5
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $R_0$.

$\endgroup$
  • $\begingroup$ @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. $\endgroup$ – Dune Aug 6 '14 at 7:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.