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Let $L$ be a Galois extension of $F$ and let $p(x) ∈ F[x]$ be irreducible. I wish to show that all irreducible factors of $p(x) ∈ L[x]$ have the same degree. Note that we do not assume that $p(x)$ has a root in $L$. By Normal Extension Equivalent to Same Degree Irreducible Factors (Hungerford, Exercise V.3.24) I know an algebraic extension $F$ of $L$ is normal over $L$ if and only if for every irreducible $p\in L[x],p$ factors in $F[x]$ as a product of irreducible factors which have the same degree.

So can I just say since every Galois extension is normal, all of the irreducible factors of $p$ in $F[X]$ are linear, i.e., $p$ splits over $F$. And given an irreducible $p\in L[X]$, $p$ has a root in $L$ if and only if $p$ admits a linear irreducible factor in $L[X]$, so all irreducible factors of $p(x) ∈ L[x]$ have the same degree?

Homaniac
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1 Answers1

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It's not necessarily the case that $p$ splits over $L$. But let $q(x)$ be an irreducible factor of $p(x)$ over $L$. Applying an element of the Galois group $\sigma\in G=\text{Gal}(L/F)$ to the coefficients of $q(x)$ gives a polynomial $q^\sigma(x)$ which is also irreducible over $L$, also a factor of $p(x)$, and of the same degree as $q(x)$. Either $q(x)=q^\sigma(x)$ or else $q(x)$ and $q^\sigma(x)$ are coprime. By taking all the elements of $G$ one generates a bunch of polynomials $q_1=q,\ldots,q_m$ say. Then $Q(x)=q_1(x)\cdots q_k(x)$ is a factor of $p(x)$, but is stable under $G$, so has coefficients in $F$. As $p(x)$ is irreducible over $F$, $p(x)=Q(x)$. So the $q_i$ are the irreducible factors of $p(x)$ over $L$, and they all have the same degree.

Angina Seng
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