Let $ K$ be a finite extension of $F$ and assume $K$ is a splitting field over $F$. Is it given that for any element $ \alpha \in K, \alpha \not\in F $ that there exists a polynomial $ f(x) \in F[x] $ in which $ \alpha $ is a root?
It relates to the following exercise: Let $ K_1 $ and $ K_2 $ be finite extensions of $ F $ contained in the field $ K $, and assume both are splitting fields over $ F $. Prove that $ K_1 \cap K_2 $ is a splitting field over $ F $.
I don't want an answer to the exercise, I'm just posting it for reference. The thing is, how can I be certain that $ K_1 $ and $ K_2 $ do not share only a set of elements which are not algebraic (a part from the obviosly shared $ F $)? Because in that case $ K_1 \cap K_2 $ would only split the same polynomials as $ F $ and since $ F $ is a proper subset, $ K_1\cap K_2 $ cannot be a splitting field.
Suppose $K$ has dimension $n$ over $F$, then the $n+1$ elements $1,a,a^2 \dots a^n$ are linearly dependent over $F$. An explicit dependence gives a polynomial over $F$ which is satisfied by $a$.