Prove that if $a$ is algebraic over $F$, then $F(a) = F(a^{-1})$.

Question

Let $E$ be an extension of a field $F$ and let $a \in E$ be nonzero.
Given that $a$ is algebraic over $F$ if and only if $a^{-1}$ is algebraic over $F$.
Prove that if $a$ is algebraic over $F$, then $F(a) = F(a^{-1})$.

$\textbf{My Attempt:}$
Assume that $a$ is algebraic over $F$.
Notice that $F(a) = \{ c_0 + \cdots + c_{d-1} a^{d-1} ~|~ \text{each $c_i \in F$}\}$
where $d = [F(a) : F] = $ degree of minimal polynomial of $a$ over $F$.
Since, given that $a$ is algebraic over $F$ if and only if $a^{-1}$ is algebraic over $F$.
Then, $a^{-1}$ is algebraic over $F$.
Then, by definition $a^{-1} \in F(a)$.
Then, $a^{-1} = \frac{c_n}{c_0} a^{n-1} + \frac{c_{n-1}}{c_0} a^{n-2} + \cdots + \frac{c_1}{c_0}$, for each $\frac{c_i}{c_0} \in F$.
Then, $c_0 a^{-1} = c_n a^{n-1} + c_{n-1} a^{n-2} + \cdots + c_1 \implies c_n a^{n} + c_{n-1} a^{n-1} + \cdots + c_1 a + c_0 = 0$.
So, $F(a^{-1}) = \{\frac{c_n}{c_0} a^{n-1} + \frac{c_{n-1}}{c_0} a^{n-2} + \cdots + \frac{c_1}{c_0} ~|~ \text{each $\frac{c_i}{c_0} \in F$}\}$.
So, $F(a) = F(a^{-1})$.

$\textbf{Don't think the prove is correct, are there ways to fix to make it look better ?}$

Answer 1

You could prove it this way:

$0\neq a\in F(a)\Rightarrow a^{-1}\in F(a)$, which means $F(a^{-1})\subseteq F(a)$. Now do the same argument for $a^{-1}\in F(a^{-1})$ to get $F(a)\subseteq F(a^{-1})$ and you're done.

By the way, this result is true for any non-zero element you'd care to adjoin to $F$.

Answer 2

Hint

Your proof is almost correct. I say almost because one missing argument is that if

$$p(x) = a_d x^d +\dots +a_1 x +a_0$$ is irreducible over $\mathbb F$ then

$$\overline{p}(x)= a_0 x^d + \dots +a_d$$ is also irreducible.

With that, you can prove that

$$[\mathbb F(a): \mathbb F] = [\mathbb F(a^{-1}): \mathbb F] .$$