Let $L:K$ be a field extension, and let $p(x) ∈K[x]$ be irreducible and has no zeros in $L$ show that all irreducible factors of $p(x)$ in $L[x]$ have the same degree.
Assume that $L:K$ is normal and separable extension >new addition<
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Actually I have solution for this -as below- but I am writing here in wish you have some hints or advices to solve this in a different method. for instance to use Möbius function
My solution is: say $p=p_1...p_n$ where $p_j ∈ L[x]$ and are all irreducible over $L$. Let $a$ be a zero for $p_1$ and $b$ be a zero for $p_i$ then there exists an isomorphism $σ: L(a) → L(b)$ such that $σ(a)=b$. Now $p_i(σ(a))=0$ and hence $σp_i(a)=0$ then $p_1$ divides $σp_i$, and since $p_i$ is irreducible it follows that $p_1=σp_i$; this gives $\deg (p_1)=\deg (p_i)$. It has been proved by (Gerry Myerson) that the question is wrong unless we add $L:K$ is normal and separable extension