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Every irreducible polynomial $f$ over a perfect field $F$ is separable.

Can you check my proof? Assume that $f$ is irreducible but inseparable. So we have $f=\sum_i h_ix^i$ and $f^p=\sum_i h_i^px^{ip}$. Now I use Frobenius mapping $\phi (f)=f^p=(\sum_i h_ix^i)^p=\sum_i h_i^px^{ip}$ so it is reducible. Contradiction. Should I also show for char(F)=0?

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  • $\begingroup$ In characteristic 0 every irreducible polynomial is separable, so you don't have to check anything. $\endgroup$ – Crostul Jul 6 '15 at 10:14
  • $\begingroup$ Oh yes, that's true. Is my thinking correct? $\endgroup$ – MrFrodo Jul 6 '15 at 10:16
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    $\begingroup$ You're proving that $f^p$ is reducible, which is obvious. $\endgroup$ – egreg Jul 6 '15 at 10:43
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In characteristic $0$, an irreducible polynomial has distinct roots (in a suitable extension field). Indeed, $\gcd(f,f')$ is a divisor of $f$, so it is either $1$ or $f$ (up to multiplication by nonzero constants).

Suppose it is $f$; then $f'=0$, which means $f$ has degree $0$: impossible.

Now, let the characteristic be $p$. If $f$ is inseparable and irreducible, then $f(x)=g(x^p)$, for some polynomial $g$; let $g(x)=a_0+a_1x+\dots+a_nx^n$. Since $F$ is perfect, we have $a_i=b_i^p$, for some $b_i\in F$ $(i=0,1,\dots,n)$. Thus $$ f(x)=\sum_{i=0}^n b_i^px^{ip}=\biggl(\,\sum_{i=0}^n b_ix^i\biggr)^{\!p} $$ contradicting the assumption that $f$ is irreducible.

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