Can there be a set of numbers, which have properties like those of quaternions, but of dimension 3? [duplicate]

We all know what the complex numbers are : basically $$R^2$$ with a specified product formula, which is $$(a,b).(c,d)=(ac-bd,ad+bc).$$Hamilton defined the quaternions, which are a wonderful set of numbers, defined to be four-dimensional numbers, which too have a nice formula for products. One can prove the four square theorem using the Hurwitz integers and these numbers have a whole lot of marvellous properties. Can we define three dimensional numbers, who too have nice multiplication and interesting properties?

Also, when I think about algebraic number theory, I imagine rings and fields that are subsets of the complex numbers, whereas here we see how the quaternions have applications in number theory, and they are an extension of the complex numbers. What can we do when we don't restrict ourselves strictly to the complex numbers?

Note: one of the reasons the quaternions are a bit uncomfortable is that they don't commute with each other. However, we still have nice factorization properties when examining the Hurwitz integers. How do they arise? Can we define them as something like the ring of integers of a field?

EDIT:here are some links that give related information: Is there an algebraic closure for the quaternions? and Proving that $\mathbb R^3$ cannot be made into a real division algebra (and that extending complex multiplication would not work).

marked as duplicate by Henning Makholm, Dietrich Burde, janmarqz, user147263, draks ...Nov 26 '15 at 22:06

• the simple answer is: $\nexists$ about 3d such a structure – janmarqz Nov 26 '15 at 20:06
• On the other hand, if you can live with having zero divisors (and so losing any hope of division being possible), you can adjoin a formal element $\varepsilon$ with the property that $\varepsilon^3=0$ and work with elements of the form $a+b\varepsilon+c\varepsilon^2$ -- that will at least give you a commutative ring. – Henning Makholm Nov 26 '15 at 20:16
• @BogdanSimeonov: The slickest way to introduce $\varepsilon$ is to take the quotient of the polynomial ring $\mathbb R[X]$ with the ideal generated by $X^3$, giving $\mathbb R[X]/\langle X^3\rangle$. The residue class containing $X$ then works as $\varepsilon$. But if you don't know quotient ring, we can also simply postulate the multiplication operation $(a,b,c)*(d,e,f) = (ad,ae+bd,af+be+cd)$ and check the various ring laws directly. – Henning Makholm Nov 26 '15 at 20:25
• Your definition of the complex numbers is good and fine, though there are also other possibilities, such as $\mathbb R[X]/\langle X^2+1\rangle$ or the ring of 2×2 matrices of the form $\begin{pmatrix} a & b \\ -b & a \end{pmatrix}$. – Henning Makholm Nov 26 '15 at 20:26