I see this theorem in several places
Let F be a field and $f(x)\in F[x]$ have degree $n\geq2$.Let $E/F$ be a splitting field of $f(x)$. Then $[E:F]$ divides $n!$
The proof involves some induction. It assumes the $\alpha$ is a root of $f(x)$, and so we can form the extension $F(\alpha)\cong F[x]/ \langle f(x) \rangle$. Then it is stated that we can factor $f(x)=(x-\alpha)g(x)$ with $g(x)\in F(\alpha)$.
I can't see why $g(x)\in F(\alpha)$ is true. Can you help me with that?
[Edit: Copied from an answer accidentally posted by the same (?) user. JL]
Let me give an example:
Suppose $F=\mathbb{Q}$ and that we have $f(x)=x^3-3$. then $f(x)$ can be decomposed as $f(x)=(x-\sqrt[3]{3})(x^2+\sqrt[3]{3}+(\sqrt[3]{3})^2)$
And so we have $g(x)=x^2+\sqrt[3]{3}+(\sqrt[3]{3})^2$
If we denote $E=F(\sqrt[3]{3})$ then we have $g(x)\in E[x]$
The claim is that $g(x)\in E[x]$ is always true. I just can't see why this is generally true
If $\alpha$ is a root of any polynomial $f(x)$ then $x-\alpha$ divides $f(x)$.
The reason for that is that we can use the "euclidean division algorithm" to express $f(x) = Q(x)(x-\alpha) + R(x)$ for some polynomials $Q,R$, with $R$ having degree smaller than the divisor $x-\alpha$.
That means that $R(x)$ is a constant $R(x) = c$. Now using the fact $\alpha$ is a root we find that $0 = Q(\alpha)(\alpha-\alpha) + c$ forces $c = 0$ which proves that $x-\alpha$ divides $f(x)$.