Suppose $S$ is an associative semiring whose underling commutative monoid is free (in particular, cancellative) and that its Grothendieck ring $G(S)$ is a unital ring. Can we conclude that $S$ must be unital, and if not, is there a nice counter-example?


Alternatively, are there additional suppositions we can put on $S$ which would allow us to conclude the unitality of $S$ from that of $G(S)$?


The context is that I am struggling to disprove that a particular ring I have constructed is unital, and I have proven it is of the form $G(S)$ for $S$ which is not unital. I am hoping this helps me somehow.


Cross-posted on [math.stackexchange](https://math.stackexchange.com/questions/4128541/if-the-grothendieck-ring-of-a-semiring-on-a-free-commutative-monoid-is-unital-i).



If the Grothendieck ring of a semiring on a free commutative monoid is unital, is the original semiring unital?
The answer is no. Let $S$ be a finite meet semilattice without maximum. For concreteness, take $S$ to be the proper subsets of $\{1,2\}$ under intersection. Let $\mathbb NS$ be the semigroup semiring of $S$. Then the underlying additive monoid is free on $S$. The Grothendieck ring is the semigroup ring $\mathbb ZS$ and this ring has an identity, but the identity cannot be expressed as a non-negative linear combination of elements of $S$. One writes it using inclusion-exclusion (or Mobius inversion).


For my particular $S$, the proper subsets of $\{1,2\}$, the identity element of $\mathbb ZS$ is $e=\{1\}+\{2\}-\emptyset$, which does not belong to $\mathbb NS$. Recalling that my semigroup multiplication is intersection, you can just check that $e$ multiplied by an element of $S$ returns that element.


For a more general meet semilattice $S$, the identity is $\sum\_{s\in S}\sum\_{t\leq s}t\mu(t,s)$ where $\mu$ is the Mobius function of $S$.


I think for general semirings there is no hope of a reasonable condition. For example if $S$ is a finite left regular band (that is a semigroup satisfying the identities $xyx=xy$ and $x^2=x$), I showed with Margolis and Saliola that $\mathbb ZS$ has an identity if and only if certain posets associated to $S$ have connected Hasse diagrams. But if $S$ is not a monoid, the identity will not belong to $\mathbb NS$. This is already a hard result. The meet semilattice case is just a commutative left regular band.


I also showed that for any finite inverse semigroup $S$, the semigroup ring $\mathbb ZS$ has an identity, but $\mathbb NS$ will not have an identity unless $S$ has an identity. This also generalizes the meet semilattice case.


**Update:** Here are some more observations. First, if $S$ is a semigroup, then the semigroup ring $\mathbb NS$ is unital if and only if $S$ is a monoid. That is because you have the surjective augmentation homomorphism $\mathbb NS\to\mathbb N$ summing up the coefficients and since the only nonzero idempotent of $\mathbb N$ is $1$, we must have that the support of an identity element has one element of $S$ with coefficient $1$ and so $S$ is a monoid.


Second, let $S$ be a meet semilattice (possibly infinite) viewed as a semigroup under meet. Then $\mathbb NS$ has an identity iff $S$ has a maximum. But $\mathbb ZS$ is isomorphic to the locally constant functions with compact support from $\widehat{S}\to \mathbb Z$ under pointwise operations. Here $\widehat{S}$ is the space of all non-zero semigroup homomorphisms $\phi\colon S\to \{0,1\}$ with the topology of pointwise convergence, where $\{0,1\}$ is a semigroup under usual multiplication. Thus $\mathbb ZS$ has an identity if and only if $\widehat{S}$ is compact. This is obviously a topological issue.


