Question

On page 13 of the book "An introduction to invariants and moduli" of Mukai http://catdir.loc.gov/catdir/samples/cam033/2002023422.pdf there is a mistake, in the end of the proof of Proposition 1.9. It seems to me that this proof can not be fixed, without using the notion of Noetherian rings and Hilbert basis theorem.

The question is: Can this proof be fixed, without using commutative algebra -- i.e., by the elementary reasoning that Mukai is using there?

I reproduce here the proof from the book for completeness. $S$ is the ring of polynomials, $G$ a group, $S^G$ is the ring of invariants

Proposition. If $S^G$ is generated by homogeneous polynomials $f_1,...,f_r$ of degrees $d_1,...,d_r$, then the Hilbert series of $S^G$ is the power series expansion at $t=0$ of a rational function $$P(t)=\frac{F(t)}{(1-t^{d_1})...(1-t^{d_r})}$$ for some $F(t)\in \mathbb Z[t]$.

Proof. We use induction on $r$, observing that when $r=1$, the ring $S^G$ is just $\mathbb C[f_1]$ with the Hilbert series $$P(t)=1+t^{d_1}+t^{2d_1}+...=\frac{1}{1-t^{d_1}}.$$ For $r>1$ consider the injective complex linear map $S^G\to S^G$ defined by $h\to f_rh$. Denote the image by $R\subset S^G$ and consider the Hilbert series for the graded rings $R$ and $S^G/R$. Since $R$ and $S^G/R$ are generated by homogeneous elements, we have $$P_{S^G}(t)=P_{R}(t)+P_{S^G/R}(t).$$ On the other hand, $dim(S^G\cap S_d)=dim(R\cap S_{d+d_r})$, so that $P_R(t)=t^{d_r}P_{S^G}(t)$, and hence $$P_{S^G}(t)=\frac{P_{S^G/R}(t)}{1-t^{d_r}}.$$ But $S^G/R$ is isomorphic to the subring of $S$ generated by the polynomials $f_1,...,f_{r-1}$, and hence by the induction hypothesis $P_{S^G/R}(t)=F(t)/(1-t^{d_1})...(1-t^{d_{r-1}})$ for some $F(t)\in \mathbb Z[t]$...

Mistake: It is not true that $S^G/R$ is isomorphic to the subring of $S$ generated by polynomials $f_1,...,f_{r-1}$. For example consider $\mathbb C^2$ with action $(x,y)\to (-x,-y)$. Then let $f_1=x^2$, $f_2=y^2$, $f_3=xy$.

Motiviation of this question. Of course this proposition is a partial case of Hilbert-Serre theorem, proven for example at the end of Atiyah-Macdonald. But the point of the introduction in the above book is that one does not use any result of commutative algebra.

Answer 1

This is not an answer to the question, I just decided to give for completeness a standard proof of the above statement that uses (a version of) Hilbert-Serre theorem. In this proof we need to use Hilbert basis theorem. In the above statement $S^G$ is clearly a finitely generated graded module over the ring of polynomials $\mathbb C[x_1,...,x_r]$, so it is sufficient to prove:

Theorem (Hilbert, Serre). Suppose that $S=\sum ^{\infty}_{j=0}S_j$ is a commutative graded ring with $A_0=\mathbb C$, finitely generated over $\mathbb C$ by homogeneous elements $x_1,...,x_r$ in positive degrees $d_1,...,d_r$. Suppose that $M=\sum_{j=0}^{\infty} M_j$ is a finitely generated graded $S$-module (i.e., we have $S_iS_j\subset S_{i+j}$ and $S_iM_j\subset M_{i+j}$). Then the Hilbert series $P(M,t)$ is of the form $$\sum_{j=0}^{\infty}dim(M_j)t^j=P(M,t)=\frac{F(t)}{\Pi_{j=1}^r(1-t^{d_j})}, \;\; F(t)\in \mathbb Z[t].$$

Proof. We work by induction on $r$. If $r=0$ then $P(M,t)$ is a polynomial with integer coefficients, so suppose $r>0$. Denote by $M'$ and $M''$ the kernel and cokernel of the multiplication by $x_r$, we have an exact sequence for each $j$ $$0\to M'_j\to M_j \to^{x_r}M_{j+d_r}\to M''_{j+d_r}\to 0.$$ Now $M'$ and $M''$ are finitely generated graded modules for $K[x_1,...,x_{r-1}]$, and so by induction their Hilbert series have the given form. From the above exact sequence we have $$t^{d_r}P(M',t)-t^{d_r}P(M,t)+P(M,t)-P(M'',t)=0.$$ Thus $$P(M,t)=\frac{P(M'',t)-t^{d_r}P(M',t)}{1-t^{d_r}}$$ has the given form.

Where did we use Hilbert basis theorem? We use it when we say that $M'$ is finitely generated.