Let $F$ be a field of characteristic $p>0$ and $a\in F$.

I have an easy question which I'm stuck on.

If the polynomial $X^p-a$ has no zeros in $F$ then is it irreducible over $F$?

Thank you!


If $X^p-a$ has no roots in $F$, then consider the splitting field $K$ of $X^p-a$. Over $K$, $X^p-a=(X-b)^p$ for some $b \in K$. Hence, $X^p-a$ is reducible over $F$ if and only if $(X-b)^n\in F[X]$ for some integer $n$ with $1\leq n <p$, in which case $b^p \in F$ and $b^n\in F$, whence $b=b^{\gcd(p,n)}\in F$ as well, which is a contradiction. Hence, $X^p-a$ is irreducible over $F$.

P.S.: Can you show that $a$ must be transcendental with respect to the prime field $\mathbb{F}_p$ of $F$?


Every element in a field of characteristic $p$ has up to 1 $p$th root. This is because $X^p$ is a ring endomorphism (the Frobenius endomorphism). This implies that the polynomial must factorize into $(X - b)^p$ in some extension field. Therefore if the polynomial can be reduced in $F[X]$ into $fg$ (where $f$ is of the lowest possible degree) then $f$ must equal $(X - b)^k$ and $g$ must equal $(X - b)^{p-k}$ for some $k$ for which $1 < k < p$. But then we can get ${g \over f}$ and through Euclid's algorithm we find that $(X - b)^{\gcd(k,p-k)} \in F[X]$. By the minimality of $f$, $\gcd(k, p - k) = k \therefore k \mid p - k \therefore k \mid p \therefore k = 1$ or $k = p$.

  • $\begingroup$ Every element in a field of characteristic p has 1 pth root is false unless the field is finite. $\endgroup$ – Bernard Jul 22 '15 at 20:09
  • $\begingroup$ @Bernard up to 1? $\endgroup$ – jkabrg Jul 22 '15 at 20:10
  • $\begingroup$ I mean there are elements that have no $p$-th root (e. g. $X$ in $\mathbf F_p[X]$). $\endgroup$ – Bernard Jul 22 '15 at 20:12

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