Edit: Let $\mathbb{F}_p$ be the finite field with $p$ elements and let $f(x)\in\mathbb{F}[x]$ be an irreducible polynomial. In addion, suppose that $x_1,\ldots,x_n$ are the roots of $f(x)$ in a splitting field $\mathbb E$ and consider the Frobenius mapping: $$ \phi:\mathbb F_p\rightarrow \mathbb F_p\qquad \phi(x)=x^p $$ How $\phi$ permutes the roots of $f(x)$? Is it possible, for instance, that $(\phi(x_1),\phi(x_2),\phi(x_3),\phi(x_4))=(x_2,x_1,x_4,x_3)$ ? Thanks!


The Frobenius map does not permute the roots of $f(x)$.
For example if $a\in \mathbb F_4$ satisfies $a^2+a+1=0$, the polynomial $f(x)=x+a$ has $a$ as unique root, but then $\phi (a)=a^2=a+1\neq a$ is not a root of $f(x)$.
If the the base field is $\mathbb F_p$, then the Frobenius map indeed permutes the roots of the irreducible polynomial $f(x)\in \mathbb F_p$.
However the Galois group $\langle \phi\rangle $of the splitting field is cyclic and in particular you cannot have $(\phi(x_1),\phi(x_2),\phi(x_3),\phi(x_4))=(x_2,x_1,x_4,x_3)$.

  • $\begingroup$ I see...Thanks! But what if my field is simply $\mathbb{F}_p$? (i edit my question). $\endgroup$ – boaz May 13 '17 at 22:13
  • $\begingroup$ Thanks @Georges Elencwajg for your answer. $\endgroup$ – boaz May 14 '17 at 6:33
  • $\begingroup$ You are welcome, dear boaz. $\endgroup$ – Georges Elencwajg May 21 '17 at 17:06

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