Prove that $a$ is algebraic over $F$ if and only if $a^{-1}$ is algebraic over $F$. Let $E$ be an extension of a field $F$ and let $a \in E$ be nonzero.
Prove that $a$ is algebraic over $F$ if and only if $a^{-1}$ is algebraic over $F$.
$\textbf{My Attempt:}$
prove of "if":
Assume that $a$ is algebraic over $F$.
Then, by definition there is a non-zero $f(x) \in F[x]$ such that $f(a) = c_n a^n + c_{n-1} a^{n-1} + \cdots + c_1 a + c_0 = 0$
where each $c_i \in F$ and for some $n \geq 0$.
Then, $c_n a^n + c_{n-1} a^{n-1} + \cdots + c_1 a + c_0 = 0 \implies c_n a^n + c_{n-1} a^{n-1} + \cdots + c_1 a = c_0$.
Then, $c_n a^{n-1} + c_{n-1} a^{n-2} + \cdots + c_1 = c_0 a^{-1} \implies a^{-1} = \frac{c_n}{c_0} a^{n-1} + \frac{c_{n-1}}{c_0} a^{n-2} + \cdots + \frac{c_1}{c_0}$.
Since, each $c_i \in F$. So does each $\frac{c_i}{c_0} \in F$.
Notice that $F(a) = \{ b_0 + \cdot + b_{n-1} a^{n-1} ~|~ b_0, ..., b_{n-1} \in F \}$.
So, $a^{-1} \in F(a)$. Which means $a^{-1}$ is algebraic over $F$.
prove of "only if":
Assume that $a^{-1}$ is algebraic over $F$.
Then, there is a non-zero $f(x) \in F[x]$ such that $f(a^{-1}) = 0$.
Which $a^{-1} = \frac{c_n}{c_0} a^{n-1} + \frac{c_{n-1}}{c_0} a^{n-2} + \cdots + \frac{c_1}{c_0}$, for each $\frac{c_i}{c_0} \in F$.
Then, $c_0 a^{-1} = c_n a^{n-1} + c_{n-1} a^{n-2} + \cdots + c_1 \implies c_n a^{n} + c_{n-1} a^{n-1} + \cdots + c_1 a + c_0 = 0$.
Where $F[a] = \{ c_n a^{n} + c_{n-1} a^{n-1} + \cdots + c_1 a + c_0 ~|~ c_n, ..., c_0 \in F \}$.
Which means $a \in F[a]$. So, $a$ is algebraic over $F$.
$\textbf{I don't think my proof is correct, but I don't know where to fix it!}$
 A: Well, the first thing to notice might be that the "if" and "only if" directions are the same: you prove one of them, and in that proof change all $a$ to $a^{-1}$, you will get a proof the other direction.
In your proof, you tried you divide by $c_0$. But $c_0$ can be zero, so that's why your proof is not correct. Nice try though!
Let us do your "only if" direction (which, in fact, is the "if" direction, but this doesn't matter.) Assume $a^{-1}$ is algebraic over $F$. Then, by definition there is a non-zero $f(x) \in F[x]$ such that $$f(a^{-1})=c_n{a}^{-n}+c_{n−1}a^{-n+1}+⋯+c_1a^{-1}+c_0=0$$
Just as you noted in your proof. Now multiply through by $a^n$, we get
$$c_n + c_{n-1}a + \cdots + c_1a^{n-1}+c_0a^n = 0$$
Now, we are still facing the problem that $c_0$ might be zero, but it doesn't matter now, since some $c_k$ is not zero: remember $f$ is a non-zero polynomial. Also it can't be the case that $c_n \neq 0$ but all others are $0$, because that would just mean $c_n = 0$. So, you disregard all the zero coefficients, and you are left with a non-zero polynomial relation in $a$ (although the degree might be lower), and you are done.
