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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?

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  • $\begingroup$ 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? $\endgroup$
    – Arturo Magidin
    Commented Feb 2 at 18:12
  • $\begingroup$ ... 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? $\endgroup$
    – Arturo Magidin
    Commented Feb 2 at 18:25
  • $\begingroup$ 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 :) $\endgroup$
    – Aberone
    Commented Feb 2 at 18:26
  • $\begingroup$ @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! $\endgroup$
    – Aberone
    Commented Feb 2 at 18:30

2 Answers 2

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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.

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    $\begingroup$ 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 $\mathbb{R}[x,y]$) $\endgroup$
    – Arturo Magidin
    Commented Feb 2 at 18:52
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    $\begingroup$ 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 $a\cdot(0,1,0,\ldots)\cdot b$. $\endgroup$
    – Arturo Magidin
    Commented Feb 4 at 1:33
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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.

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    $\begingroup$ In fact, those examples have isomorphic multiplicative structures as well. $\endgroup$
    – Arturo Magidin
    Commented Feb 2 at 18:36

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