Say we have a field $F$ and an extension $K/F$, such that the polynomial $f(x) \in F[x]$ splits completely in $K$. $K$ may not be a splitting field for $F$ because it could contain a proper subfield containing $F$ in which $f(x)$ also splits completely, but is it always true that $F(\alpha_1,\ldots,\alpha_k) \subset K$ is a splitting field for $f(x)$, where $\alpha_1,\ldots,\alpha_k$ are the roots of $f(x)$ over $K$? I'm fairly sure this is true if the roots all have multiplicity 1, but I can't determine if it's true in general.
Splitting Field as Subfield Generated by Roots
extension-fieldfield-theorysplitting-field
Best Answer
The answer is yes!
Let $L\subseteq K$ be a splitting field for $f$. Then $f$ has the following factorization in $L$: $$ f(x) = (x-\alpha_1)^{r_1}\dots (x-\alpha_k)^{r_k} $$ for some positive integers $r_1,\dots,r_k$. This mean that $(x-\alpha_i)^{r_i}\in L[x]$. Expanding and looking to the coeficient of $x^{r_1-1}$ you obtain that $\alpha_i\in L$ for every $i$, and thus $F(\alpha_1,\dots,\alpha_k)\subseteq L$. But $L$ is a splitting field for $f$ and $f$ factors over $F(\alpha_1,\dots,\alpha_k)\subseteq K$, then $L\subseteq F(\alpha_1,\dots,\alpha_k)$. Hence we obtain $L=F(\alpha_1,\dots,\alpha_k)$.