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Assume that $F$ is a field and $\operatorname{char}(F)=p$. Let $a$ be an element in $F$ without $p$th root, then the polynomial $$x^{p^n}-a$$ is irreducible and inseparable over $F$ for all $n$.

I have proved the inseparable part by considering the derivative of the polynomial, but I'm having trouble with the irreducible part. Any help?

Srivatsan
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CC_Azusa
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1 Answers1

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Let $\alpha$ be a $p^n$-th root of $a$ in some extension of $F$. Then $x^{p^n} - a = (x - \alpha)^{p^n}$. If you have a non-trivial monic factor of this polynomial in $F[x]$, then it is of the form $(x - \alpha)^k$ for some $0 < k < p^n$. Can you get a contradiction out of the coefficients of this polynomial? It might be good to write $k = p^rs$ with $s$ not divisible by $p$.

Dylan Moreland
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  • why can we write $k$ as $p^r s$? – XXX Nov 12 '22 at 15:09
  • @XXX If $s=1$, then $a$ would have a $p$th root, $\alpha^k$, in $F$. – 19021605 Jun 24 '25 at 11:30
  • Following the hint: By https://math.stackexchange.com/questions/590885/proving-p-nmid-binomprmpr-where-p-nmid-m?noredirect=1&lq=1 and looking at the coefficient of $x^{p^rs-p^r}$ we have $\alpha^{p^r}\in F$. So we have $\alpha^{p^m}\in F$ for all $r\leq m\leq n$. This implies (since $r<n$) that $a$ is a $p$th root in $F$. – 19021605 Jun 24 '25 at 11:59