If $[K:F]$ is a prime, then $K=F(a)$ for every $a\in K - F$.

Question

This question originates from Pinter's Abstract Algebra, Chapter 28, Exercise D3.

Prove: If $[K:F]$ is a prime, then $K=F(a)$ for every $a\in K - F$.

For every element $a\in K - F$,

1. $a$ is algebric over $F$. Therefore $F(a)$ is a field extension of $F$. There are only three possibilities: (i) $F(a) = K$,(ii) $F(a) \subset K$, or (iii) $K \subset F(a)$.
2. $F(a)\ne F$, as given.
3. Given $[K:F]$ is prime, there is no field properly between $F$ and $K$. Hence (ii) is not possible.
4. $K\not\subset F(a)$ as $a$ is a linear combination of the basis of $K$. So (iii) is not possible.

Hence $K=F(a)$ for every $a\in K - F$.

Is this reasonable?

Answer 1

How do $3$ and $4$ prove that $F(a)$ must be equal to $K$? You didn't even mention that it is a subfield.

Here is a solution. $F(a)$ is by definition the smallest subfield of $K$ which contains the field $F$ and the element $a$. So $F(a)\subseteq K$. Then, by the tower law $[K:F]=[K:F(a)][F(a):F]$. Since $[K:F]=p$ is a prime number we conclude that $[F(a):F]=1$ or $[F(a):F]=p$. But this extension degree can't be $1$ because $a\in K\setminus F$, hence $F(a)\ne F$. So $[F(a):F]=p$, and hence $[K:F(a)]=1$. This means $K=F(a)$.