A field extension of prime degree

Question

Suppose that $E$ is an extension of $F$ of prime degree. Show that $~~\forall~ a \in E : ~ F(a)=F$ or $F(a)=E$

Attempt: Suppose that $E$ is an extension of a field $F$ of prime degree, $p$. Therefore $p = [E :F] = [E : F(a)][F(a) : F]$. Since $p$ is a prime number, we see that either $[E : F(a)] = 1$ or $[F(a) : F] = 1$.

Now, $[E : F(a)] = 1 \implies $ there is only one element $x \in E$ which forms a basis and every element in $E$ is generated by $x$ i.e. $E = \{x~c~|~ c \in F(a)\}$

$[F(a) : F] = 1 \implies $ there is only one element $y \in F(a)$ which forms a basis and every element in $F(a)$ is generated by $y$ i.e. $F(a) = \{y~d~|~ d \in F\}$

Have I inferred it correctly? How do I move ahead?

Thank you for your help.

Answer 1

Of course, any nonzero vector of a $1$-dimensional subspace generates that subspace.

Therefore in $F(a) = \{yd \mid d \in F\}$ you may as well choose $y=1\in F(a)$ as the generator. Thus $F(a) = \{d \mid d \in F\}=F$.

The general observation to be made is $[E:F]=1$ iff $E=F$. See if you can't convince yourself of this.

You will move ahead by applying this to both $[F(a):F]$ and $[E:F(a)]$

Answer 2

Alternative argument: If $a\in F$, clearly $F(a)=F$. Otherwise the degree of the minimal polynomial of $a$ over $F$ is $[F(a):F]$, which divides $[E:F]=p$. But $p$ is a prime, and $a\notin F$, which means that $[F(a):F]\ne 1$, so we must have $[F(a):F]=[E:F]=p$. Now, since $F(a)/F$ is a subextension of $E/F$ with the same extension degree, we get $E=F(a)$.