9
$\begingroup$

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.

  • 0
    @Nils Why don't you ask it as a new question?2012-05-10

1 Answers 1

9

It suffices to prove that every irreducible polynomial in $K[X]$ splits in E(why?). Let $f(X)$ be an irreducible polynomial in $K[X]$. By the assumption, $f(X)$ has a root in $E$. Since $E$ is separable over $K$, $f(X)$ has no multiple root in an algebraic closure of $E$. Let $\alpha_1, ..., \alpha_n$ be all the roots of $f(X)$ in an algebraic closure of $E$. Put L = $K(\alpha_1, ..., \alpha_n)$. Since $L/K$ is separable, by the primitive element theorem, there exists an element $\omega$ in $L$ such that $L = K(\omega)$. Let $g(X)$ be the minimal polynomial of $\omega$ in $K[X]$. By the assumption, $g(X)$ has a root $\lambda$ in E. Since $L/K$ is a Galois extension, $L = K(\omega) = K(\lambda)$. Hence $L ⊂ E$ as desired.

  • 0
    Of course, thanks alot!2012-05-10