# 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