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Let $F$ be a field of characteristic $p$, and consider the polynomial $f(x) = x^p -a$ in $F[x]$. I want to show that $f(x)$ is either irreducible or splits over $F$. It's easy to show that if there is a root, then $f(x)$ splits. If $f(x)$ does not have a root, then it cannot split, so I should be able to show that if $f(x)$ is reducible, then it must have a root.

I found a hint by this earlier post and was able to continue in the following manner:

Suppose $f(x)$ is reducible. Then we can write $f(x) = g(x) h(x)$. Let $E$ be the splitting field for the polynomial $g(x)$. We note in this field that $g(x)$ and therefore $f(x)$ has a root, $\beta$ and so we can write $f(x) = (x-\beta)^p$.

This means $g(x) = (x-\beta)^k$. Since all of the coefficients are over $F$, we must have that $k\beta \in F$. However, I don't see how this can show that $\beta$ is in $F$.

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    $\begingroup$ This is false, Let $F=\Bbb Q(\sqrt[3]{2})$ and $f(x)=x^3-2$. The polynomial has a root in $F$ but does not split. $\endgroup$ – Adam Hughes Dec 5 '16 at 18:39
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    $\begingroup$ Is $F$ of characteristic $p$? $\endgroup$ – André 3000 Dec 5 '16 at 18:40
  • $\begingroup$ Sorry, $F$ is of characteristic $p$. I will edit that into the post. $\endgroup$ – Nitin Dec 5 '16 at 18:44
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The step you appear to missing is the following. Here $k$ is an integer in the range $0<k<p$ (otherwise $g(x)$ would not be a proper factor of $f(x)$). Therefore $$ k\beta=(k\cdot1_F)\beta. $$ Here $k\cdot1_F$ is an invertible element of $F$. So if $k\beta\in F$, then you can infer that $\beta\in F$.

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  • $\begingroup$ But I think this post has the best answer. $\endgroup$ – Jyrki Lahtonen Dec 5 '16 at 18:56
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Let $K$ be an extension of $F$ where $X^p-a$ has a root $b$. You have $(X-b)^p=X^p-a$. This implies that if $X^p-a$ is reducible, $X^p-a$ is the product of irreducible polynomials which are of the form $(X-b)^i$, this implies they are equal thus $X^p-a=f^l$. If $m$ is the degree if $f$, $p=lm$, we deduce that $l=1$ or $l=p$ and $l=p$ since $X^p-a$ is reducible. We deduce that $b$ is in $F$.

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  • $\begingroup$ Why are the degrees of the factors equal? This happens e.g. when we consider the factorization of an irreducible polynomial over $K$ over a Galois extension $L$, but why does that apply here? $\endgroup$ – Jyrki Lahtonen Dec 5 '16 at 19:18
  • $\begingroup$ Because you can divide the polynomial with bigger degree by the one with smaller degree. $\endgroup$ – Tsemo Aristide Dec 5 '16 at 19:24
  • $\begingroup$ Ok. That will work :-) $\endgroup$ – Jyrki Lahtonen Dec 5 '16 at 20:33

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