Zeroes of the polynomial $f(x)$ over the field $F$ of order 256. 

Let $F$ be a field  with 256 elements and $f \in F[x]$be a polynomial with all roots in $F$. Then


(1) $f \neq x^{15} -1$.
(2) $f \neq x^{63} - 1$
(3) $f \neq x^2 + x + 1$
(4) if $f$ has no multiple root then $f$ is a factor of $x^{256}-x$.
Let $K$ be the spllitinf Field of the polynomial $x^{256}-x$ over $\mathbb Z_2$ and this polynomial is seperable over $\mathbb Z_2$ .Thus $K$ is the set of all the zeroes of the $x^{256}-x$ , thus  $F \cong K$ . Basically we have to check zeroes of the polynomial lies in $F$ or not.But Zeroes of the polynomial $x^{15}-1$ is $\beta =  e^{\frac{2k\pi i}{15}}$, where $k=0,1,\cdots , 14$,. Similarly $\gamma =e^{\frac{2k\pi i}{63}}$, where $k=0,1,\cdots , 62$ is the root of $x^{63}-1$ and $\omega$ is the cube root of unity.
Clearly $\beta$ and $\omega \in K$. Please tell me about (4). Thank you.
 A: *

*The multiplicative group of $F$ is cyclic of order $255=3 \cdot 5 \cdot 17$, a multiple of $15$. Therefore, $x^{15}-1$ has exactly $15$ roots in $F$ and $f$ can be equal to $x^{15}-1$.

*$x^{63}=1$ and $x^{255}=1$ imply $x^3=1$ because $3=\gcd(63,255)$.  Therefore, $x^{63}-1$ has exactly $3$ roots in $F$ and so $f$ cannot be equal to $x^{63}-1$ because $x^{63}-1$ has no multiple roots.

*Since $x^3-1=(x-1)(x^2+x+1)$, the two roots of $x^2+x+1$ are the two elements of order $3$. Therefore, $f$ can be equal to $x^2+x+1$.

*Every element of $F$ is a root of $x^{256}-x$. Therefore, if $f$ has no multiple roots then $f$ is a factor of $x^{256}-x$.
A: Observe that $\;F=\Bbb F_{2^8}\cong\Bbb F_2[x]/\langle f(x)\rangle\;$ , with $\;f(x)\;$ some polynomial of degree $\;8\;$ irreducible over $\;\Bbb F_2\;$ ,  and then $\;F\;$ is the set of all the zeros in some algebraic closure $\;\overline{\Bbb F_2}\;$ of the prime field $\;\Bbb F_2\;$ of the polynomial $\;x^{2^8}-x\in\Bbb F_2[x]\;$.
Also, observe that $\;\Bbb F_{2^m}\;$ is a subfield of $\;F\;$ iff $\;m\,\mid\,8\iff m=1,2,4,8\;$ , so we get  that $\;\Bbb F_{2^m}\;,\;\;m=3,5,7\;$ is not a subfield of $\;F\;$ .
By the above,  why you wrote in the second parraph "We know that..." cannot be true as that polynomial there isn't reducible and thus the ideal generated by it isn't maximal and the quotient ring isn't even a field then!
Try now to tackle your question. Just observe, for example, that $\;\alpha^{63}=1\iff\alpha\in\Bbb F_{2^7}\;$, by group theory, so this element $\;\alpha\;$ cannot belong to our field $\;\Bbb F\;$  by the above.
