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We consider a semimodule over commutative semiring. It's a well-known fact that the category of $R-$modules (over a ring) is an abelian category. Does this result generalize to the category of $R-$semimodules (when $R$ is a semiring, but not necessary a ring, i.e. a tropical semiring $\mathbb{T}$)?

I have been interested in tropical semirings and matrices over this semiring and I asked myself the question above. I have to admit that I don't have any idea. I would be grateful for an answer and some bibliography to further study this subject.

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No, it is not abelian. For example, semimodules over the semiring of natural numbers are just commutative monoids, and the category of commutative monoids is not abelian. One way to see why is to note that it is not balanced, since the inclusion $\mathbb{N}\to \mathbb{Z}$ is a monomorphism and an epimorphism of commutative monoids, but not an isomorphism. Alternatively, one can note that the addition map $\mathbb{N}\times \mathbb{N}\to \mathbb{N}$ has zero kernel but is not a monomorphism.

In fact I believe the category $\mathbf{SMod}_T$ of semimodules over a semiring $T$ is abelian if and only if $T$ is a ring (and then $\mathbf{SMod}_T=\mathbf{Mod}_T$). Indeed, every category of semimodules is enriched over commutative monoids (i.e. every $\operatorname{Hom}$-set has a natural structure of commutative monoid). Moreover this enrichment is unique, thanks to the Eckmann-Hilton argument and the fact that it is determined by the canonical isomorphism $X+Y\to X\times Y$ (see this blog post for an explanation). But abelian categories are enriched over abelian groups again in a unique way; so if $\mathbf{SMod}_T$ is abelian, the two enrichment must coincide, so the $\operatorname{Hom}$-monoids must all be groups. In particular, $\operatorname{Hom}_{\mathbf{SMod}_T}(T,T)$ must be a group. But it's not too hard to show (as in the case of rings) that this is actually isomorphic to the additive monoid of $T$, so $T$ is in fact a ring.

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