I study an undergraduate course for Galois theory, although the following question concerns elementary number theory. The following is given in our lecture notes:
Let $\mathbb{K} \subseteq \mathbb{M} \subseteq \mathbb{F}$ be an ascending chain of fields where $\mathbb{K} \subseteq \mathbb{F}$ is normal. In general it does not follow that $\mathbb{K} \subseteq \mathbb{M}$ is normal.
Consider for example
$\mathbb{Q} \subseteq \mathbb{Q}(\sqrt[3]{2}) \subseteq \mathbb{A}$
where $\mathbb{A} = \mathbb{A}_{\mathbb{C}/\mathbb{Q}}$ is the field of the algebraic elements in $\mathbb{C}$ over $\mathbb{Q}$. Here $\mathbb{Q} \subseteq \mathbb{A} $ is normal but $\mathbb{Q} \subseteq \mathbb{Q}(\sqrt[3]{2})$ is not normal.
I don't understand why $\mathbb{Q} \subseteq \mathbb{A}$ is normal. I do not have a counter example; I believe that it's true, but I can't think how it is so obvious as to not give this any justification. Is this something simple that I'm missing or should it be a little work?
By Wikipedia's definition, a field extension $K \subset L$ is normal iff every irreducible polynomial $f(X) \in K[X]$ with at least one root in $L$ splits completely in $L$. Since $\mathbb A$ is defined to be the algebraic closure of $\mathbb Q$, every polynomial $f(X) \in \mathbb Q[X]$ splits completely in $\mathbb A$. (We don't even need to assume the hypothesis that $f(X)$ has at least one root in $\mathbb A$ - even this statement is automatically true.)
Maybe you're more familiar with the definition that says that $K \subset L$ is a normal extension iff $L$ is the splitting field of some polynomial $f(X) \in K[X]$. This definition is equivalent to the Wikipedia one for field extensions of finite degree, but it doesn't work for our example because $\mathbb Q \subset \mathbb A$ is an infinite field extension.