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?
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It would be useful if you explained what sort of example youhave in mind. It is very hard to write an answer that is useful to you if we don't know what you want (and know) You asked for "the startegies", and at that level of generality there is a lot of things that are not going to be useful for you... – Mariano Suárez-Álvarez Feb 08 '12 at 21:59
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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.
Georges Elencwajg
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I appreciate your response, but my question is motivated by a homework question which does not assume the knowledge required to understand your answer (i.e. I don't know what you're talking about yet). I was hoping for a more algebraic answer, but perhaps there isn't one; maybe it's too case-specific. – Matt Feb 08 '12 at 20:16
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Ah, I'm sorry to hear that, @Matt. I hope this answer may be of use to you in the future or to some users now. I would guess that there are not many examples of monogenic algebras besides those corresponding to $f=Y-\phi(X)$ or $f=X-\psi (Y)$. – Georges Elencwajg Feb 08 '12 at 20:27
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@Matt: I have added an explanation for the non-isomorphism of the algebras $k[T]$ and $k[X,X^{-1}]$ at the end of my answer. – Georges Elencwajg Feb 08 '12 at 21:52