# The lattice points of a f.g. rational cone form a f.g. monoid.

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:

Lemma: Let $R$ be a $\mathbb{Z}^n$-graded noetherian ring. Then $R$ is a finitely generated $R_0$-algebra.

Proof: For $n=0$ there is nothing to show. For $n>0$ consider the $\mathbb{Z}$-grading on $R$ w.r.t. the last coordinate. Then by Lemma 4.2 $R$ is finitely generated over its subring $R'$ of $0$-homogeneous elements. By the same lemma $R'$ is noetherian, and obviously $\mathbb{Z}^{n-1}$-graded. So by induction we conclude that $R'$ is finitely generated over $R'_0 = R_0$. Hence by transitivity of finite generatedness $R$ is a finitely generated $R_0$-algebra.

Corollary: Any noetherian submonoid of $\mathbb{Z}^n$ is finitely generated.

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 $R_0$.

• @user26857: This is precisely the theorem I was talking about. It is Theorem 4.11 in [Polytopes, Rings, and K-Theory] and is proven after Lemma 4.12. – Dune Aug 6 '14 at 7:17