In my exercise, K is a field of characteristic p, and $K \subset L$ is a finite extension. $ x \in L$ and I must prove that $K[x^p] = K[x]$ if and only if x's minimal polynomial is separable.

However I "managed to prove" that $K(x^p) = K(x)$ is always true , but I don't see where is the flaw.

My proof: $f(x^p) = f(x)^p$ for $f \in K[X]$ (characteristic p) so f(x) = 0 if and only if $f(x^p)=0$ and x and $x^p$ have the same minimal polynomials.

Since $K(x^p) \subset K(x)$ and since they have the same dimension (degree of their shared minimal polynomial) they are equal.

What is wrong? Thanks!

  • 3
    $\begingroup$ It's not necessarily true that $f(x^p)=f(x)^p$. Be careful about coefficients. $\endgroup$
    – jgon
    Jan 7, 2018 at 6:53
  • $\begingroup$ @jgon what is such an example of $f$? $\endgroup$
    – Kenny Lau
    Jan 7, 2018 at 7:01
  • 1
    $\begingroup$ Any $f$ with a coefficient not in the characteristic subfield should be an example. $\endgroup$
    – jgon
    Jan 7, 2018 at 7:02

1 Answer 1


To elaborate on my comment, it's not generally true that $f(x)^p=f(x^p)$. Let $f(x)=\sum_i a_ix^i$. Then $$f(x)^p = \sum_i a_i^p x^{ip},$$ but $$f(x^p) = \sum_i a_i x^{ip}.$$ These are not in general equal, since $a_i^p\ne a_i$ unless $a_i$ is in the characteristic subfield.

This is the core error in your proof.


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