# significance of no multiple roots $x^q-x$

From Wikipedia on Finite Fields:

"The polynomial $X^q-X$ factors into linear factors over a field of order q. More precisely, this polynomial is the product of all monic polynomials of degree one over a field of order q. This implies that, if $q = p^n$ that $X^q − X$ is the product of all monic irreducible polynomials over $GF(p)$, whose degree divides $n$. In fact, if $P$ is an irreducible factor over $GF(p)$ of $X^q − X$, its degree divides $n$, as its splitting field is contained in $GF(p^n)$. Conversely, if $P$ is an irreducible monic polynomial over $GF(p)$ of degree $d$ dividing $n$, it defines a field extension of degree $d$, which is contained in $GF(p^n)$, and all roots of $P$ belong to $GF(p^n)$, and are roots of $X^q − X$; thus $P$ divides $X^q − X$. As $X^q − X$ does not have any multiple factor, it is thus the product of all the irreducible monic polynomials that divide it. "

I'm confused about the significance of the fact that $X^q-X$ does not have multiple roots.

It mean that in the splitting field, $$x^q-x=(x-a_1)\cdot...\cdot (x-a_{q})$$ with $a_i\neq a_j$ whenever $i\neq j$.
• For example, the product of all irreducible polynomials that divide $x^7$ is $x$. – Mariano Suárez-Álvarez Mar 3 '16 at 8:20
Let $X^q-X= \prod_i p_i(x)^{e_i}$ be the unique decomposition into monic irreducible factors over $\mathbb{F}_q$. If $e_i >1$ for any $i$, then $X^q-X$ has multiple roots - the roots of $p_i(x)$ are repeated. Therefore, since $X^q-X$ is does not have multiple roots (i.e is separable) over $\mathbb{F}_q$, the unique decomposition must be precisely the product of all the monic irreducibles dividing $X^q-X$.