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My question is motivated from studying logarithmic algebraic geometry. (For a detailed introduction to the subject, see this book by Arthur Ogus.) However, I believe my question is accessible to anyone who knows basic algebra.

An integral monoid is a finitely generated commutative monoid $M$ such that the cancellative rule holds: if $a + b = a + c$, then $b = c$. A face $F$ of a monoid $M$ is a submonoid such that if $a+b \in F$, then both $a \in F$ and $b \in F$.

My question is the following. Suppose $F$ is a face of an integral monoid $M$. Is there a unique submonoid $E \subseteq M$ such that:

  1. $F \subseteq E$,
  2. there exists some $E' \subseteq M$ such that $M$ is the (internal) direct sum of $E$ and $E'$, and
  3. if $E''$ is some other submonoid satisfying (1) and (2), then $E \subseteq E''$ (that is, $E$ is minimal with respect to inclusion)?

Thanks!

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    If I were looking for counterexamples, I'd start with finite local rings. Their unique maximal right ideal is certainly a face of the integral monoid (the whole ring), and it might resist complementation. There could be a reason to rule this out, but I haven't seen it yet.2012-08-22

2 Answers 2

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The answer is "yes" in the special case of sharp monoids; that is, monoids where the only invertible element is 0. For a sharp integral monoid $P$, there is a finite collection of indecomposible elements, where an element $x\in P$ is indecomposible if whenever $x = y+z$, then either $y=0$ or $z=0$.

In this case, a face $F \subseteq P$ determines a subset of indecomposible elements (namely, the indecomposible elements contained in $F$). Thus, it is clear that $P$ has only finitely many faces.

Since any submoniod $E$ satisfying (2) in a sharp monoid is a face, it suffices to show the following: if we have two submonoid $E_1, E_2$ satisfying (1) and (2), then their intersection $E_1 \cap E_2$ also satisfies (1) and (2).

Write the indecomposible elements of $P$ as $f_1,\ldots,f_n,e_1,\ldots,e_m$. Let $f_1,\ldots, f_n, e_1,\ldots, e_\ell$ be the indecomposible elements associated to $F_1$, and $f_1,\ldots f_n,e_p,\ldots, e_m$ be the indecomposible elements associated to $F_2$, where $1 \leq \ell < p \leq m$. Let $F_0 = F_1 \cap F_2$, and let $E_0$ be the submonoid generated by $e_1, \ldots, e_m$. Clearly $F_0 + E_0 = P$.

It suffices to show that if $ \sum_{i=1}^n n_i f_i + \sum_{i=1}^m m_i e_i = \sum_{i=1}^n n_i' f_i + \sum_{i=1}^m m_i' e_i$ then both $ \sum_{i=1}^n n_i f_i = \sum_{i=1}^n n_i' f_i $ and $ \sum_{i=1}^m m_i e_i = \sum_{i=1}^m m_i' e_i. $ By the cancellative rule, one implies the other.

Because $E_1 \oplus E_1' \cong P$, we can conclude that $ \sum_{i=1}^n n_i f_i + \sum_{i=1}^\ell m_i e_i = \sum_{i=1}^n n_i' f_i + \sum_{i=1}^\ell m_i' e_i. $ Because $E_2 \oplus E_2' \cong P$, we can conclude that $ \sum_{i=1}^p m_i e_i = \sum_{i=1}^p m_i' e_i. $ If we add $\sum_{i=\ell+1}^p (m_i + m_i') e_i$ to the top equation, we have $ \sum_{i=1}^n n_i f_i + \sum_{i=1}^p m_i e_i + \sum_{i=\ell+1}^p m_i' e_i = \sum_{i=1}^n n_i' f_i + \sum_{i=1}^p m_i' e_i + \sum_{i=\ell+1}^p m_i e_i. $ By the second equation and the cancellative law, we see that $ \sum_{i=1}^n n_i f_i + \sum_{i=\ell+1}^p m_i' e_i = \sum_{i=1}^n n_i' f_i + \sum_{i=\ell+1}^p m_i e_i . $ Using either direct sum decomposition, we now can conclude that $ \sum_{i=1}^n n_i f_i = \sum_{i=1}^n n_i' f_i, $ which is what we had to show.

Hence, the submonoid satisfying (1), (2), and (3) above is the intersection of all the monoids satisfying (1) and (2).

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Consider the monoid $M\subseteq\Bbb Z^2$ generated by $(1,0),(1,1),(1,2).$ The submonoid $E=\langle (1,0)\rangle$ is a face of $M.$ Any $E'$ such that $E+E'=M$ must contain $(1,1)$ and $(1,2),$ so we must have $E'\supseteq\langle(1,1),(1,2)\rangle.$ Note that $(2,2)=2(1,1)=(1,0)+(1,2)$ is not uniquely expressed as a sum $f\oplus g$ for $f\in E,g\in E',$ thus, we cannot have $M=E\oplus E'$ in this case.


Edit: Following the comment. Suppose $E\neq F$. If $E$ contains $(1,1)$, then $E'$ must contain $(1,2)$ (otherwise $E=M$), and we have the same problem. If $E$ does not contain $(1,1)$, then $E'$ must. Further, if $E'$ contains $(1,2)$ then we are in the previous situation. So we are left with $(1,0),(1,2)\in E$ and $(1,1)\in E'.$ But again, $2(1,1)$ and $(1,0)+(1,2)$ are distinct sums giving $(2,2).$ So in this case we find that the only submonoid possible with the desired properties is $M.$

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    I agree with your remarks about a "descending argument," except that heretofore I have been unable to show that (2) holds for the intersection (I suppose this is really the only tricky part).2012-08-24