Let $E/K$ be a separable, algebraic extension such that every noncostant polynomial in $K[x]$ has a root in $E$, then $E$ is an algebraic closure of $K$. Could you help me to solve this exercise? (there is this hint: use the primitive element theorem).
EDIT: well it's enough to prove $E$ is algebraically closed. So take $f(x)\in E[x]$ I want to prove that it has a root in $E$. One of the thing that I was trying is to prove that the minimal polynomial of $\alpha$ over $K$ divides $f(x)$, but I couldn't, I think it's not true.