Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 15 / 2 / b / Solution

Consider the subalgebraIt is closed under multiplication because every nonconstant term of its elements is divisible by $x$. Evaluation at $x=0$ defines a homomorphism $A\to \mathbb{C}$ with kernel $I=x\mathbb{C}[x,y]$. Its square is $I^2=x^2\mathbb{C}[x,y]$, soas a complex vector space. In particular, the classes of $x{y}^j$, $j\ge 0$, are linearly independent and span it.

If $A$ were a finitely generated algebra, take a finite list of generators and subtract their scalar images under evaluation. The resulting generators $z_1,\dots ,z_r$ lie in $I$ and still generate $A$ over $\mathbb{C}$. Every element of $I$ is a polynomial in the $z_i$ with zero constant term. Modulo $I^2$, all terms of degree at least two vanish, leaving a linear combination of the classes of $z_1,\dots ,z_r$. This would make $I/I^2$ finite-dimensional, contradicting the displayed infinite basis. Thus this subalgebra is not finitely generated. This is the infinite cotangent space obstructs finite algebra generation argument; it proves failure of finite generation even when the proposed generators are arbitrary polynomials, rather than just monomials.