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Definitions

By a commutative $\textit{semiring}$ (with 1 and without 0), I mean a triple $(S,+,\cdot)$ where $(S,\cdot)$ is a commutative monoid, $(S,+)$ is a commutative semigroup, and $\cdot$ distributes over $+$. If $(S,\cdot)$ is in fact a group, the triple $(S,+,\cdot)$ is called a $\textit{semifield}$.

Given two semirings $S_1, S_2$, we say that $S_1$ $\textit{embeds into}$ $S_2$ if there exists an injective semiring homomorphism $f: S_1 \to S_2$.

Examples:

i) the semiring $(\mathbb{Z}_{>0}, +, \cdot)$ embeds into the semifield $(\mathbb{Q}_{>0}, +, \cdot)$, via the obvious inclusion on sets $\mathbb{Z}_{>0} \subset \mathbb{Q}_{>0}$.

ii) the tropical semiring $(\mathbb{Z}_{\geq 0}, \max, +)$ embeds into the tropical semifield $(\mathbb{Z}, \max, +)$. Here $\max$ associates to a pair of integers $(a,b)$ its maximum $\max(a,b)$.

Question

Are there "interesting" examples of semirings that do not embed into any semifield?

Remarks

  1. If one replaces "semirings" and "semifield" by "monoids" and "group" respectively, I am aware that this is a well-known question (whose answer reduces to determining whether the monoid in question is cancellative). I am curious to know how the constraint of having a second binary operation with distributive properties with respect to the monoid structure affects the problem.

  2. One can consider the more familiar setting of semifields which embed into fields. For instance, the semifield $(\mathbb{Q}_{>0}, +, \cdot)$ embeds into the field of rational numbers. The tropical semifield $(\mathbb{Z}, \max, +)$, on the other hand, cannot embed into any field. This can be seen from the fact that $\max(a,a) = a$ for all $a \in \mathbb{Z}$ (i.e. every element is idempotent), whilst the only idempotent elements in a field are the additive identity element 0 and the multiplicative identity element 1.

Daniel Donnelly
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Antoine de Saint Germain
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    This definition doesn't seem quite right because, as stated, fields would not be semifields since $(F, \cdot)$ is not a group due to the zero element. – CJ Dowd Jul 05 '24 at 05:41
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    Indeed, with this definition of a semifield, fields are not semifields. The motivation comes from the theory of total positivity, where one crucially requires semifields not have a zero element. – Antoine de Saint Germain Jul 05 '24 at 08:06

2 Answers2

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Let $X$ be a infinite set and let $S$ be the collection of all cofinite subsets of $X$, i.e. those subsets $A \subseteq X$ such that $X \setminus A$ is finite. Make $S$ into a semiring by defining multiplication to be set intersection and addition to be set union. $S$ is a semiring under your definition because $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$, the multiplicative unit is $X$, and there is no additive identity because the empty set is not in $S$. Then this semiring cannot be embedded into a semifield because it lacks cancellation: if $A \subseteq B, C$, then $A \cap B = A = A \cap C$, but we need not have $B = C$.

CJ Dowd
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  • This is a fantastic example, thank you! – Antoine de Saint Germain Jul 05 '24 at 09:20
  • Well, you don't need cofinite subsets nor an infinite set. For each set $X$, $(\mathcal{P}(X), \cup, \cap)$ is a semiring. – J.-E. Pin Jul 23 '24 at 15:55
  • @J.-E.Pin Based on their comment about fields not being semifields, OP's definition requires that a semiring does not have a zero element (rather than merely not requiring the existence of a zero element). We need to exclude the empty set in this construction for that reason, and cofiniteness was the cleanest way I saw to do this. – CJ Dowd Jul 23 '24 at 20:56
  • Thanks a lot for your answer to my comment. You are perfectly right. – J.-E. Pin Jul 24 '24 at 15:45
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If your semiring multiplication is $\text{LCM}(x,y)$ essentially then taking the Grothendieck group of such a structure will result in trivial Grothendieck group, and since Grothendieck group is the "optimal" such group by certain UMP, then that answers your question.

One example I alone can come up with off the top of my head is:

$$ \Bbb{Z}[[d\mid \cdot ] : d \in \Bbb{N}] $$

I.e. the ring generated by the monoid of all Iverson brackets $[d \mid \cdot ]: d \in \Bbb{N}$ where $[d\mid x]:= 1 \text{ if } d \text{ divides } x \text{ else } 0$, are simply functions $\Bbb{Z} \to \{0,1\}$. It is known that a finite sum over such functions:

$$ 0= \sum_{d \in D}\lambda_d [d\mid x], \ \lambda_d \in K $$

achieve $\lambda_d = 0$ because of lineary independence amongst these "basis" functions. They in facto form a Schauder basis for all of $\Bbb{Z} \to \Bbb{Z}$.

However, when you multiply in this ring or in the monoid $M = \{ [d\mid \cdot]: d \in \Bbb{N}\}$, these divisorial functions multiply as $[d\mid \cdot][c \mid \cdot] = [\text{LCM}(c,d)\mid \cdot]$.

Since the Grothendieck group of that monoid is trivial, that means I think that the monoid of the entire ring $Z[M]$ is at least partially needing to be quotiented out before you can embed this into a field. Every ring is a semiring, and every field is a semifield. So I didn't quite answer your question, but is this close enough?

I just want to add that there is a "kind-of related" group phenomenon in which you consider $N = \{ [d \mid \cdot] : d \text{ is square-free}\}$. Then $[c\mid \cdot]\star [d\mid \cdot] := [\frac{\text{LCM}(c,d)}{\text{GCD}(c,d)}\mid \cdot]$ givens a boolean group.

Daniel Donnelly
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