Let $F$ be a field.
I already know that there is an algebraic closure $\overline{F}$ of $F$ which satisfy that every $f(x)\in \overline{F}[x]$ is splits in $\overline{F}[x]$.
Is there a smallest extension field $E$ of $F$ such that every $f(x)\in {\mathbf F}[x]$ is splits in $E[x]$?
Or such extension field $E$ must be the algebraic closure of $F$?
Let $E$ be the smallest extension of $F$, where every polynomial in $F[X]$ splits. Then $E$ is algebraically closed, i.e., every polynomial in $E[X]$ splits as well: Consider $f\in E[X]$. Its finitely many coefficients are in a finite algebraic extension $K$ of $F$ (presumably smaller than $E$). If we multiply together all the (finitely many) conjugates of $f$, we obtain a polynomial that is in fact in $F[X]$, so splits by assumption; consequently its factor $f$ splits as well.
This argument uses Galois theory; if you don't have enough of it available, you may have to gnaw yourself through the base of $K/F$ step by step ...