Question: Let $L$ be the extension of the field $K$ such that $[L:K]=p$, where $p$ is a prime number, and $\alpha \in L$. Prove that $K(\alpha)=K$ or $K(\alpha)=L.$
Proof: From $$ \alpha \in L \implies K \subseteq K(\alpha) \subseteq L, $$ we have $$ p=[L:K]=[L:K(\alpha)]\cdot [K(\alpha):K] $$ from which we have two cases, one being that $[L:K(\alpha)]=1$ and $[K(\alpha):K]=p$, or vice-versa. In each of the cases I then conclude either $K(\alpha)=K$ or $K(\alpha)=L.$
However, I am unsure if it is enough to state that i.e. $[L:K(\alpha)]=1$ implies that $L=K(\alpha)$. Is it?
To see it fully, formally and explicitly, recall the definition of $[E:F]$ for any field extension $E$ over $F$: it is the dimension of $E$ as a vector space over $F$:
$[E:F] = \dim_F E. \tag{1}$
If $[E:F]$ = 1, then there is a basis for $E$ as a vector space consisting of precisely one element $e \in E$; then any element of $E$ may be written $\lambda e$ for some $\lambda \in F$; in particular the elements of $F$ may be so written; thus for each $\mu \in F$ there is $\lambda \in F$ with
$\mu = \lambda e, \tag{2}$
or
$e = \lambda^{-1} \mu \in F; \tag{3}$
since the basis element $e \in F$, we have
$E = \{ \alpha e \mid \alpha \in F \} = eF \subset F; \tag{4}$
but $F \subset E$ is given; hence $E = F$.
The present case is thus resolved by taking $L = E$, $K(\alpha) = F$; we conclude $L = K(\alpha)$.