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It is well known that if $F_d(x)$ is the product of monic irreducible polynomials in $\mathbb{Z}/(p)[x]$ of degree $d$, then $$ x^{p^n}-x=\prod_{d\mid n}F_d(x). $$

It is also known that $x^{p^n}-x$ has no repeated roots, as can be seen by formal differentiation. But, is it possible to see that $\prod_{d\mid n}F_d(x)$ has no repeated roots directly without using the above theorem?

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up vote 3 down vote accepted

Yes. It is sufficient to show that no irreducible polynomial over $\mathbb F_p$ has a repeated root and that no two irreducible polynomials have a common root.

Let $f \in \mathbb F_p[X]$ be an irreducible polynomial. If $x \in \overline{\mathbb F_p}$ is a repeated root of $f$ then it is also a root of $f'$. Since $f$ is irreducible, it is the minimal polynomial of $x$, hence divides $f'$. But $f'$ has lower degree than $f$, so $f' = 0$. This means that $f$ involves only $p$-powers of $X$, i.e. $f(X) = g(X^p)$ for an $g \in \mathbb F_p[X]$. But $g(X^p) = g(X)^p$ in $\mathbb F_p[X]$, contradicting the irreducibility of $f$.

Now let $f,g \in \mathbb F_p[X]$ two different monic irreducible polynomials with a common root $x$. Then both $f$ and $g$ are the minimal polynomial of $x$, and since they are both monic they are in fact equal.

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