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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<


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

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$$p(x)=x^8-2$$ is irreducible over the rationals and has no zeros in the real field $$L={\bf Q}(\root4\of2)$$ where it factors as $$p(x)=(x^2-\root4\of2)(x^2+\root4\of2)(x^4+\sqrt2)$$ The first two factors are clearly irreducible over $L$. The third factor factors over the reals as $$x^4+\sqrt2=(x^2-2^{5/8}x+\root4\of2)(x^2+2^{5/8}x+\root4\of2)$$ but $2^{5/8}$ is not in $L$, so $x^4+\sqrt2$ is also irreducible over $L$.

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I agree... but if we consider the extension is normal do you think the question will be correct ? – Rahman Nov 23 '12 at 13:41
Yes, and I think that question was asked and answered here recently. See…? (though I think there has been a more recent occurence). – Gerry Myerson Nov 23 '12 at 21:55

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