Can we show that the equality
$$∀A,B∈R:\ \ A*B=A\mathcal{J}^{-1}B$$
necessarily follows, given a ring $R$ with addition $(+)$, multiplication $(·)$ with respect to the identity $\mathcal{I}$ and another multiplication $(∗)$ with respect to the identity $\mathcal{J}≠\mathcal{I}$? It is implied that $\mathcal{J}$ has an inverse with respect to $(·)$:
$$∃!\mathcal{J}^{-1}∈R:\ \ \mathcal{J}^{-1}\mathcal{J}=\mathcal{I}$$
I'm working on a set on which I'm trying to define a "multiplication" with respect to the "addition" that already exists. There seems to be quite a few ways to do this but I noticed that all those that I know of can be reduced to any single one of them by the formula I gave at the beginning, thus making them all "equivalent" in a sense. I suspect that this is an abstarct-algebraic property induced by my set being a ring rather than by its specific structure. It is easy to show that, given a multiplication $(·)$ with unity $\mathcal{I}$ and an invertible element $\mathcal{J}$, the formula $A*B=A\mathcal{J}^{-1}B$ does indeed define another multiplication $(∗)$ with unity $\mathcal{J}$ (checking the axioms is straighforward); but so far I've struggled to prove the converse, mostly because the two multiplications don't seem to interact with each other at all. I don't even know whether this is in fact true but couldn't construct counterexamples in the more familiar rings either.
P.S. My algebraic structure of interest is also commutative with respect to $(·)$ so it's ok if the proof has to rely on that (but not all elements it contains are invertible, I would have called it a field otherwise; neither do I know anything about zero divisors in it). Also, if the hypothesis is not true, maybe it can somehow be strenghtened to work?
Best Answer
Let $R=\{(a,b)\mid a,b\in\mathbb{R}\}$ with coordinatewise addition. We can turn this into a ring in two ways:
The first makes $R$ isomorphic to $\mathbb{C}$, with unity $(1,0)$. The second makes $R$ isomorphic to $\mathbb{R}\times\mathbb{R}$, with unity $(1,1)$, which is a unit in $(R,+,\cdot)$, with inverse $(\frac{1}{2},-\frac{1}{2})$.
Is $*$ given by $(a,b)*(c,d) = (a,b)\cdot(1,1)^{-1}(c,d)$?
For example, take $(a,b)=(c,d)=(0,1)$. Then $(a,b)*(c,d) = (0,1)$. But $$\begin{align*} (a,b)\cdot(1,1)^{-1}(c,d) &= (0,1)\cdot \left(\frac{1}{2},-\frac{1}{2}\right)\cdot (0,1)\\ &= (-1,0)\cdot\left(\frac{1}{2},-\frac{1}{2}\right)\\ &= \left(-\frac{1}{2},\frac{1}{2}\right)\neq (0,1). \end{align*}$$
You can construct an example where neither of them are fields by adding a dummy third coordinate, so taking underlying set of $R$ to be $\mathbb{R}^3$, defining $\cdot$ by $(a,b,c)\cdot(x,y,z) = (ax-by,ay+bx,cz)$ and $*$ by $(a,b,c)*(x,y,z) = (ax,by,cz)$.
You can construct a finite example by considering for example the field of order $q^2$ ($q$ a prime power) and $\mathbb{F}_q\times\mathbb{F}_q$ with coordinatewise multiplication, essentially the same example as above.