## The function field of a plane curve

My colleague and me were confronted with a question concerning the introductory chapter of a text on algebraic geometry. The scenario of a plane curve $C=Z(f)\subseteq \mathbb A_{\Bbbk}^2$ over a field $\Bbbk$ is considered, for a nonconstant polynomial $f\in\Bbbk[x,y]$. It was stated that the function field $\mathbb K:=\mathrm{Quot}(\Bbbk[x,y]/\langle f\rangle)$ is obviously of transcendence degree $1$ over $\Bbbk$ because $f(x,y)=0$ is a new relation between $x$ and $y$. The prerequisites to this text are basic undergraduate knowledge of algebra and topology. The question was about the obviousness of the above statement. We came up with a proof that only really requires some linear algebra. I found it rather cute. Do you want to see it?