P1:Let $f$ be a polynomial over finite field $F$ and $n$ be the degree of $f$. Suppose $f$ has $n$ distince roots $\alpha_i$, over its splitting field $E$, $i=1,2,\ldots,n$ and the orders of $\alpha_i$ are the same. Can we say that $f$ is irreducible. If it is true, show it.
P2:Moreover, if $f$ is square-free. Then the number of distince irreducible factors equals the number of distince orders of its roots over $E$. If it is true, show it.
Example:$F=\mathbb{F}_3$, $f=x^5 + 2x^4 + x^3 + 2x^2 + 2=gh$, where $h=x^3 + 2x + 1$, $g=x^2 + 2x + 2$, are two irreducible polynomial over $F$. Then $E=\mathbb{F}_{3^6}$.
All the roots of $f$ over $E$ are $g^{28}$, $g^{84}$, $g^{91}$, $g^{252}$, $g^{273}$, where $g$ is a generator of $E$. The orders of these roots are 26,26,8,26,8.
So the number of distince irreducible factors is 2 and the number of distince orders of its roots is 2, too.
Computation:Let $m$ be the number of distince irreducible factors. For $m=1,2,3$, $n=2,\ldots,10$, $q=2,\ldots,20$, Proposition P2 holds.
Thanks for any replies.
This is usually not true (and I think your "computation" is incorrect, though I am not sure exactly what computation you did). For a very simple counterexample, consider $F=\mathbb{F}_5$ and $f(x)=(x-2)(x-3)$. More generally, if this were true, then it would mean that if $E$ is any finite extension of $F$, the set of all elements of $E$ of a given order forms a single orbit under the Frobenius map (since they all would have the same minimal polynomial). You can see this is not true in many cases simply by counting. For instance, taking $F=\mathbb{F}_2$ and $E=\mathbb{F}_8$, $E$ has $6$ elements of order $7$, which must form two separate Frobenius orbits since the Frobenius map has order $3$. Explicitly, the product $\prod_{\alpha\in F\setminus E}(x-\alpha)$ will have coefficients in $F$, all its roots have order $7$, but it is not irreducible over $F$ (it has two irreducible factors, obtained by taking the product $(x-\alpha)(x-\alpha^2)(x-\alpha^4)$ for any particular $\alpha\in E\setminus F$).