Inquiries around a new number system

Question

We can suppose that we will create a new number system with essentially two imaginaries that do not interact. (Besides this, all quantities are taken to be integers) For example, we have an $i_1$ and an $i_2$. Then we could say

$$(a+b i_1)(c+d i_1) = ac + (ad + bc)i_1-bd$$

and, similarly for $i_2$:

$$(a+b i_2)(c+d i_2) = ac + (ad + bc)i_2-bd$$

However, for a system with $i_1$ AND $i_2$:

$$(a+b i_1+c i_2)(d + e i_1 + f i_2)=$$
$$(ad + ae i_1 + af i_2) + (bd i_1 - be + 0i_1i_2) + (cd i_2 + 0i_1i_2 -cf)$$

Above, the key thing to note is that $i_1\cdot i_2 = 0$.

QUESTIONS

I'm wondering if there is any idea or statement in math that says that I simply cannot do this. Without having worked in general systems of numbers, I'm wondering what ideas I should know about when I try to create a system like this.

Can I create this system if my main purposes are to carry out addition, subtraction, and multiplication with these numbers? Also, if I pose the additional constraint that all of these calculations are carried out modulo a prime, will this affect the system?

Answer 1

Um, scratch the conclusions of my earlier answer. You get into more trouble than mere zero divisors:

$$1 = (-1)\cdot (-1) = i_1^2\cdot i_2^2 = (i_1 i_2)^2 = 0^2 = 0 $$

so everything collapses!

My observation that what you get is a quotient ring was technically right, but I failed to notice that the ideal $\langle X^2+1,Y^2+1,XY\rangle$ is just $\langle 1\rangle$ because $$1 = 1\cdot(X^2+1) - X^2\cdot(Y^2+1) + XY\cdot(XY) $$so the quotient ring you get is actually the zero ring.

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If you drop the rule that $i_1\cdot i_2=0$ and instead just accept that your objects can also contain some multiple of $i_1 i_2$, then you stand a better chance.

In fact, if one decides that multiplication is not commutative and $i_1 i_2 = -i_2i_1$ but ordinary numbers commute with everything, then what one gets is exactly the quaternions!

Answer 2

BEWARE. This answer is as wrong as it is possible for it to be without actually containing any technically false statements. See my other answer. I'm leaving this up because the references to ring theory and quaternions are probably still helpful.

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If $i_1 i_2=0$, then $i_1$ and $i_2$ are zero divisors, and neither of them can have multiplicative inverses (assuming other usual rules such as associativity and $0x=0$ hold), so what you get isn't a field.

You get even more zero divisors from the fact that $i_1$ and $i_2$ have the same square without being each other's negatives: $(i_1-i_2)(i_1+i_2)=i_1^2-i_2^2=0$.

It is, however, a ring, which allows you to do addition, subtraction and multiplication as you please. Division won't always be possible, like in the integers. There's a quite rich and well developed theory about such structures (look for "ring theory" or "commutative algebra"); what you have is a particular instance of one of the most common way of producing new rings, namely as quotients of polynomial rings. More specifically, in standard notation your ring can be constructed as $\mathbb R[X,Y]/\langle X^2+1, Y^2+1, XY\rangle$.

You can certainly use modular arithmetic as a base instead of $\mathbb R$. In standard notation that will give you $\mathbb Z[X,Y]/\langle p, X^2+1, Y^2+1, XY\rangle$. Note that you may get yet more zero divisors if $-1$ is already a square in $\mathbb Z$ modulo $p$.

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(Ah, I just noticed that the question already specified that the base was the integers rather than $\mathbb R$. Then of course it is not a particular problem that you cannot always divide, because you can't do that even in $\mathbb Z$ itself. Note, however, that even division with remainder (which works well in $\mathbb Z$) won't have any nice generalization to any setting where there are zero divisors).

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You might also look into quaternions which manages to be almost a field by having three imaginary units $i$, $j$, $k$, where the product of any two is either the third one or its negative. Then inverses always exist, but on the other hand multiplication is no longer commutative.

Answer 3

If you drop the rule that $i_1 i_2=0$ but maintain commutativity, you will get bicomplex numbers.