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Let $F$ be of characteristic $p \in \mathbb{N}$ and $f(x) \in F[x]$ be an irreducible polynomial s.t. $f(x) = \sum_i a_i x^{n_i} $ where for each $i$, $a_i = 0$ or $p\,\,|\,\, n_i$. That is, we can write $n_i = p^{m_i} q_i$ for $(p, q_i)= 1$. If we further set $m = min\{ m_i\}$, then we can write

$$ f(x) = \sum_i a_i x^{n_i} = \sum_i a_i (x^{p^m})^{s_i} \text{ where } s_i p^m = n_i $$

Question: If we let $y = x^{p^m}$, why is $f(y) = f(x^{p^m}) = \sum_i a_i (x^{p^m})^{s_i} $ separable?

Edited Attempt:

  1. $f(y)$ is separable iff $gcd(f(y), f'(y)) = 1$.

  2. $f'(y) = f'(x^{p^m}) = \sum_i s_i a_i (x^{p^m})^{s_i-1} \ne 0$ since none of the $s_i$ have $p$ as a factor so there is no way for all of the terms of this sum to be zero.

  3. But then $deg(f'(y)) < deg(f(y))$ implies that $f(y) \ne f'(y)$ so that if these two polynomials shared a common factor, it would be have to be a polynomial of degree less than $deg(f(y))$.

  4. But $f(y)$ is irreducible, so the only such factor possible that $f(y)$ and $f'(y)$ share is a field coefficient. Namely, $1$ is a common factor of both of them. Of course, there may be other field coefficients that divide both of $f(y)$ and $f'(y)$, but $1$ being among them means $1$ is a gcd of these two polynomials by definition (meaning there may be other gcd's as well, but $1$ is at least one of them).

  5. Then $gcd(f(y),f'(y)) = 1$ implies that $f(y)$ is separable, as desired.

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    $\begingroup$ In (2), you must take derivation with respect to $y$, as you are interesting in the separability of $f(y)$ as a polynomial in $y$. $\endgroup$ – Cantlog Oct 11 '13 at 20:38
  • $\begingroup$ I edited the proof above using your comment. I think the proof is done now. $\endgroup$ – user1770201 Oct 11 '13 at 21:54
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  1. $f(y)$ is separable iff $gcd(f(y), f'(y)) = 1$.

  2. $f'(y) = f'(x^{p^m}) = \sum_i s_i a_i (x^{p^m})^{s_i-1} \ne 0$ since none of the $s_i$ have $p$ as a factor so there is no way for all of the terms of this sum to be zero.

  3. But then $deg(f'(y)) < deg(f(y))$ implies that $f(y) \ne f'(y)$ so that if these two polynomials shared a common factor, it would be have to be a polynomial of degree less than $deg(f(y))$.

  4. But $f(y)$ is irreducible, so the only such factor possible that $f(y)$ and $f'(y)$ share is a field coefficient. Namely, $1$ is a common factor of both of them. Of course, there may be other field coefficients that divide both of $f(y)$ and $f'(y)$, but $1$ being among them means $1$ is a gcd of these two polynomials by definition (meaning there may be other gcd's as well, but $1$ is at least one of them).

  5. Then $gcd(f(y),f'(y)) = 1$ implies that $f(y)$ is separable, as desired.

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