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.
First, let us write $R:=\Bbbk[x,y]/\langle f\rangle$ and denote by $\pi:\Bbbk[x,y]\twoheadrightarrow R$ the canonical projection. Given an element $\alpha\in R$ with the property that $\{\,\alpha^k\mid k\in\mathbb N\,\}$ is linearly independent over $\Bbbk$, it follows from the definition that $\alpha\in\mathbb K$ is transcendental over $\Bbbk$.
Lemma. An element $\alpha=\pi(g)$ has this property if and only if $\deg(g)>0$ and $f$ does not divide any power of $g$.
Proof. For "$\Leftarrow$", note that the $g^k$ are linearly independent in $\Bbbk[x,y]$ since they have pairwise different degrees. If $f$ does not divide any of them, then their $\Bbbk$-span has trivial intersection with $\ker(\pi)$ and thus, the images $\alpha^k=\pi(g)^k=\pi(g^k)$ will be linearly independent. For "$\Rightarrow$", assume that the $\alpha^k$ are linearly independent. It follows immediately that $\alpha\notin\Bbbk$, so $\deg(g)>0$. If the $g^k$ were not linearly independent, then a nontrivial linear relation $0=\sum_{k=1}^\infty \lambda_k g^k$ with $\lambda_k\in\Bbbk$ would project via $\pi$ to a linear relation of the $\alpha^k$. Let $k\in\mathbb N$ be such that $\lambda_k\ne 0$. Since the $\alpha^k$ are linearly independent, we have $\pi(\lambda_k)=0$. This would imply that $\lambda_k\in\langle f\rangle$, a contradiction to the assumption that $f$ is nonconstant.
Now since $\{ x,y \}$ was a transcendence basis of $\mathbb L:=\Bbbk(x,y)$ over $\Bbbk$, it is clear that the transcendence degree of $\mathbb K$ could be at most one. To show that it is equal to one, we only need to find a $g\in\Bbbk[x,y]$ such that $f$ does not divide any power of $g$. But that is certainly easy to accomplish: If $f=x^n$, we take $g=y$. Otherwise, we take $g=x$. The statement follows because the polynomial ring is factorial.