During this weekend I thought about an extension of real field (an algebra over $\Bbb{R}$) quit similar to complex numbers which uses $j: j^3=-1$ instead of $i :i^2=-1$ and defined as $$\Bbb{E}=\{x-yj+yj^2 : x, y\in\Bbb{R}\,\,\,\, \text{and}\,\,\,\, j^3=-1\}$$ with addition and multiplication defined in the canonical way.

The reason to define these numbers with this form is for any $z=x-yj+yj^2\in\Bbb{E},$ the number defined by $\bar{z}=(x+y)+yj-yj^2\in\Bbb{E}$ act as the conjugate of $z.$
That is
1. $z+\bar{z}, z\bar{z}\in\Bbb{R}$ and $\overline{\bar{z}}=z.$
2. $\overline{w+z}=\bar{w}+\bar{z}$
3. $\overline{w.z}=\bar{w}.\bar{z}$

However $z\bar{z}=x^2+xy-2y^2=(x+y/2)^2-(3y/2)^2$ is not always positive as it does for complex numbers.

Also obvious candidates $0_{\Bbb{E}}=0-0.j+0.j^2$ and $1_{\Bbb{E}}=1-0.j+0.j^2$ served as additive identity and multiplicative identity. Therefore $\Bbb{E}$ is form a commutative ring with an identity.

If we define the norm of these numbers as $N(z)=\sqrt{|x^2+xy-2y^2|},$ where $z=x-yj+yj^2,$ then it is multiplicative $N(wz)=N(w)N(z)$ and all the numbers with non-zero norm are invertible with $z^{-1}=\overline{z}(N(z))^{-2}.$

Here I have two questions:

  1. Has anyone studied this Ring before?
    (If so and if it is possible, please give me a reference of it.)

  2. We can easily identify $\Bbb{E}$ with $\Bbb{R}^2$ by $x-yj+yj^2\mapsto (x,y).$
    Now the question is; Is there any natural geometrical explanation for $\Bbb{E}$?

  • 2
    $\begingroup$ Let $a := j^2 - j$. Then $a^2 = 2 + a$. So, $a^2 - a - 2 = (a - 2)(a + 1) = 0$. Thus $a = 2$ or $a = -1$, assuming no zero divisors. $\endgroup$
    – J126
    Feb 4, 2018 at 23:10
  • $\begingroup$ @JoeJohnson126 yes, but the beauty of the thing is that there is no way to choose between $2$ and $-1$, therefore this structure is more than just $\mathbb{R}$. $\endgroup$ Feb 4, 2018 at 23:18
  • $\begingroup$ $\mathbb{R}[X]/(X-1)$ is just $\mathbb{R}$, but $\mathbb{R}[X]/(X-1)(X-2)$ is bigger... $\endgroup$ Feb 4, 2018 at 23:19
  • $\begingroup$ Yes, it's bigger, but not so much so. $\mathbb{R}[X]/(X + 1)(X - 2)$ is just $\mathbb{R}^2$. See below. $\endgroup$ Feb 4, 2018 at 23:21
  • 1
    $\begingroup$ @ArnaudMortier I didn't mean to imply it was just $\mathbb{R}$. That's why I added the proviso "assuming zero divisors". It's just something I realized that I thought might help. It is somewhat related to the isomorphism given in the answer below. $\endgroup$
    – J126
    Feb 4, 2018 at 23:21

1 Answer 1


This ring is generated as an algebra over $\mathbb{R}$ by $k = j^2 - j$, which satisfies the minimal polynomial equation $(k + 1)(k - 2) = 0$. Thus, this is isomorphic to $\mathbb{R}^2$ under the unique $\mathbb{R}$-algebra homomorphism which sends $j^2 - j$ to $(-1, 2)$ [i.e., sending $(x, y)$ in your terms to $(x - y, x + 2y)$ in $\mathbb{R}^2$].

Your conjugation operation is the one which interchanges the two components of an element of $\mathbb{R}^2$.

(Analogues of this presentation could be done with any arbitrary pair of distinct values, not just $-1$ and $2$.)


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