What is a field with bounded whole and unbounded fractional part called?

Question

Given numbers of the form:

$W + \frac n d$ where $d \gt n \ge 0, W \ge 0$, and all are integers,

when defining addition and multiplication on these numbers, I want $W$ to be bound by a positive integer $B$ such that $W$ without the fractional part would act like a finite field of size B (integers modulo B), but with the numerator and denominator of the fractional part not similarly bounded (but note that the fractional part is always less than 1).

Is there a name or a field of study for something like this?

I apologize for the strange wording of the question. If someone knows a better way to phrase it feel free to edit. Also feel free to modify the tags if any are not appropriate. I was uncertain which ones applied.

Answer 1

There is an easy way to define this as a set: just consider the set $S=\mathbb{Q}\cap [0,B)$. We can also easily make it an abelian group, by just adding normally and then subtracting a multiple of $B$ to land in $S$. More abstractly, we can identify $S$ with the quotient group $\mathbb{Q}/B\mathbb{Z}$.

Multiplication, however, presents a problem. We could define multiplication the same way as addition: multiply two numbers, then subtract a multiple of $B$ to land in $S$. However, this operation does not distribute over addition! For instance, suppose $B=1$ (you can construct similar examples for any value of $B$); then $$\left(\frac12+\frac12\right)\cdot \frac12=0\cdot\frac12=0$$ but $$\frac12\cdot\frac12+\frac12\cdot\frac12=\frac14+\frac14=\frac12.$$

For $B>1$, this multiplication is also not associative. For instance, if $B=2$, then $$\left(\frac32\cdot\frac32\right)\cdot\frac12=\frac14\cdot\frac12=\frac18$$ but $$\frac32\cdot\left(\frac32\cdot\frac12\right)=\frac32\cdot\frac34=\frac98.$$

So this multiplication definitely doesn't make $S$ a ring, let alone a field. On an abstract level, this is just because $B\mathbb{Z}$ is not an ideal of the ring $\mathbb{Q}$, even though it is an ideal of $\mathbb{Z}$. There was no problem with addition because $B\mathbb{Z}$ is an additive subgroup of $\mathbb{Q}$ (a subgroup always remains a subgroup if you enlarge the ambient group, but ideals can stop being ideals if you enlarge the ring, because you have more elements you have to be closed under multiplying by).

In fact, you can show that there is no multiplication which (together with the previously defined addition) makes $S$ a ring (with unit). To show this, suppose $S$ were a ring, with unit $u$. Since $u$ is a rational number, there is some positive integer $n$ such that $nu$ is an integer divisible by $B$. Thus when you add $u$ to itself $n$ times in $S$, you get $0$. But by distributivity, this implies that for any $x\in S$, $$x+x+\dots+x=ux+ux+\dots+ux=(u+u+\dots+u)x=0x=0,$$ where there are $n$ terms in the sum. But this is false for, say $x=1/N$ for any $N<n$.

Answer 2

Define $\phi:\mathbb{Q}\to \mathbb{Q}\cap [0,1)$ as the fractional part of a rational.

Define $\kappa:\mathbb{Q}\to \mathbb{Z}$ as the floor of a rational.

Define $\tau:\mathbb{Z}\to \mathbb{Z}/B\mathbb{Z}$ as the natural projection of each integer into its congruence class modulo $B$.

Now, what you are studying is the function $$(\tau\circ \kappa,\phi):\mathbb{Q}\to (\mathbb{Z}/B\mathbb{Z})\times \left(\mathbb{Q}\cap [0,1)\right)$$ defined as $$(\tau\circ \kappa,\phi)(x)=(\tau(\kappa(x)),\phi(x))$$

Unfortunately, $\kappa$ is not a ring homomorphism, as $$\kappa(2/3)+\kappa(2/3)\neq \kappa(2/3+2/3)$$ Consequently, even though the codomain is a ring, the function you are studying is not a ring homomorphism, merely a bijection.