Algebraic function fields I am trying to understand what an algebraic function field is, so i was looking for some examples. 
The example on Wiki says: 
Given a polynomial ring $k[X,Y]$. Consider the ideal generated by the irreducible polynomial $I=Y^{2}-X^{3}$, by definition $L=Frac(k[X,Y]/(Y^{2}-X^{3}))$ is the field of fractions of the quotient field $k[X,Y]/(Y^{2}-X^{3})$. My question is how to show explicitely that the algebraic function field has a transcende basis of one element and what is the basis? In the example are given two algebraic function fields of different degree.
Here is the link to the article.
I am sorry if the question is too elementary. Thank you in advance! 
 A: a) The example in Wikipedia is rather uninteresting because their function field $L=\operatorname{Frac}(k[X,Y]/(Y^2-X^3))=k(x,y)$ is isomorphic to the  purely transcendental extension field $k(T)$ of $k$, the isomorphism  being $$f:k(T)\to L: T\mapsto \frac yx  $$ and the inverse isomorphism is characterized by $$f^{-1}(x)=T^2,\quad      f^{-1}(y)=T^3          $$
In other words  $L$ is purely transcendental with transcendency basis $\frac yx$ over $k$.   
b) About the simplest non-trivial example is the field $$K=\operatorname{Frac}(k[X,Y]/(Y^2-X^3+1))=k(x,y)$$  It is a degree $2$ extension $k(x)\subset k(x)(y)$ of the purely transcendental extension $k(x)$ of $k$ but $k(x,y)$ itself is not purely transcendental over $k$.
This is astonishingly difficult to prove with bare hands.  
c) But why use bare hands when you have a powerful tank for attacking function fields: algebraic geometry!
Indeed if you know its basics the assertion in b) becomes obvious since the curve $y^2=x^3-1$ (or rather its projective completion) has genus $g=1$, so that its function field $K$ cannot be rational.
A great reference for an initiation to these considerations is Miranda's Algebraic Curves and Riemann Surfaces
