Let $\mathscr S$ be the class of all semigroups with zero. For $(S,\times,0)\in\mathscr S,$ I want to count additive operations $+$ on $S$ such that $(S,+,\times,0)$ is a ring (possibly without unity). For $S\in\mathscr S,$ let $\Sigma_S$ be the set of all such additive operations on $S$. Let, for $+_1,+_2\in \Sigma_S,$ define $+_1\sim +_2$ to mean that $$(S,+_1,\times,0)\cong (S,+_2,\times,0).$$

Let $$\Sigma'_S:=\Sigma_S/\sim.$$

Let $$\kappa_S:=\operatorname{card}(\Sigma_S)$$ and $$\kappa'_S:=\operatorname{card}(\Sigma'_S).$$

Are there upper bounds to the values of $\kappa_S$ and $\kappa'_S$ for $S\in \mathscr S?$ Do they admit all non-negative integer values? If not, which non-negative integer values do they admit?

These questions are not similar to anything I know and I don't even know how to search for answers in literature. I'm especially curious what the behavior of $\kappa_S$ and $\kappa'_S$ is for finite semigroups $S.$

EDIT I have found this question on MO. Arturo Magidin gives an example there of two non-isomorphic rings having isomorphic multiplicative structures. If I understand correctly, this is proven by using the unique factorization property in polynomial rings over rings with the unique factorization property.

So my question isn't trivial, which I didn't know at the time of asking: there is a semigroup with zero for which there are at least two additive operations making the semigroup two non-isomorphic rings.

EDIT An answer to the following questions wouldn't be off-topic:

Is there a semigroup $S$ with $0$ such that $\kappa_S$ is infinite? Is there a semigroup $S$ with $0$ such that $\kappa'_S$ is infinite?

(If the answer is "yes", this doesn't answer any of my previous questions, but I would be interested in knowing it nonetheless.)

  • $\begingroup$ Your isomorphism for the multiplicative semigroup of $F[X]$ is incorrect; you want to take the nonzero elements of $F$, and then the direct sum, and then add $0$ to it; Otherwise, $(0,r)$ and $(0,s)$ (with $r,s\in\oplus\mathbb{N}$) will be distinct, but you don't want them to be distinct (they both should correspond to $0$). $\endgroup$ – Arturo Magidin Mar 10 '12 at 23:05
  • $\begingroup$ @ArturoMagidin You're right, thank you. I'll just delete part of what I wrote. You explained it better in your post. I can't think of a way to correct what I wrote without making the formula even more involved than it already is. $\endgroup$ – user23211 Mar 10 '12 at 23:12
  • $\begingroup$ Of course, those examples do much more: not only are they two nonisomorphic rings with isomorphic multiplicative monoid structure: they are nonisomorphic rings with isomorphic multiplicative monoid and additive group structures. $\endgroup$ – Arturo Magidin Mar 10 '12 at 23:42
  • $\begingroup$ @ArturoMagidin Yes, I understood that. I just extracted from your example what seemed relevant to my question. I must say that this example is very impressive and I'm certainly going to boast my knowledge of it in front of my friends on Monday. :-) $\endgroup$ – user23211 Mar 10 '12 at 23:45
  • $\begingroup$ It's possible that mathoverflow might be a better place for this question? $\endgroup$ – Tara B Mar 14 '12 at 0:31

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