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

The multiplicative algebraic group ${\mathbb{G}}_m$ has coordinate ring $\mathbb{C}[t,t^{-1}]$. An algebraic morphism $f:{\mathbb{G}}_m\to {\mathbb{G}}_m$ therefore sends the invertible coordinate on the target to a unit of this Laurent polynomial ring.

Every unit of $\mathbb{C}[t,t^{-1}]$ has the form $c{t}^i$, with $c\in {\mathbb{C}}^{\ast }$ and $i\in \mathbb{Z}$. To prove this, suppose Laurent polynomials $P,Q$ satisfy $PQ=1$. Write their least and greatest active exponents as $a\le b$ and $c\le d$. The least and greatest exponents in their product are $a+c$ and $b+d$, since their endpoint coefficients are nonzero. Both must be zero. Consequentlyso each factor has only one active exponent. This proves the asserted unit description.

Thus $f(t)=c{t}^i$. A group homomorphism sends the identity to the identity, so $f(1)=1$ and $c=1$. Conversely, for every integer $i$, including negative integers and zero, $t\mapsto t^i$ is regular on ${\mathbb{G}}_m$ and satisfies $(st{)}^i=s^i{t}^i$. We obtainThese are exactly the characters of an algebraic torus in the one-dimensional case. The algebraic-morphism requirement is essential to the argument: the map must come from the Laurent coordinate ring.

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.