Is there a name for $x/x$ in a non-unital algebra?

Question

I'm looking into an algebraic structure on $\mathbb{R}^3$ in which multiplication is defined (commutative), as well as division. That is, if $a\cdot b = c$, then $c/b=a$, if $c$ and the modulus of $b$ are non-zero. If $x=(a,0,0)$ or anything else with two zeros, then $x/x=x/a$, which means $(1,0,0)$ acts as a sort of "relative identity" for elements of the form $(a,0,0)$.

However, there is no "universal" multiplicative identity. In general, $x/x$ just refers to the unique element $z\in\mathbb{R}^3$ for which $x\cdot z=x$ (if the modulus of $x$ is non-zero), but $x/x$ is not necessarily equal to $y/y$. Although it may be worth noting that the sum of the components of $x/x$ is generally $1$. See comments for a more precise definition of this system.

Is there standard algebra terminology to refer to the value of $x/x$, when there is no multiplicative identity?

Also, are there other terms here that I should be describing differently (since I'm not studying this topic in any sort of structured way and having words for things has been a challenge).

Answer 1

Commenting more extensively on your structure. The goal is to make some of its properties possibly a bit more transparent. Not sure I succeed, and I'm not really answering the question either :-(

The space $A=\Bbb{R}^3$ is a direct sum of two subspaces, namely $$ I_1=\{(a,a,a)\mid a\in\Bbb{R}\} $$ and $$ I_2=\{(a,b,c)\mid a+b+c=0\}. $$ Both $I_1$ and $I_2$ are "ideals" of $A$ in the sense that for all $u\in A$ we have $ui_1\in I_1$ for all $i_1\in I_1$, and $ui_2\in I_2$ for all $i_2\in I_2$. Consequently $i_1i_2=0_A$ for all $i_1\in I_1, i_2\in I_2$. We can thus conclude that $A\cong I_1\oplus I_2$ also as algebras.

Here $I_1$ is isomorphic with the field $\Bbb{R}$. This is because we easily see that $(a,a,a)(b,b,b)=(3ab,3ab,3ab)$, meaning that $\phi:\Bbb{R}\to I_1$, $$\phi(t)=(\frac t3,\frac t3,\frac t3)$$ is an isomorphism.

In $I_2$ let's single out the elements $e=\dfrac13(2,-1,-1)$ and $v=\dfrac1{\sqrt3}(0,1,-1)$. Your product (denoting it $\star$) formula shows that $$ \begin{aligned} e\star e&=\dfrac19(6,-3,-3)&=&&e,\\ e\star v&=\dfrac1{3\sqrt3}(0,-3,3)&=&&-v,\\ v\star v&=\dfrac13(-2,1,1)&=&&-e, \end{aligned} $$ showing that the basis $\{e,v\}$ is simple for calculations. For all real numbers $a,b,c,d$ we have $$ (ae+bv)\star(ce+dv)=(ac-bd)e-(ad+bc)v. $$ In other words, if we equate the complex number $z=x+yi$ with $xe+yv\in I_2$, your product looks like $$ z_1\star z_2=\overline{z_1z_2}; $$ the usual complex multiplication followed by complex conjugation.

So it looks like your algebra is isomorphic to $\Bbb{R}\times\Bbb{C}$ with a product defined like $$(x_1,z_1)\star(x_2,z_2)=(x_1x_2,\overline{z_1z_2}).$$

I don't recall seeing anything like this. Your algebra is non-associative for sure: $$ e\star(v\star v)=e\star(-e)=-e,$$ but $$ (e\star v)\star v=(-v)\star v=e. $$

Anyway, the connection to complex multiplication makes it obvious that also in the "ideal" $I_2$ we have "unique division" for the property follows from that of the arithmetic of complex numbers. Let us denote by $\psi:I_2\to\Bbb{C}$ the linear isomorphism $$\psi(xe+yv)=x+iy.$$ If $u_1,u_2,u_3\in I_2$ are arbitrary, $u_2\neq0$, and $z_i=\psi(u_i), i=1,2,3$ are the corresponding complex numbers, then the equation $$ u_2\star u_3=u_1 $$ is equivalent to the equation of complex numbers $$ \overline{z_2z_3}=z_1, $$ which has $$ z_3=\frac{\overline{z_1}}{z_2} $$ as its unique solution. Thus the element $$ u_3=\psi^{-1}\left(\frac{\overline{\psi(u_1)}}{\psi(u_2)}\right) $$ is the only solution of the equation $u_2\star u_3=u_1$ in $I_2$, and "entitled" to be called $u_1/u_2$.

In particular, for an arbitrary element $x=t+u\in A$ such that $t\in I_1$ and $u\in I_2$ are both non-zero, we have $$ \frac x x=\phi(1)+\psi^{-1}\left(\frac{\overline{\psi(u)}}{\psi(u)}\right). $$ Here $z(u):=\overline{\psi(u))}/\psi(u)$ is on the unit circle, so $z(u)=\cos\alpha+i\sin\alpha$ for some $\alpha$ depending on $u$. Pulling it all back to the original notation of $A$, the elements $x/x$ all have the form $$ \begin{aligned} \phi(1)+\psi^{-1}(z(u))&=(\frac13,\frac13,\frac13)+\cos\alpha e+\sin\alpha v\\ &=\frac13(1+2\cos\alpha,1-\cos\alpha+\sqrt3\sin\alpha,1-\cos\alpha+\sqrt3\sin\alpha)\\ &=\frac13\left(1+2\cos\alpha,1+2\cos(\alpha-2\pi/3),1+2\cos(\alpha+2\pi/3)\right). \end{aligned} $$ If $x$ happens to be an element of either $I_1$ or $I_2$, then $x/x$ is not unique (=not well-defined).

This gives us the following characterization:

The elements of the form $x/x$ consist exactly of the points $(a,b,c)\in\Bbb{R}^3$ on the intersection of the plane $a+b+c=1$ and the unit sphere $a^2+b^2+c^2=1$.