Proving that the inverse of an algebraic element is algebraic Let $E$ be an extension of a field $F$. Suppose $c \in E$ is algebraic over $F$ where $c \neq 0$. I want to prove that $c^{-1}$ is also algebraic over $F$. 
I feel like I'm missing something obvious here and that it shouldn't be very difficult to do. I know that $c^{-1} \in F(c)$, but that's about it. Do I actually need to construct a polynomial over $F$ with root $c^{-1}$ (presumably from the minimum polynomial of $c$)? Any hints would be appreciated.
 A: Note: If you just want algebraicity, please don't read this and see DonAntonio's answer.

If $c\neq 0$ is algebraic over $F$, then  $c^{-1}$ is algebraic over $F$. Moreover,  $F[c^{-1}]=F[c]=F(c)$ and $c^{-1}$ has the same degree as $c$.

Recall, that an element $d\in E$ is algebraic over $F$ if and only if $F[d]$ is a finite-dimensional vector space over $F$, whose dimension is the degree of $d$, i.e. the degree of its minimal polynomial. Indeed, it suffices to consider the evaluation $F[X]\longrightarrow F[d]$ and to mod out, in case $d$ is algebraic, by its kernel which is a principal ideal generated by the minimal polynomial.
Let $P(x)$ be the minimal polynomial of $c$. i.e. the monic polynomial of minimal degree such that $P(c)=0$. Then write $P(x)=x^n+\ldots+a_1x+a_0$. 
If $a_0=0$, we get
$$
c^n+\ldots+a_1c=c(c^{n-1}+\ldots+a_1)=0\quad\Rightarrow\quad c^{n-1}+\ldots+a_1=0.
$$
This contradicts the minimality of $P$. 
So $a_0\neq 0$ and we have
$$
a_0=-c^n-\ldots-a_1c\qquad\Rightarrow\qquad c^{-1}=-a_0^{-1}c^{n-1}-\ldots-a_0^{-1}a_1\in F[c].
$$
Therefore $F[c^{-1}]\subseteq F[c]$. So $F[c^{-1}]$ has finite dimension. It follows that $c^{-1}$ is algebraic over $F$. 
Applying the latter to $c^{-1}$ instead of $c$, we get $F[c]\subseteq F[c^{-1}]$,  whence the equality. 
The fact that $F[c]= F[c^{-1}]$ is equal to $F(c)$ and that it is therefore a field follows easily from similar considerations: every nonzero element $d=Q(c)$ in $F[c]$ has $F[d]\subseteq F[c]$ finite dimensional, so it is algebraic and its inverse belongs to $F[d]$, hence to $F[c]$. So $F(c)\subseteq F[c]\subseteq F(c)$.
A: Hints: for $\,c\neq 0\,$ (of course)
$$0=f(c)=\sum_{k=0}^na_kc^k=c^n\sum_{k=0}^n\frac{a_k}{c^n}c^k\ldots$$
