Exercise 6.20

Proof. Let $\alpha \in F$, with ${\dim <!--\; FUNCTION\; APPLICATION\; -->}_{\mathbb{Q}}F=n$. Any subset of $n+1$ vectors in $F$ is linearly dependent, so $\{1,\alpha ,{\alpha }^2,\cdots ,{\alpha }^n\}$ is linearly dependent.

Thus there exists $(a_0,\dots ,a_n)\in {\mathbb{Q}}^{n+1},(a_0,\dots ,a_n)\ne (0,0,\dots ,0)$ such that $a_0+a_1\alpha +\cdots +a_n{\alpha }^n=0$.

Let $f(x)=a_0+a_{1}x+\cdots +a_n{x}^n$. Then $f(x)\in \mathbb{Q}[x],f(x)\ne 0$ and $f(\alpha )=0,\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(f(x))\le n$. So every element of $F$ is algebraic of degree at most $n$. □