Integral extensions with finitely generated k-algebras

Question

I have $k$ a field, and I am assuming that the finitely generated $k$-algebra $K = k[x_1,x_2]$ is also a field. I am trying to prove Zariski's lemma in this case, by seeing first that $K$ is an integral extension of the ring $k[x_1,x_1^{-1}]$, but after many attempts I couldn't find any result. Any ideas?

Answer 1

$k[x_1,x_2]$ is the homomorphic image of $k[s,t]$ where $s$ and $t$ are indeterminates, with $s\mapsto x_1$ and $t\mapsto x_2$. A polynomial in the kernel of this homomorphism with some nonzero power of $t$ considered as a polynomial in $t$ is not necessarily monic, but it can be made into a monic polynomial by multiplying by some $k$-multiple of a negative power of $s$. Thus $x_2$ is integral over $k[x_1,x_1^{-1}]$.

To expand on Krish's helpful comment, $x_1$ and $x_2$ can't be algebraically independent over $k$ because the algebra is a field, implying that such a polynomial exists; to construct one somewhat explicitly, note that since $x_2$ has a multiplicative inverse there is a polynomial $p(s,t)$ such that $$x_2p(x_1,x_2)-1=0$$ so $tp(s,t)-1$ is in the kernel.