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If $k$ is a field of characteristic $p > 0$ and $f (x) = x^{2p} − x ^p + t \in k(t)[x]$, how can we show that $f (x)$ is an irreducible polynomial in $k(t)[x]$ and that $f (x)$ is inseparable?

If $f(x)$ is irreducible then $D_xf(x) = 0 \implies( f, D_xf) = ( f, 0) = f$. Hence, if $f(x)$ is irreducible then $( f, D_xf) \neq 1 \implies f(x)$ is inseparable.

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  • $\begingroup$ Homework problem? If it is, people will not go beyond giving you a hint. $\endgroup$
    – Lubin
    Mar 4 '13 at 15:34
  • $\begingroup$ @Lubin no sir it is not a homework . a hint will do. $\endgroup$
    – jim
    Mar 4 '13 at 15:38
  • $\begingroup$ I think that $f'=0$ suffices to show it is inseparable $\endgroup$
    – Belgi
    Mar 4 '13 at 15:45
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Here’s a hint: notice that $k(t)=k(\tau)$ when $\tau=1/t$. If $\rho$ is a root of $f$, what polynomial in $k(\tau)[x]$ does $1/\rho$ satisfy?

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  • $\begingroup$ sir i have posted an answer could you tell me is it right? $\endgroup$
    – jim
    Mar 5 '13 at 3:05
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$K$ is a field $\implies K[t]$ is a U.F.D.

By gauss lemma: $f(x)$ is irreducible in $K(t)[x] \iff f(x)$ is irreducible in $K[t][x]$

then $K[t][x]= K[x][t]$

$f(x)$ is irreducible in $K[x][t]$ being a linear polynomial in $K[x][t]$

hence $f(x)$ is irreducible in $K(t)[x] $

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  • $\begingroup$ Nice answer. In Gauss Lemma, $f$ should be primitive. $\endgroup$
    – user18119
    Mar 5 '13 at 20:14
  • $\begingroup$ Very nice indeed, and much quicker than what I had in mind! $\endgroup$
    – Lubin
    Mar 8 '13 at 20:17

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