So I read a definition of separable polynomial like this:

"A separable polynomial is an irreducible polynomial which does not have multiple root in its splitting field."

So, it is equivalent that $f$ and $f'$ have common root, right? My question is, does it count if $f'= 0$, I mean in a field of characteristic $p>0$, $f'$ could be $0$ but $f$ only needs to be in $K(X^p)$, then how do we decide if $f$ is separable?

  • $\begingroup$ If $f'=0$, then $f$ is automatically inseparable. $\endgroup$ – Lubin Apr 10 '17 at 15:24
  • $\begingroup$ can we prove that or it's just a convention? I mean how do we know $x^p+1\in F_p[x]$ is separable or not? $\endgroup$ – chí trung châu Apr 10 '17 at 15:27
  • $\begingroup$ For that specific example, $x^p+1=(x+1)^p$. That should give you a clue to the general proof. $\endgroup$ – Lubin Apr 10 '17 at 15:34
  • $\begingroup$ If $K^p=K$, then yes, I have a solution. But if it's not the case, let $a\neq x^p$ for all $x\in K$. Then what about $x^p+a$? $\endgroup$ – chí trung châu Apr 10 '17 at 15:37
  • $\begingroup$ But then the splitting field is $K(a^{1/p})$, and the condition is that there should be repeated roots in the splitting field, and we get $x^p+a=(x+a^{1/p})^p$. $\endgroup$ – Lubin Apr 10 '17 at 18:28

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