If every polynomial in $F[x]$ splits then there exists no nontrivial algebraic extension

Question

Im trying to prove the statement of the title:

If every polynomial in $F[x]$ splits then $F$ has no nontrivial algebraic extension

I was thinking about arguing as follows: if there existed an algebraic extension $K=F(\alpha)$ then $\alpha$ would have to be transcendental. But i)I don't know if this would prove the claim ii)I don't even know how to continue along that line of reasoning

Answer 1

Let $E$ be an algebraic extension of $F$ and let $\alpha \in E$.

Then $F(\alpha) \cong F[x]/(f(x))$, where $f(x)$ is the minimal polynomial of $\alpha$.

Since $f$ is irreducible, it must have degree $1$ and so $\alpha \in F$.

Thus $E \subseteq F$ and so $E=F$.