# Uniqueness of multiplication in a ring

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?

• So, you have an abelian group $(R,+)$, and you have two (commutative) multiplications $\cdot$ and $*$ that make $(R,+)$ into a commutative ring with unity, and you are asking if it is always the case that there exists a unit $u$ in $(R,+,\cdot)$ such that $a*b = a\cdot u\cdot b$? (there is no reason to invert $u$ if you already state it is a unt). Is that accurate? Commented Feb 2 at 18:12
• ... I think there is an additional condition... you require $u$ to be the unity relative to $*$, which is why you have the inverse... is that right? Commented Feb 2 at 18:25
• It's almost accurate. I'm asking whether, if the multiplication $(∗)$ has an element $\mathcal{J}∈R$ as its identity, and $\mathcal{J}$ turns out to be a unit with respect to $(·)$, it is true that $a∗b=a\mathcal{J}^{-1}b$. Your version of the question is slightly broader, although it would be also nice to know whether "any possible multiplication is produced by some unit (with respect to a known multiplication) in $R$". But I'm somehow not hoping much for the latter :) Commented Feb 2 at 18:26
• @ArturoMagidin Yes that's what I meant by "another multiplication $(∗)$ with respect to the identity $\mathcal{J}$" in the question, $\mathcal{J}$ must be the unity relative to $(∗)$ (but relative to $(·)$, it is not, since $\mathcal{J}≠\mathcal{I}$). Sorry if I wasn't clear enough! Commented Feb 2 at 18:30

Let $$R=\{(a,b)\mid a,b\in\mathbb{R}\}$$ with coordinatewise addition. We can turn this into a ring in two ways:

1. Define multiplication on $$R$$ by $$(a,b)\cdot(c,d) = (ac-bd,ad+bc)$$.
2. Define multiplication on $$R$$ coordinatewise: $$(a,b)*(c,d)=(ac,bd)$$.

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.

• In the examples, while the unity of $(R,+,*)$ is invertible in $(R,+,\cdot)$, the unity of $(R,+,\cdot)$ is not invertible in $(R,+,*)$. I'll think about whether one can have an example where you also have the unity of $(R,+,\cdot)$ invertible in $(R,+,*)$, and have them be different (it is easy to find examples where they are the same element, e.g. $\mathbb{R}[x]$ and $\mathbb{R}[x,y]$) Commented Feb 2 at 18:52
• I think one can find such an example by considering $\mathbb{R}[x,x^{-1}]$ and $\mathbb{R}[x,x^{-1},y]$. Both have underlying additive group $\oplus_{i=0}^{\infty}\mathbb{R}$, and one can arrange it so that $\cdot$ "thinks" that $(1,0,0,0,\ldots)$ is $1$ and $*$ thinks it's $x$, while $\cdot$ thinks that $(0,1,0,\ldots)$ is $x$ and $*$ thinks it's $1$. Then verify that $y*y$ is not equal to $a\cdot(0,1,0,\ldots)\cdot b$. Commented Feb 4 at 1:33

Let $$R_1 = \langle R; +, -, 0, \cdot, \mathcal{I}\rangle$$ be a ring containing an invertible element $$\mathcal{J}\in R_1^{\times}$$, and let $$R_2 = \langle R; +, -, 0, \ast, \mathcal{J}\rangle$$ be the ring defined with new multiplication $$A\ast B = A\mathcal{J}^{-1}B$$ and new unit element $$\mathcal J$$. The function $$h\colon R_1\to R_2\colon x\mapsto x\mathcal{J}$$ is an isomorphism of rings.

This shows that if you can find two rings $$R_1, R_2$$ defined on the same set $$R$$ which have the same underlying additive group, but where $$R_1\not\cong R_2$$, then the multiplication of the ring $$R_2$$ cannot be of the form $$A\ast B = A\mathcal{J}^{-1}B$$ for any invertible element $$\mathcal{J}$$.

For an example of two nonisomorphic rings with isomorphic additive structures, see this question.

• In fact, those examples have isomorphic multiplicative structures as well. Commented Feb 2 at 18:36