$\textbf{Question}$: Would you say that normal extension $\Longleftrightarrow$ splitting field extension is intuitive or counterintuitive?
A splitting field extension $L:K$ is exactly large enough to allow a polynomial $f \in K[x]$ (or, more generally, a set $S$ of polynomials in $K[x]$) to split. It is constructed by introducing the roots $\alpha_1,...,\alpha_n$ of $f$ (or, more generally, the roots of the polynomials in $S$) to $K$, and so $L = K(\alpha_1,...,\alpha_n)$.
A normal extension $L:K$, on the other hand, requires that for each $\beta \in L$, the minimal polynomial $m_\beta \in K[x]$ splits.
($\Rightarrow$) The implication is straightforward to prove. Take $S$ = {$ m_\beta \in K[x]$ | $\beta \in L$}.
($\Leftarrow$) To me, the converse seems almost too much to ask. Specifically, pick a polynomial $f \in K[x]$, form the splitting field extension $L:K$, and take any $\beta \in L$, not necessarily a root of $f$. Then all the conjugate roots of $\beta$ are guaranteed to be in $L$ also? It seems reasonable to expect a counterexample. But, of course, the proof of the converse, which takes noticeably more work, makes the converse undeniable.
Apparently, a rational expression over $K$ in the roots of $f$ must have conjugate roots that are also rational expressions over $K$ in the roots of $f$. Is this intuitive?