Exercise 6.5.1

Proof.

$\text{∙}$ Suppose that $L=F(\alpha )$. As ${\alpha }_i\in L$, ${\alpha }_i\in F(\alpha )$. By definition of $F(\alpha )$, there exist ${𝜃}_i\in F(x)$ such that ${\alpha }_i={𝜃}_i(\alpha )$.

$\text{∙}$ Conversely, suppose that for all $i$, $1\le i\le n$, ${\alpha }_i={𝜃}_i(\alpha ),{𝜃}_i\in F(x)$. Thus ${\alpha }_i\in F(\alpha )$. Consequently $L=F({\alpha }_1,\dots ,{\alpha }_n)\subset F(\alpha )$.

As $F(\alpha )=F({\alpha }_1)\subset F({\alpha }_1,\dots ,{\alpha }_n)$, $L=F({\alpha }_1,\dots ,{\alpha }_n)=F(\alpha )$.

Conclusion: $L=F(\alpha )⟺{\alpha }_i={𝜃}_i(\alpha ),{𝜃}_i\in F(x)(1\le i\le n)$.

Moreover, as $\alpha$ is algebraic over $F$, $F(\alpha )=F[\alpha ]$, therefore every ${\alpha }_i\in F(\alpha )=F[\alpha ]$ is of the form ${\alpha }_i={𝜃}_i(\alpha )$, where the ${𝜃}_i\in F[x]$ are polynomials. □