Exercise 4.1.1

Proof. Suppose that $\alpha \in L\setminus \{0\}$ is algebraic over a subfield $F$ of $L$. Then there exists a polynomial $p=\underset{k=0}{\overset{d}{\sum <!--\; FUNCTION\; APPLICATION\; -->}}{a}_k{x}^k\in F[x]$, with $a_d\ne 0$, whose $\alpha$ is a root:

Dividing by ${\alpha }^d$, we obtain $\underset{k=0}{\overset{d}{\sum <!--\; FUNCTION\; APPLICATION\; -->}}{a}_k{\left(\frac{1}{\alpha }\right)}^{d-k}=0$, which we can write $\underset{i=0}{\overset{d}{\sum <!--\; FUNCTION\; APPLICATION\; -->}}{a}_{d-i}{\left(\frac{1}{\alpha }\right)}^i=0$.

So $1∕\alpha$ is a root of the polynomial $q=\underset{i=0}{\overset{d}{\sum <!--\; FUNCTION\; APPLICATION\; -->}}{a}_{d-i}{x}^i\in F[x]$, and $q\ne 0$ since $a_d\ne 0$, thus $1∕\alpha$ is algebraic over $F$. □