I am trying to prove Zariski's lemma in the following case:
let $k$ be a field and $k[x]$ be an algebra generated by $x$, such that $k[x]$ is a field. Then the extension $[k(x):k]$ is finite.
I am not sure what to do here at all. Suppose I want to find a basis for $k(x)$ as a $k$-vector space, I know that every $f\in k[x]$ can be written as $$ f=a_n x^n+...+a_1 x+a_0 $$ for $a_i\in k$ and some $n$. But how do I get a basis?
Since $x$ is invertible, there is an element $p(x) \in k[x]$ such that $xp(x) = 1$. Then $x$ is a root of $q(X) = Xp(X)-1$, and is therefore algebraic. (We could also argue that if $x$ is not algebraic, $k[x]$ is not a field.)
If the minimal polynomial of $x$ has degree $d$, then the monomials $1,x,\ldots , x^{d-1}$ are a $k$-basis for $k[x]=k(x)$.