficient of $x^p$ on both sides you can conclude that constant must be 1.
As for the case of multivariate polynomials, you might think of the ideal of all polynomials in $\mathbb{F}_p[x,y,z]$ which are zero when evaluated at any point in $\mathbb{F}_p^3$. This would be equal to $\bigcap_{a,b,c \in \mathbb{F}_p} \langle x-a, y-b, z-c \rangle$. I'm not sure off the top of my head what exactly this ideal is - though a reasonable conjecture would be $\langle x^p - x, y^p - y, z^p - z \rangle$.