I am trying to prove the fact that every finite field is perfect. Hence, every irreducible polynomial is separable (does not have a repeated root).

This is easy to prove when in a field of characteristic $p $, the irreducible polynomial under consideration has at least one element $ax^b $ such that $p\not | b $. We just differentiate, and prove that the two cannot have any factors in common.

We now take polynomials of the form $\Sigma a_n x^{kp} $ for some whole number $k $. For an example, I took the field to be $Z_3$ and the irreducible polynomial to be $x^6+x^3-1$. I reasoned that if it has a repeated root, then in the polynomial ring $\frac {Z_3}{(x^6+x^3-1)}[t] $, the polynomial $t^6+t^3-1$ will be divisible by $t-x $, And the quotient too will be divisible by $t-x $. However, on dividing $t^6+t^3-1$ by $t-x $, I am getting $x^6$ as a remainder. This is not equal to zero, as we had earlier assumed $x $ satisfies the irreducible polynomial $x^6+x^3-1$. Where am I going wrong?

  • $\begingroup$ In the ring $\Bbb{Z}_3[x]$ you have $$x^6+x^3-1=(x^2+x-1)^3,$$ so it cannot be irreducible. $\endgroup$ – Jyrki Lahtonen Aug 11 '15 at 21:50

The punchline is the following:

Polynomials of the form $\sum a_n x^{np}$ are never irreducible, since we have $$\sum a_n x^{np} = \left(\sum \sqrt[p]{a_n} x^{n}\right)^p.$$

For $\sqrt[p]{a_n}$ to make sense, we need the Frobenius to be surjective, which is the case in a finite field.

All in all we are left with the case you describe in your first lines. So the proof is done.


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