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If $\mathcal{C}$ is a category with finite coproducts, we may associate to it a commutative monoid $\mathcal{C}/\cong$ of isomorphism-classes of objects, with addition induced by the coproduct and zero induced by the initial object. It has the property that $a+b=0 \Rightarrow a=b=0$.

Question. Conversely, let $M$ be a commutative monoid with the property that $a+b=0 \Rightarrow a=b=0$ for $a,b \in M$ (one also says that $M$ is reduced). Is there a category with finite coproducts $\mathcal{C}$ such that $\mathcal{C}/\cong$ is isomorphic to $M$?

Let us call $M$ realizable if $M$ is of the desired form. Then:

  • $\mathbb{N}$ is realized by the category of finite sets.
  • $\mathbb{N}/(1=2)$ is realized by the preorder $\{0 < 1 \}$.
  • $\mathbb{N}/(1=3)$ is realizable, see SE/834653.
  • If $M,N$ are realizable, then $M \times N$ is also realizable.
  • If $S$ is a set, then the free commutative monoid $\bigoplus_{s \in S} \mathbb{N}$ on $S$ is realizable. In fact, there is some (commutative) ring $R$ such that $S$ is isomorphic to the set of isomorphism-classes of simple $R$-modules. Now consider the category of those $R$-modules which are finite direct sums of simple $R$-modules.
  • If $M$ is a submonoid of a realizable commutative monoid $N$, then $M$ is also realizable: If $N$ is $\mathcal{C}/\cong$, consider the full subcategory of $\mathcal{C}$ of those objects whose isomorphism-class belongs to $M$.
  • A Theorem by Grillet states that every finitely generated cancellative torsion free reduced commutative monoid embeds into some $\mathbb{N}^n$. Hence, these are realizable.
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I think every idempotent commutative monoid is realizable: these are essentially the same as join-semilattices. –  Qiaochu Yuan Jun 15 '14 at 17:04
@Qiaochu: That's right. One defines $a \leq b$ iff $\exists c (a+c=b)$ iff $a+b=b$. This gives a partial order with $\mathrm{sup}(a,b)=a+b$. –  Martin Brandenburg Jun 15 '14 at 17:35
+1 nice question. The world needs more questions like this. –  goblin Nov 26 '14 at 23:19

1 Answer 1

up vote 4 down vote accepted

In George Bergman's paper "Coproducts and some universal ring constructions" (Trans. A.M.S. 200 (1974), 33-88) he proves in Theorems 6.2 and 6.4 that if $A$ is a commutative monoid satisfying both

(i) For all $a,b\in A$, $a+b=0\Rightarrow a=b=0$, and

(ii) There is an element $1\in A$ such that for every $a\in A$ there are $b\in A$ and $n\in\mathbb{N}$ such that $a+b=n.1$,

then there is a ring $R$ such that $A$ is isomorphic to the monoid of finitely generated projective right $R$-modules under direct sum. In fact, $R$ can be chosen to be left and right hereditary (in the paper referred to this is only proved for $A$ finitely generated, with "hereditary" weakened to "semihereditary" for arbitrary $A$, but "semihereditary" was strengthened to "hereditary" in a later paper of Bergman and Dicks).

Condition (ii) is necessary for the monoid of finitely generated projectives for a ring, since choosing $1$ to be the free module of rank one, this condition is satisfied. However, any commutative monoid $A$ satisfying (i) embeds in one satisfying both (i) and (ii), for example by adjoining an element $1$ such that $1+a=1$ for all $a\in A\cup\{1\}$. Hence any commutative monoid satisfying (i) is the monoid associated to some additive category.

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Thank you! If we choose $1$ in the way you described, the ring $R$ even has the property $R \cong R^2$. –  Martin Brandenburg Jun 16 '14 at 8:20
@MartinBrandenburg: Yes. I haven't really read the paper, but I suspect the proof uses essentially the method I sketch-suggested in my answer to your previous question (at least, the keywords "universal localization" are there) and there might well be a very straightforward "rings with several objects" generalization that gets rid of condition (ii) without needing the "adjoin $1$" trick. I'm still wondering if there's a simpler construction using non-additive categories. –  Jeremy Rickard Jun 16 '14 at 12:42

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