# A structure which looks almost like a semi-ring.

Today I have encountered an interesting structure, similar to that of a ring or a semi-ring.

It is a structure $$(S, +, \cdot, 1)$$, where $$S$$ is a set, $$+, \cdot$$ are binary operations, and $$1\in S$$.

$$(S, \cdot, 1)$$ is a commutative monoid, $$(S, +)$$ is a commutative semigroup, and $$+$$ is distributive with respect to $$\cdot$$, i. e. $$a(b+c) = ab+ac$$ for any $$a, b, c\in S$$.

Does this structure have any names in the literature?

• Why are you especially interested in dropping the requirement that $(S,+)$ has a neutral element? As far as I can tell this is the only thing making the structure you describe differ from a semiring/rig. Apr 13, 2020 at 19:09
• @mrtaurho because $(S, +)$ didn't have a neutral element to begin with in what I am dealing with. Also, it's not exactly the only thing that makes it different from a semi-ring. Note that in a semi-ring there is a requirement that $0x = x0 = 0$. Apr 13, 2020 at 19:13
• That's exactly my question: why asking for this special case? Do you have a particular example in mind which sparked your interest in this structure. I think it's odd as I can't think of a "trivial" (as e.g. the trivial ring $\{0\}$) structure satisfying these axioms without invoking a $0$. The absorption axiom has to be forced as it can't be deduced in the usual way from the other axioms (we're missing the conceps of negatives). Also, of course, we don't require $0x=0=x0$ if there is no $0$ to begin with :) Apr 13, 2020 at 19:17
• I am puzzled by the last comment, I am probably missunderstanding what you mean: exponets of powers don't behave nicely when adding two natural numbers. Apr 13, 2020 at 19:23
• Again, unless I am missunderstanding what you mean, the exponents of powers when adding or multiplying two natural numbers do not have this structure. or do you mean the powers of primes when multiplying/exponentiating natural numbers? Also, primes do appear at the power 0, don't they? Apr 13, 2020 at 19:29

Consider such a set $$S$$. Define $$R= S \cup \{ 0_R \}$$ with the operations extended by $$0_R+x =x \\ 0_R\cdot x= 0_R$$
Then $$R$$ becomes a commutative semi-ring without zero divisors (i.e. $$xy=0$$ implies $$x=0$$ or $$y=0$$).
Converesely, let $$R$$ be any commutative semi-ring without zero divisors. Then $$S= R \backslash \{ 0 \}$$ satisfies your given conditions.