When is a face of a monoid contained in a minimal direct summand? 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!
 A: 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 $M.$
A: 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). 
