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Over a perfect field $k$, the notions of separable and square-free polynomial are equivalent. Indeed if $b^2|P$, then $P$ has a repeated root in $\bar k$. Conversely, if $p$ has a repeated root in $\bar k$, then it cannot be irreducible, since $k$ is perfect, and (why?) $P'\neq 0$ so that $\operatorname{gcd}(P, P')\neq 0$ and we can set $b$ to $\operatorname{gcd}(P, P')$--that is, $\operatorname{gcd}(P, P')^2|P$.

So I am looking for a square free polynomial over an imperfect field that is not separable.

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  • $\begingroup$ The problem is that in characteristic $p$, squarefreeness depends on the constant field. Take $X^p-t\in\Bbb F_p(t)[X]$. The polynomial is squarefree over the field $\Bbb F_p(t)$ (because irreducible), but not over its algebraic closure. $\endgroup$ – Lubin Jan 6 '16 at 18:39
  • $\begingroup$ @Lubin Hi Lubin, thanks! Also, do you know why $P'\neq 0$ in the above condition? I know it should be... $\endgroup$ – Rodrigo Jan 7 '16 at 1:29
  • $\begingroup$ In your question, I think that your argument for the Converse should go like this: if $P$ has a repeated root $\alpha$, then $\alpha$ is also a root of $P'$. If the field is perfect, then the irreducible $P$ must have $P'$ nonzero (i.e. not the zero polynomial) because otherwise $P$ would be an inseparable polynomial. So we have $P$ and $P'$ with $\gcd(P,P')\ne1$, and thus this gcd is of degree less than that of $P$, which shows that $P$ is not irreducible after all. (I’ll bet there’s a more efficient way of saying this!) $\endgroup$ – Lubin Jan 7 '16 at 5:06
  • $\begingroup$ @Lubin, but the definition of separable polynomial is not that $P'$ is nonzero. Instead it is that $P$ has different roots, so, in the converse, we are assuming that $P$ is inseparable from the start. $\endgroup$ – Rodrigo Jan 7 '16 at 18:19
  • $\begingroup$ Ah, but an irreducible polynomial is inseparable if and only if $P'=0$. $\endgroup$ – Lubin Jan 8 '16 at 3:08

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