Galois extension preserves irreducibility Consider a Galois extension $K/F$. Let $f\in F[x]$ be irreducible of prime degree. If $f$ has no roots in $K$, prove that $f$ is irreducible in $K[x]$.
 A: I will show that if $f$ is reducible in $K[x]$ then it has a root in $K$. 
Let $p(x)\in K[x]$ be a irreducubla over $K[x]$ and $p(x)\mid f(x)$. 
Let $\sigma\in Aut(K/F)$ then $\sigma(p(x))\mid \sigma (f(x))=f(x)$ set $r(x)=\sigma_0(p(x))\sigma_1(p(x))...\sigma_k(p(x))$ such that we takes all different $\sigma(p(x))$.
Notice that $\sigma(r(x))=r(x) \ \ for \ \ all \ \ \sigma\in Aut(K/F) \implies r(x)\in F(x)$ and $r(x)\mid f(x)$ and $f$ is irreducable so $r(x)=f(x)$. 
Now let $\deg(p(x))=n$ since $\deg(p(x)=\deg(\sigma_i(p(x)))\implies \deg(r(x))=\deg(f(x))=kn$ but since $\deg(f(x))$ is prime we must have $n=1$ so $f$ has a root in $K$.
A: Here is another proof which may be longer but perhaps is more conceptual.
It uses several lemmas: 
Lemma 1. If $E/F$ is a field extension of prime degree, then it has no proper subextensions.
Pf. If $F\subseteq E'\subseteq E$, then $[E:E']$ divides the prime number $[E:F]$, so either $[E:E']=1$ and $E=E'$ or $[E:E']=[E:F]$ and $F=E$. 
Lemma 2. If $K/F$ is Galois and if $E/F$ is a field extension, then $KE/E$ is Galois, and the restriction-of-automorphism map induces an isomorphism $Gal(KE/E)\cong Gal(K/E\cap K)$. 
The proof is standard, e.g it appears in Lang's Algebra, hence I omit it.
Corollary 3. If $K/F$ is Galois and $E/F$ is of prime degree, then either $E\subseteq K$ or $[KE:E]=[K:F]$.
Pf. By Lemma 2, $Gal(KE/E) \cong Gal(K/E\cap K)$, so $[KE:E]=[K:E\cap K]$. By Lemma 1, either $E\cap K = F$ and then $[KE:E]=[K:F]$ or $E\cap K = E$ and then $E\subseteq K$. 
Now answering the question: Apply Cor. 3 to $E=F(\alpha)$, where $\alpha$ is a root of $f(x)$. Then either $E\subseteq K$ or $[EK:K]=[E:K]=\deg f$. The former contradicts the assumption that $\alpha\not\in K$ and the latter implies that $f$ is irreducible over $K$, since $EK=K(\alpha)$. QED
