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.

  • $\begingroup$ 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. $\endgroup$
    – deaton.dg
    May 19 at 17:49
  • $\begingroup$ Which result did you want a reference to? I mentioned several. $\endgroup$ May 19 at 18:21
  • $\begingroup$ 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 $\endgroup$ May 19 at 18:28
  • $\begingroup$ Also you might look at 4.10 and 4.15 of arxiv.org/pdf/1508.05446.pdf#page72 $\endgroup$ May 19 at 19:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.