Suppose that $a$ is algebraic over a field $F$. Show that $a$ and $1+a^{-1}$ have the same degree over $F$.
Not really sure where to start for this one.
I know that I have to show that $[F(a):F]=[F(1+a^{-1}):F]$, and I assume that being told that $a$ is algebraic over $F$ gives me some info, but I'm not sure how to use this info.
$a$ algebraic over $F$ means that $a$ is a zero of some non-zero polynomial with coefficients in $F$.
So there exists some minimal polynomial of the form $p(x)=c_0+c_1 x+\ldots+ c_{n-1} x^{n-1}$ for $c_i\in F$ such that $p(a)=0$. We also know that $F(a)$ is isomorphic to $F[x]/\langle p(x)\rangle$.
Guidance how to approach this problem would be appreciated. Thank you.
It is obvious that $F(1+a^{-1})\subseteq F(a)$. And since $$\frac1{(1+a^{-1})-1}=a$$ we have that $F(a)\subseteq F(1+a^{-1})$.
Thus $F(a)=F(1+a^{-1})$ They have the same index because they are the same extension.