PROBLEM
Let $p$ be a prime not equal to the characteristic of $F$. Show that if $a\in F$ then $x^p-a$ is either irreducible in $F[x]$ or has a root in $F$.
PROOF
Let $f=x^p-a$ be as above and let $r$ be a root of $f$ in $E_f$, the extension with roots of $f$. The roots of $f$ are $r\xi^i$ where $\xi$ is a primitive root of unity. Note that $f'(x)=px^{p-1}\neq 0$ since the characteristic of $F$ is assumed not to be $p$. So $(f,f')=1$ which implies $f$ is seperable and $E_f=F(r,\xi)$ is Galois. We have two cases now: either $r\in F(\xi)$ or $r\not\in F(\xi)$.
If $r\not\int F(\xi)$ then $|Gal(E_f\backslash F(\xi))|=[E_f: F(\xi)]>1$. So there exists $\alpha\neq 1\in Gal(E_f\backslash F(\xi))$, $E_f=F(\xi,r)$. So by definition of $\alpha$ there must be another root $r\xi^i$ for $i\neq 0$. Then $\alpha^k(v)=r\xi^{ki}$ implies $\alpha$ has order $p$. So $p=|\alpha| \leq [E_f:F(\xi)]\leq deg(m_r(x))\leq deg(f)=p$. So $f(x)$ is the minimum polynomial and is hence irreducible.
Now consider the case where $r\in F(\xi)$. Let $G$ be $Gal(F(\xi)\backslash F)$ and define $H$ to be a subgroup of $U(\mathbb{Z}_p)$. Note that $H$ is cyclic. Let $G=<\alpha>$ $\alpha_k(r)=r\xi^\ell$ so $\alpha_k(r)\alpha_k(\xi^i)=r\xi^\ell\xi^{ki}=r\xi^{ki+\ell}$. Let $ki+\ell=i$ so $i(1-k)=\ell$. Then $k=1$, $i=1$ implies that $f(x)$ has a root in $F$, else $i=\ell(1+k)^{-\ell}$ so $\alpha_k(r\xi^i)=r\xi^i$ for $\xi^i$ fixed by $\alpha_k$. But $G=<\alpha_k>$, $r\xi^i\in Inv(G)=F$. So $f(x)$ has a root in $F$.
Let $p$ be a prime not equal to the characteristic of $F$. Show that if $a\in F$ then $x^p-a$ is either irreducible in $F[x]$ or has a root in $F$.
PROOF
Let $f=x^p-a$ be as above and let $r$ be a root of $f$ in $E_f$, the extension with roots of $f$. The roots of $f$ are $r\xi^i$ where $\xi$ is a primitive root of unity. Note that $f'(x)=px^{p-1}\neq 0$ since the characteristic of $F$ is assumed not to be $p$. So $(f,f')=1$ which implies $f$ is seperable and $E_f=F(r,\xi)$ is Galois. We have two cases now: either $r\in F(\xi)$ or $r\not\in F(\xi)$.
If $r\not\int F(\xi)$ then $|Gal(E_f\backslash F(\xi))|=[E_f: F(\xi)]>1$. So there exists $\alpha\neq 1\in Gal(E_f\backslash F(\xi))$, $E_f=F(\xi,r)$. So by definition of $\alpha$ there must be another root $r\xi^i$ for $i\neq 0$. Then $\alpha^k(v)=r\xi^{ki}$ implies $\alpha$ has order $p$. So $p=|\alpha| \leq [E_f:F(\xi)]\leq deg(m_r(x))\leq deg(f)=p$. So $f(x)$ is the minimum polynomial and is hence irreducible.
Now consider the case where $r\in F(\xi)$. Let $G$ be $Gal(F(\xi)\backslash F)$ and define $H$ to be a subgroup of $U(\mathbb{Z}_p)$. Note that $H$ is cyclic. Let $G=<\alpha>$ $\alpha_k(r)=r\xi^\ell$ so $\alpha_k(r)\alpha_k(\xi^i)=r\xi^\ell\xi^{ki}=r\xi^{ki+\ell}$. Let $ki+\ell=i$ so $i(1-k)=\ell$. Then $k=1$, $i=1$ implies that $f(x)$ has a root in $F$, else $i=\ell(1+k)^{-\ell}$ so $\alpha_k(r\xi^i)=r\xi^i$ for $\xi^i$ fixed by $\alpha_k$. But $G=<\alpha_k>$, $r\xi^i\in Inv(G)=F$. So $f(x)$ has a root in $F$.
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