In their book "Polytopes, Rings, and K-Theory" Bruns and Gubeladze sketch an alternative approach to Gordan's Lemma, which is stated in the headline (f.g. = finitely generated). I don't understand the final conclusion here.
The key tool in their reasoning is the following lemma:
Lemma 4.12. Let $R$ be a $\mathbb{Z}$-graded noetherian ring, $R_+ = \bigoplus_{k \geq 0} R_k$, and $R_- = \bigoplus_{k \leq 0} R_k$. Then
(a) $R_0$ is a noetherian ring, and each graded component $R_i$ is a finitely generated $R_0$-module;
(b) $R$, $R_+$ and $R_-$ are finitely generated $R_0$-algebras.
Now in order to prove Gordan's lemma the authors argue as follows:
Consider an affine monoid¹ $M$ and a rational hyperplane $H$ through the origin. Choosing a linear form defining $H$, we define a $\mathbb{Z}$-grading on $M$, and obtain that $M \cap H^+$ is an affine monoid. By induction on the number of support hyperplanes of a rational cone $C$ it follows that $M \cap C$ is affine.
If I am understanding right the authors want to use the fact that a monoid $M$ is finitely generated if and only if $k[M]$ is a finitely generated $k$-algebra (for any fixed field $k$). It is clear that $k[M \cap H^+]$ is equivalent to $R_+$ for the $k$-algebra $R = k[M]$ graded with respect to the linear form defining $H$. But now the above lemma only guarantees that $k[M \cap H^+]$ is finitely generated as $R_0$-algebra, and we don't have $k = R_0$ in general (actually $R_0 = k[M \cap H]$). How can we conclude that $k[M \cap H^+]$ is a finitely generated $k$-algebra?
¹: For convenience: An affine monoid is a finitely generated submonoid of $\mathbb{Z}^n$ for some $n$, and it is considered to be embedded in $\mathbb{R}^n$.
In short: I think the whole problem can be reduced to the following questions: If $M \subset \mathbb{Z}^n$ is a finitely generated submonoid and $H \subseteq \mathbb{R}^n$ is a rational hyperplane through 0, then why is the intersection $M \cap H$ also finitely generated? How do we get generators of $M \cap H$ from generators of $M$ (algorithmically)?
Since Bruns and Gubeladze did not mention this questions at all, there is either an obvious solution which I miss, or the authors completely overlooked this problem which seems less likely to me.
A related problem: Actually, Lemma 4.12. is just a special case of the consecutive theorem, where the noetherian ring $R$ is graded by a finitely generated abelian group $G$ (instead of $\mathbb{Z}$), and $R_+$ is replaced by the direct sum of the homogeneous components $A = \bigoplus_{m \in M} R_m$ belonging to some finitely generated submonoid $M \leq G$. The statement of the theorem is analogous. In particular, $A$ is claimed to be finitely generated as an algebra over $R_0$.
In the proof the authors begin with the special case, where $G = \mathbb{Z}^n$ and $M$ consists of all lattice points of some rational cone. Here the whole reasoning is:
In this case M is cut out by finitely many halfspaces and the claim follows by an iterated application of Lemma 4.12.
It is the same argument as above, I am missing here: In the end we want to show that $A$ is finitely generated over $R_0$. But this "final algebra" $R_0$ with respect to the $G$-grading may be something completely different from the "intermediate algebras" $R_0$ with respect to some $\mathbb{Z}$-gradings to which we want to apply Lemma 4.12. I have the feeling that we are loosing generators of $A$ in this way.
I asked Prof. Bruns now. He agreed with me that those proofs are lacking an argument. Fortunately, they can be easily repaired by the following lemma:
Of course by Hilbert's Basis Theorem the converse is also true. In fact the statement of the corollary even holds for all commutative cancellative noetherian monoids [Gilmer, Commutative Semigroup Rings].
Now we can fix both proofs. In the proof of Gordan's Lemma we still have to show that $k[M \cap H]$ is finitely generated over $k$. This follows immediately from the corollary since we already know that $k[M \cap H]$ is noetherian, so $M \cap H$ is finitely generated. In the proof of the consecutive theorem we start with a $\mathbb{Z}^n$-graded ring $R$ and successively remove the negative component, or restrict to the 0-homogeneous component w.r.t. some $\mathbb{Z}$-grading. In each step the resulting subring is noetherian and still $\mathbb{Z}^n$-graded (although some homogeneous components may be zero), so by the lemma all those subrings are finitely generated over $R_0$.