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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)?


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There's a unique minimal submonoid containing $F$, but why do you think there should be one that splits $M$? – Kevin Carlson Aug 21 '12 at 19:50
I mean to say that $E$ is the minimal submonoid that satisfies (1) and (2). For example, $E = M$ works, but there might also be a smaller one. – Michael Kasa Aug 21 '12 at 20:15
Are these monoids commutative? I ask because you're using + to denote the monoid operation. – Harry Altman Aug 21 '12 at 22:21
Yes! Thanks for the clarification. – Michael Kasa Aug 21 '12 at 22:27
You should probably edit the clarifications into the original question. Also: I'm guessing that you actually want M to be the internal direct sum of E and E', rather than just being abstractly isomorphic to their direct sum? (These are not the same in general). – Harry Altman Aug 22 '12 at 0:27

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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So here you show that for $F = <(1,0)>$, $E=F$ does not satisfy property (2). Keep in mind that I do not require $E=F$, and I do not require that $E$ is a strict submonoid of $M$. Indeed, the entire monoid $E=M$ obviously satisfies (1) and (2), and I have yet to find a smaller submonoid that violates (3). (I suspect that the answer to my question in this case is $E=M$.) – Michael Kasa Aug 22 '12 at 15:29
@MichaelKasa, I think you're correct that $E=M$ in this case, and I edited the question to prove this. I suppose I didn't help you prove/disprove the question here, but merely gave an example where the decomposition must be trivial. However, given that $M$ always satisfies 1 and 2 makes me feel like there should be some sort of "descending argument", like taking the intersection over the family of submonoids satisfying 1&2, which we know is non-empty. – Andrew Aug 22 '12 at 17:11
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). – Michael Kasa Aug 24 '12 at 3:27
up vote 0 down vote accepted

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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