Show $α^{ −1}$ is algebraic over $ F $ of degree $n$. 
Let $E, F$ be distinct fields such that $E$ is a field extension of $F$. Show that if $\alpha \in E \setminus F$ is algebraic over $F$ of degree $n \in  \{2, 3, \cdots\}$, then $α^{ −1}$ is algebraic over $ F $ of degree $n$.


Since $\alpha \in E \setminus F$ is algebraic over $F$ of degree $n \in  \{2, 3, \cdots\}$, then there exists a polynomial in $F$,$f(x)= a_1x^2+a_2x^3 \cdots+a_{n-2}x^{n-1}+a_{n-1}x^n$ , such that  $f(\alpha)=0$. So $$f(\alpha)= a_1\alpha^2+ a_2\alpha^3+\cdots+a_{n-2}\alpha^{n-1}+a_{n-1}\alpha^n=0$$
Then $$a_1\alpha^2 =-\left( a_2\alpha^3+\cdots+a_{n-2}\alpha^{n-1}+a_{n-1}\alpha^n\right)$$ $$ \alpha=-\frac{\left( a_2\alpha^3+\cdots+a_{n-2}\alpha^{n-1}+a_{n-1}\alpha^n\right)}{a_1\alpha}$$ $$\alpha^{-1}=\frac{-a_1 }{\left( a_2\alpha^2+\cdots+a_{n-2}\alpha^{n-2}+a_{n-1}\alpha^{n-1}\right)}$$

At this step, can I say $\alpha^{-1}\in E\setminus F$ such that it is an algebraic over $ F $ of degree $n$ ? 
 A: (Why does your $f$ start with $a_2$ instead of $a_0$?)
All we know from $\alpha$ being algebraic is that an equality
$$\tag 1\sum_{k=0}^na_k\alpha^k=0$$
holds, where $a_k\in F$ and at least one $a_k$ is $\ne 0$. Since $\alpha\notin F$, certainly $\alpha\ne 0$ so that we obtain 
$$\sum_{k=0}^na_{n-k}(1/\alpha)^k=\alpha^{-n}\sum_{k=0}^na_k\alpha^k=0$$
showing that $1/\alpha$ is algebraic. Also, if $n=\deg\alpha$ in $(1)$, we obtain $\deg(1/\alpha)\le \deg \alpha$. By the seame rasoning $\deg(\alpha\le \deg(1/\alpha)$, hence the degrees are equal.
A: Notice that $\frac{1}{\alpha}$ belongs to $F(\alpha)$ so $F(\frac{1}{\alpha}) \subseteq F(\alpha)$. Similarly $\alpha =\frac{1}{\frac{1}{\alpha}}$ so $F(\alpha)=F(\frac{1}{\alpha})$. Hence $[F(\frac{1}{\alpha}):F]=n$ So $\frac{1}{\alpha}$ is algebraic of degree n. (I hope I am not mistaken), 
A: Hint If
$$a_n\alpha^n+...+a_1\alpha +a_0=0$$
then 
$$a_n+a_{n-1}\alpha^{-1}+...+a_1\alpha^{-n-1} +a_0\alpha^{-n}=0$$
To complete the proof use the above idea to show that the minimal polynomial for $\alpha$ and $\alpha^{-1}$ have the same degree.
A: Assume we have a polynomial such that $\sum\limits_{k=0}^n a_k\alpha^k$. Multiply by $\alpha^{-n}$ keeping in mind $\alpha\neq 0$. So we get
$$\sum_{k=0}^n a_k\alpha^{k-n}=\sum_{j=0}^n a_{n-j}\alpha^{-j}$$
thanks to change of index $j=n-k$. Whence $\alpha^{-1}$ is algebraic.
$deg(\alpha^{-1})\leq deg(\alpha)$ because the minimal polynomial of $\alpha^{-1}$ divides the polynomial above of degree $n$. 
Therefore $deg((\alpha^{-1})^{-1})\leq deg(\alpha^{-1})$ and we've concluded
