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Let $(M,\times)$ be a monoid with zero. Let $\Sigma(M,\times)$ be the set of binary operations $+$ on $M$ such that $(M,+,\times)$ is a ring. Let $\sim$ be an equivalence relation on $\Sigma(M,\times)$ defined by $$+_1\sim+_2\iff(M,+_1,\times)\cong(M,+_2,\times).$$

Let's denote the quotient set $\Sigma(M,\times)/\sim$ by $\Sigma'(M,\times)$ and consider the number $$\mathrm{add}(M,\times)=|\Sigma'(M,\times)|.$$

I have several questions about the behavior of this function, none of which I know how to approach.

$(1)$ For an integer $n\geq 0$, is there always a monoid $(M,\times)$ such that $n=\mathrm{add}(M,\times)?$

$(2)$ The same question with the requirement that $M$ be finite.

$(3)$ The two previous questions are equivalent if $\mathrm{add}(M,\times)\geq\aleph_0$ for $M$ infinite. Is it true? It is false by Todd Trimble's answer. $(\mathbb Z,+)$ with a zero element adjoined is another example.

$(4)$ Is there an upper bound to the values of $\mathrm{add}(M,\times)$ over all finite monoids with $0?$ What about all monoids with zero?

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Somewhat tangentially, but I'm curious: for $M$ countably infinite, $\Sigma(M, \times)$ can be thought of as a Borel subset of $2^\omega$, so if uncountable it has cardinality $2^{\aleph_0}$. Is it obvious that $\Sigma'$ behaves similarly? I.e., if $M$ is countable and $\Sigma'(M, \times)$ is uncountable, must $\vert\Sigma'(M, \times\vert=2^{\aleph_0}$? – Noah Schweber May 1 '13 at 1:38
up vote 11 down vote accepted

I think the answer to (2) is yes (which also answers (1) and (4), of course).

First, note that $R=\mathbb{Z}/4\mathbb{Z}$ and $S=\mathbb{F}_2[x]/(x^2)$ have isomorphic multiplicative monoids (via the map $0\mapsto 0$, $1\mapsto 1$, $2\mapsto x$, $3\mapsto 1+x$).

Call this monoid $M$.

Then the direct product $M^n$ of $n$ copies of $M$ is the multiplicative monoid of $T_k=R^k\times S^{n-k}$ for $k=0,\dots,n$, giving $n+1$ rings that are pairwise non-isomorphic, since they have different additive groups.

I claim that these are the only rings with $M^n$ as multiplicative monoid. Suppose $T$ is another. Then it follows just from the multiplicative structure that $T$ is commutative and has $n$ primitive idempotents $e_1,\dots,e_n$ such that, as a ring, $T\cong Te_1\times\dots\times Te_n$, where each $Te_i$ is a ring whose multiplicative monoid is $M$.

But it's quite easy to see that $R$ and $S$ are the only rings with multiplicative monoid $M$: the subring generated by $1$ can only be $\mathbb{Z}/4\mathbb{Z}$ or $\mathbb{Z}/2\mathbb{Z}$, so either the ring is isomorphic to $R$ or is a two-dimensional $\mathbb{F}_2$-algebra generated by an element $x$ with square zero, in which case it must be isomorphic to $S$.

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That's very nice! – Todd Trimble May 1 '13 at 10:49
Beautiful, thank you! – Michał Masny May 1 '13 at 21:27

As for question 3, if the question is whether it's true that there are always countably many nonisomorphic additive group structures that extend a structure of infinite monoid with zero to a ring structure, the answer is no. In fact, there need not be any such extensions.

For example, consider a monoid given by a meet-semilattice with a top and bottom element; the multiplication is defined to be the meet. Every element is idempotent and so any ring extension would have to be a Boolean ring. There is in fact at most one way that a poset can be a Boolean algebra (so the number of ring extensions is either zero or one), but most posets are not Boolean algebras. For example, for $n \geq 2$ the poset of linear subspaces of $\mathbb{R}^n$ cannot be a Boolean algebra, because the distributive law fails.

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Oh, of course. I actually knew that (by other examples). Something went wrong in my head when I asked the question. I could delete that part of the question, or I can leave it so your answer doesn't become off topic. – Michał Masny May 1 '13 at 1:06
No, please don't delete. You can update the question if you like by referring to this information, and if you like you can use the html command "strike" which crosses out the bit you don't like but still leaves it visible. – Todd Trimble May 1 '13 at 1:14
OK, I've edited the question. – Michał Masny May 1 '13 at 1:59
given a Boolean algebra structure, the poset is recovered by setting $a\le b$ iff $a\wedge b=a$. – Omar Antolín-Camarena May 1 '13 at 2:25
Just to add to what Noah said: given a poset, meets and joins (if they exist) are uniquely determined by universal properties of cartesian product and coproduct (thinking of the poset as a category). Thus, $a \wedge b$ would be the unique element such that $x \leq a \wedge b$ iff $x \leq a$ and $x \leq b$. Negation can also be defined universally: $x \leq \neg a$ iff $x \wedge a \leq 0$ (where $0$ is the bottom element). – Todd Trimble May 1 '13 at 2:36

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