I'm taking a graduate course in groups and fields. This is a theorem I came across in my professor's notes.
Lemma If $K$ is a splitting field for some polynomial $g$ over $F$. Say, $f(x) \in F[x]$ which is irreducible over $F$ and has zero in $K$. Then $f(x)$ splits in $K$.
The proof begins with the following statement.
If $\alpha_1, \alpha_2, \ldots$ are the roots of $f(x)$, then $[K(\alpha_i):K]$ is independent of $i$.
I'm not sure why this is true.
Can someone please explain why the degree of $K(\alpha_i)$ over $K$ should be independent of which $\alpha_i$ we choose among the roots?
Suppose, the root $\alpha_1 \in K$ then $\alpha_1$ satisfies a degree $1$ polynomial with coefficients in $K$, namely $x - \alpha_1$. However, we do not yet know whether each of the other roots $\alpha_2, \alpha_3, \ldots$ also satisfy a degree polynomial with coefficients in $K$.
This is the full proof in the notes:
Let $K_i$ be the spliting field of $g$ over $F(\alpha_i)$, then $[K_i: F(\alpha_i)]$ is independent of $i$, because $F(\alpha_i)\simeq F(\alpha_j)$ for all $i,j$.
But $K_i = K(\alpha_i)$, hence $[K(\alpha_i):F] = [K(\alpha_i):F(\alpha_i)][F(\alpha_i):F]$ is independent of $i$, and therefore $[K(\alpha_i):K] = [K(\alpha_i):F]/[K:F]$ is independent of $i$.
I guess the point here is this is true even if $f(x)$ has no zero in $K$, so it provides more info than the lemma itself.