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Let $S$ be a set with at least two elements $1$ and $0$, which are distinct. Suppose $*$ is an operation on $S$ such that $*$ is associative, $x*0=0*x=0$ for all $x$, and $x*1=1*x=x$ for all $x$.

Does there exist an operation $+$ on $S$ such that $(S,+,*,0,1)$ is a ring with additive identity $0$ and multiplicative identity $1$? If not, what additional necessary and sufficient conditions are required on $*$?

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I think the trivial addition $a+b=0$ for all $a,b$ will make it a ring. – pritam Nov 29 '13 at 6:12
@pritam Obviously not. After all, 0+1=1, and I specifically said 1 is not equal to 0. – user107952 Nov 29 '13 at 6:18
up vote 2 down vote accepted

Let me first phrase the question in terms of category theory (although this isn't necessary in order to understand the mathematics here, but it's helpful). Let $\mathsf{Ring}$ denote the category of rings (which I assume to be unital) and let $\mathsf{Mon}_*$ be the category of pointed monoids (monoids equipped with an absorbing element $0$). There is a forgetful functor $U : \mathsf{Ring} \to \mathsf{Mon}_*$ which maps $(R,+,*,0,1) \mapsto (R,*,0,1)$. It only forgets $+$. Observe that $U$ is faithful, but not full. Although $U$ forgets about the addition, $U$ can be used in order to describe various ring-theoretic properties, see SE/66555. Your question is: Is $U$ essentially surjective? Does every pointed monoid come from a ring? And if not, what is the essential image?

$1.$ If $R$ is a boolean ring (i.e. $x^2=x$ for all $x \in R$), then $R$ is commutative (i.e. $xy=yx$ for all $x,y \in R$). The proof uses addition: First we observe $-x=(-x)^2=x^2=x$. We expand $x+y=(x+y)^2=x^2+xy+yx+y^2=x+xy-yx+y$, so that $xy=yx$. But not every pointed boolean monoid is commutative. The universal pointed boolean monoid generated freely by two elements $x,y$ is given by $\{0,1,x,y,xy,yx,xyx,yxy\}$ with $(xyx)y=x(yxy)=xy$ etc., and here $x,y$ do not commute.

$2.$ More generally, it is a Theorem of Jacobson that a ring is commutative when for every element $x$ there is some integer $m>1$ (possibly depending on $x$) such that $x^m=x$.

$3.$ If $R$ is a field (i.e. $1 \neq 0$, $R$ is commutative and and for every $x \in R \setminus \{0\}$ there is some $y \in R$ with $xy=1$, so this only depends on the underlying pointed monoid!), then the multiplicative structure enjoys several extra properties: For example, for every $n \geq 2$, the equation $x^n=1$ has at most $n$ solutions (and it follows that every finite subgroup of $R^*$ is cyclic). But in an arbitrary abelian group $G$, which we make into a pointed monoid via $\{0\} \sqcup G$, the equation $x^n=1$ may have more than $n$ solutions. In fact, $x^n=1$ has uncountably many solutions in $\prod_{p \in \mathbb{N}} C_n$.

This already gives some restrictions on the essential image of $U$. But there are probably a lot of other ones, and I doubt that we can find all of them.

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