# Field must be perfect or characteristic p

Problem statement: Given field $$F$$, if for any field extension $$M/F$$, $$[M:F]$$ is divisible by a fixed prime $$p$$, show that $$F$$ is either perfect or have characteristic $$p$$.

Previously in this question Extension degree must be power of prime, I see that $$[K:F]$$ is a power of $$p$$ through Galois closure. I also know that irreducible but inseparable polynomials must have certain form. But is it possible to continue from here without the notion of separable closure?

• Any field of characteristic 0 is perfect. Commented Sep 3, 2021 at 2:46
• @MartinSkilleter Huh? That does not seem to be a hint? Commented Sep 3, 2021 at 2:48
• Consider your field $F$. It either has characteristic 0 or characteristic $p$. Use my above comment to conclude. Commented Sep 3, 2021 at 2:56
• @MartinSkilleter No. It is not that simple.You may want to read the question again..Why does it have to be $p$? Commented Sep 3, 2021 at 2:58
• My apologies, you're right. I missed that the two primes were meant to be the same. Commented Sep 3, 2021 at 3:03

Suppose that $$F$$ is not perfect; then we have $$q:=\operatorname{char}F>0$$, and there exists an element $$\alpha\in F$$ that is not a $$q$$-th power. Now the polynomial $$x^q-\alpha$$ is irreducible in $$F$$, so the ring $$E:=F[x]\big/\langle x^q-\alpha\rangle$$ is a field extension of $$F$$ of degree $$q$$. On the other hand, by the problem hypotheses, $$[E:F]$$ must also be divisible by $$p$$, so this forces $$q=p$$, as desired.
• Why does there exists an $\alpha$ that is not a $q$th-power? Commented Sep 3, 2021 at 3:10
• @SmoothKen oh, apologies, that's the definition of a perfect field that I use :) (namely, a field is perfect iff either its characteristic is $0$ or its characteristic is $p>0$ and every element of it is a $p$-th power.) ... what characterization are you familiar with? Commented Sep 3, 2021 at 3:13
• precisely! :) ${}{}$ Commented Sep 3, 2021 at 3:17