How to show a given algebra is not generated by one element Suppose I have a $k$-algebra $k[x,y]/\langle f\rangle$, where $f$ is a (given, fixed) irreducible polynomial. What are the strategies for showing that this isn't generated by one element as a $k$-algebra?
 A: If your algebra $A=k[X,Y]/(f)$ is monogenic, i.e. generated by a single element, the class of a  polynomial  $P(X,Y)$, then $A$  is  of the form $A=k[P(x,y)]$ and since $P(x,y)$ is transcendental over $k$ this means that $A$ is isomorphic to a polynomial ring: $A\simeq k[T]$.  
This implies that the curve $C=V(f)\subset \mathbb A^2(k)$ or its projective closure $\bar C=V(f_{hom})\subset \mathbb P^2(k)$ is rational ( $ f_h(X,Y,Z$ being the homogeneization of $f(X,Y)$), which is relatively easy to check.
For example, if $char(k)=0$ and $f(X,Y)=X^n+Y^n-1=0$, the algebra $A$ is generated by one element if and only if $n=1$  .
Indeed the curve $\bar C$ given by $X^n+Y^n-Z^n=0$ is smooth of genus $g=\frac{(n-1)(n-2)}{2}$ and is thus not rational for $n\geq 3$ since  then  $g\geqq 1$ .
For  $n=2$ we can reduce to the rational curve $X'Y'-1=0$ by an obvious  change of  variables, but:
Warning
 Even if $C$ is rational, the algebra $A$ need  not be monogenic: think of $f=XY-1$  for which  $A=k[X,X^{-1}]$ is not monogenic.
Indeed if it were, it would be generated by a transcendental element $T$ (as mentioned above) and would thus be isomorphic to $k[T]$.
 But the algebras  $k[T]$ and $A=k[X,X^{-1}]$    are not isomorphic: the only invertible elements of $k[T]$ are the  non-zero constants, $(k[T])^*=k^*$, whereas $(k[X,X^{-1}])^*=\bigcup _{m\in \mathbb Z}k^*X^m$, a much bigger group. 
