Let $A$ be a subring of ring $B$, with $B$ integral over $A$. If $x\in A$ is a unit in $B$, then it is a unit in $A$.
I know that $f(t) = t - x$ is in $A[t]$ with $f(x) = 0$, and that there exists some monic polynomial $g(t) = t^n + a_{n-1}t^{n-1}... + a_0$ also in $A[t]$ so that $g(x^{-1}) = 0$, and as I understand it, this proof comes down to showing that $x^{-1}$ must be an element of A. But I'm not sure what kind of manipulation of these polynomials will do that for me. I tried examining $x^n g(t)$ and saw that $0 = x^ng(x^{-1}) = 1 + xa_{n-1} + ... + x^na_0$, but I can't seem to get anything useful out of this. Any hints toward the right direction would be appreciated.
Hint: If
$$x^{-n}+a_{n-1}x^{-n+1}+\cdots+a_1 x^{-1}+a_0=0$$
Then
$$x^{-1}+a_{n-1}+\cdots+a_1x^{n-2}+a_0x^{n-1}=0$$