Exercise 5.4.5

Proof. As in Exercise 4, $F\subset L=F(\alpha ,\beta )$.

Suppose that $F(\alpha +\mathit{\lambda \beta })=F(\alpha +\mathit{\mu \beta }),\lambda \ne \mu$.

(a)

Then $$\begin{array}{llll}\hfill \alpha +\mathit{\mu \beta }& \in F(\alpha +\mathit{\lambda \beta }),& \hfill & \\ \hfill \alpha +\mathit{\lambda \beta }& \in F(\alpha +\mathit{\lambda \beta }).& \hfill & \\ \hfill & & \hfill \end{array}$$

Consequently their difference is also in the subfield $F(\alpha +\mathit{\lambda \beta })$:

$$(\mu -\lambda )\beta \in F(\alpha +\mathit{\lambda \beta }).$$

As $\mu -\lambda \in F,\mu -\lambda \ne 0$,

$$\beta \in F(\alpha +\mathit{\lambda \beta }).$$

Since $\alpha =(\alpha +\mathit{\lambda \beta })-\mathit{\lambda \beta }$, with $\alpha +\mathit{\lambda \beta },\beta \in F(\alpha +\mathit{\lambda \beta })$, and $\lambda \in F$, then

$$\alpha \in F(\alpha +\mathit{\lambda \beta }).$$

(b)

$\alpha +\mathit{\lambda \beta }\in F(\alpha ,\beta )$, thus $F(\alpha +\mathit{\lambda \beta })\subset F(\alpha ,\beta )$. Moreover, by part (a), $\alpha ,\beta \in F(\alpha +\mathit{\lambda \beta })$, thus $F(\alpha ,\beta )\subset F(\alpha +\mathit{\lambda \beta })$. $$F(\alpha ,\beta )=F(\alpha +\mathit{\lambda \beta }).$$

But Example 5.4.4 shows that $F(\alpha ,\beta )$ has no primitive element : this is a contradiction.

This shows that all the fields $F(\alpha +\mathit{\lambda \beta })$, where $\lambda$ varies over all elements of $F$, are distinct. $F$ being infinite, there exists infinitely many intermediate fields between $F$ and $L$.