MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $P$ be an integral, sharp, finitely generated commutative monoid (say even torsion-free and saturated if you like), and consider the "rational cone" $P_\mathbb{Q}\subseteq P^{gp}\otimes_\mathbb{Z} \mathbb{Q}$ generated by $P$ (i.e. the submonoid of elements of the form $\frac{m}{n} p$ where $m, n \in \mathbb{N}$ and $p\in P$).

Consider the monoid algebra $R=\mathbb{Z}[P_\mathbb{Q}]$. I'm interest in the answer to the


is $R$ coherent?

By coherent here I mean that every finitely generated ideal $I\subseteq R$ is finitely presented, i.e. if $I=\langle r_1,\dots, r_k \rangle$, then the kernel of the obvious map $R^k\to R$ with image $I$ is finitely generated.

Notice that in general (and I think almost always) $R$ will not be noetherian. For example if $P=\mathbb{N}$, then $R=\mathbb{Z}[\mathbb{Q}_+]=\mathbb{Z}[t^q | q>0]$ with the obvious relations $t^qt^{q'}=t^{q+q'}$ for $q,q' \in \mathbb{Q}_+$, and this is clearly not noetherian.

To elaborate a bit, consider $\frac{1}{n} P \subseteq P_\mathbb{Q}$ and the rings $A_n=\mathbb{Z}\left[\frac{1}{n} P\right]$, with in particular $A_1=\mathbb{Z}[P]$. There are obvious maps $f_{n,m}: A_n\to A_m$ when $n | m$, induced by the inclusions $\frac{1}{n} P\subseteq \frac{1}{m} P$.

$\{A_n,f_{n,m}\}$ form a directed system of rings, and the inductive limit is $R$.

One case in which I know that $R$ is coherent is when the transition maps $f_{n,m}$ are flat (this happens for example for $P=\mathbb{N}^r$).

Notice also that $R$ depends only on $P_\mathbb{Q}$, so for example if there is another submonoid $P'\subseteq P_\mathbb{Q}$ with the same properties as $P$ and such that $P'_\mathbb{Q}=P_\mathbb{Q}$, and the transition maps for $P'$ are flat, then $R$ is coherent.
This is what happens for the monoid $P=\langle (2,0), (1,1), (0,2) \rangle_\mathbb{N} \subseteq \mathbb{Z}^2$ (for which the transition maps are not flat), and in this case $P'=\langle (2,0),(0,2)\rangle\subseteq P$ does the trick.

An example in which something like this should not happen is $P=\mathbb{N}^4/(e_1+e_2=e_3+e_4)$, and so $A_1=\mathbb{Z}[x,y,z,w]/(xy-zw)$.

In general, a sufficient condition seems to be the following: for every $n$ and every ideal $I\subseteq A_n$ there exists an $m$ such that $n|m$ and $$ Tor_1^{A_m}(I_m,R)=\varinjlim Tor_1^{A_m}(I_m,A_k)=0 $$ where $I_m=IA_m\subseteq A_m$ is the extension of $I$.


is this true/plausible?

Thanks in advance for any comment!


All of this comes from root stacks of logarithmic schemes. If the answer to the first question is affirmative, I would be happy because then an "infinity root stack" will have coherent structure sheaf, coherent sheaves on it will be exactly those which are locally of finite presentation, and those have good "approximation" properties w.r.t. inverse limits.
[the case of the question is the "universal" one, in some sense]

share|cite|improve this question
what mean the terms " integral, sharp" for a (commutative) monoid? – Buschi Sergio Nov 30 '11 at 22:04
"integral" means that the map $P\to P^{gp}$ is injective, and "sharp" that the only invertible element of $P$ is the identity. While we're at it, by "torsion-free" I mean that $P^{gp}$ is torsion-free. – Mattia Talpo Dec 1 '11 at 0:46

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.