Any finite dimensional extension is contained in a finite dimensional normal extension

Question

I am new to this website so I'm not really sure what I'm doing, but here it goes.

I have no idea how to do this. I know that an extension is normal if the irreducible polynomial of any element of the extension splits over the field. I am guessing that we can take some polynomial and extend the field by roots of it or something, but i don't know

Answer 1

Let $\Omega$ be an algebraically closed field containing $E$. Since $E$ is finite over $F$, we can write $E = F(a_1, ... , a_n)$ for some $a_i \in \Omega$, algebraic over $F$. Each $a_i$ is the root of an irreducible polynomial $f_i$ with coefficients in $F$. Obtain the required normal extension by adjoining to $F$ (or to $E$) all of the roots of all the $f_i$ (which lie in $\Omega$, since $\Omega$ is algebraically closed).