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