# If the Grothendieck ring of a semiring on a free commutative monoid is unital, is the original semiring unital?

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.

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.

• This is incredibly enlightening, thank you! Do you have a reference to your result about when $\mathbb{Z}S$ has an identity? I don’t quite have a semigroup semiring, but this sounds very interesting to me. I quickly looked but didn’t find it. May 19 at 17:49
• Which result did you want a reference to? I mentioned several. May 19 at 18:21
• The case of a finite meet semilattice can be extracted from Theorem 1 of Solomon's paper sciencedirect.com/science/article/pii/S0021980067800644 and its proof May 19 at 18:28
• Also you might look at 4.10 and 4.15 of arxiv.org/pdf/1508.05446.pdf#page72 May 19 at 19:03