Degree of extension $[K(\alpha):F(\alpha)]$

Question

Given $K$ finite normal extension over $F$, $\alpha$ algebraic over $K$, can we say anything about $[K(\alpha):F(\alpha)]$? I think it is related to $[K:F]$ somehow.

Answer 1

We get $[K(\alpha):F]=[K(\alpha):F(\alpha)][F(\alpha):F]$ and $[K(\alpha):F]=[K(\alpha):K][K:F]$; so $[K(\alpha):F(\alpha)] \cdot \displaystyle \frac{[F(\alpha):F]}{[K(\alpha):K]}=[K:F]$

Let $f$ be the minimal polynomial of $\alpha$ over $F$ and let $f=g_1g_2 \ldots g_n$ be the factorization of $f$ over $K$ into irreducibles such that $g_1(\alpha)=0$. Let $\overline{F}$ be an algebraic closure of $F$ and assume that $K \subset \overline{F}$ and $\alpha \in \overline{F}$. Let $\beta$ be a root of one of the $g_i$, then $\beta$ is also a root of $f$, so because $f$ is irreducible over $F$, there is a $\sigma \in \operatorname{Aut}_F(\overline{F})$ such that $\sigma(\alpha)=\beta$. We get $\sigma(K(\alpha))=\sigma(K)(\sigma(\alpha))= \sigma(K)(\beta)$. Now $\sigma(K)=K$, because $K$ is normal. Thus $\sigma(K(\alpha))=K(\beta)$. As $\sigma$ is $F$-linear, $[K(\alpha):F]=[K(\beta):F]$, so dividing by $[K:F]$ gives $[K(\alpha):K]=[K(\beta):K]$, this shows that the factor $g_i$ of which $\beta$ is a root has the same degree as the factor $g_1$ of which $\alpha$ is a root, so all the $g_i$ must have the same degree. This implies that $\operatorname{deg}(f)=n \cdot \operatorname{deg}(g_1)$, so $\operatorname{deg}(g_1) \mid \operatorname{deg}(f)$.

Because $[K(\alpha):K]=\operatorname{deg}(g_1)$ and $[F(\alpha):F]=\operatorname{deg}(f)$, this implies that $\displaystyle \frac{[F(\alpha):F]}{[K(\alpha):K]}$ is an integer, so $[K(\alpha):F(\alpha)] \mid [K:F]$.

If we do not assume that $K/F$ is normal, we still get $[F(\alpha):F] \geq [K(\alpha):K]$, so $[K(\alpha):F(\alpha)] \leq [K:F]$.

Answer 2

This is a case of a standard theorem in Galois Theory, the Theorem on Natural Irrationalities. It goes like this:

Theorem. Let $F\subset K\subset\Omega$ be fields, with $K$ finite and Galois over $F$. Let also $L\subset\Omega$. Then $KL$ is finite and Galois over $L$, and $\text{Gal}^{KL}_L\cong\text{Gal}^K_{K\cap L}$.

In particular, $[KL:L]=[K:K\cap L]\,\big\vert\,[K:F]$.

In your case, $L=K(\alpha)$, but note that in the general theorem, there is no hypothesis that $L$ be finite or algebraic over $F$. Maybe I should say also that there’s nothing mysterious about the Theorem: you’ll find it not hard to prove, if you’re comfortable with Galois theory.