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.

Attiyah-McDonaldsis a very funny typo. $\endgroup$ – Rasmus Mar 22 '11 at 22:11Maconaldis less funny, but still a typo! $\endgroup$ – Hailong Dao Mar 23 '11 at 0:22