Question on the normal closure of a field extension

Question

Hi all I was given the following question in my field theory class which I am stuck on: I am given $ K/F $ a finite extension of fields. I am asked to show the existence of an $ K \subset L $ such that L/F is a normal extension and if we have $ L/L'/K $ and $ L'/F $ is a normal extension then we have L'=L and I am asked to prove that L is unique up to isomorphism. I have no real idea where to begin as in where to use the assumption K/F is finite and I do not know if the proof is constructive or any other approach. Thanks for any help

Answer 1

The $L$ you're trying to find should be the smallest field extension of $K$ such that $L/F$ is normal. If you think about the definition of a normal extension, the construction of $L$ becomes clear. Since $K/F$ is finite, $K$ is finitely generated over $F$, i.e. $K = F(\alpha_1, \ldots, \alpha_l)$ for some elements $\alpha_i \in K$. For simplicity, I'll assume $l = 1$ (this holds by the primitive element theorem if $K/F$ is separable). Then consider the minimal polynomial $m(x) \in F[x]$ of $\alpha := \alpha_1$. Let $\overline{K}$ be some algebraic closure of $K$. Then the roots of $m(x)$ are all contained in $\overline{K}$, call them $\beta_1, \ldots, \beta_n$, and we can therefore define $L = K(\beta_1, \ldots, \beta_n)$. This makes sense because everything on the right-hand side is in $\overline{K}$.

As you can see, we've done the minimal amount necessary to ensure that $L/F$ is normal and that $L \supset K$. It follows that if $L/L'/K$ such that $L'/F$ is normal, then $L' = L$, because $\alpha \in L'$ implies that $m(x)$ splits over $L'$, hence $L' \supset L$.

It just remains to show that $L$ is unique up to isomorphism. But given any other $\tilde{L}/K$ satisfying the same conditions, you should be able to show that $\tilde{L}$ is the splitting field of $m(x)$ in some algebraic closure of $K$ (just show that it is no bigger!). Then the roots of $m(x)$ in $\overline{K}$ correspond to the roots of $m(x)$ in the other algebraic closure of $K$, and this correspondence defines an isomorphism $L \cong \tilde{L}$.