$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$

# 8.0: The Problem of Division

Let's consider the problem of division. To get at a notion of division in general rings, let's recap what we know about division for familiar number systems.

Definition 8.0.0:

For numbers $$x, y$$, we say that $$y$$ divides $$x$$ if there exists a number $$z$$ such that $$\frac{x}{y}=z$$ if $$x=zy$$. We call $$z$$ the quotient of $$x$$ by $$y$$.

We can try to offload the problem of division to a problem of finding multiplicative inverses. If $$y$$ has a multiplicative inverse, then division by $$y$$ is easy: we can set $$z=xy^{-1}$$, so that $$zy=xy^{-1}y=x$$. If every element other than $$0$$ has a multiplicative inverse, then $$R$$ is called a field. You should already know three examples of fields: $$\mathbb{Z}, \mathbb{R}$$, and $$\mathbb{C}$$. Part of the reason for the importance of fields is that most of the basic facts in linear algebra work for any field.

Definition 8.0.1: Field

A field $$F$$ is a commutative ring in which every element other than $$0$$ has a multiplicative inverse.

Since the field must already be a commutative, associative ring with unity, we see that the set $$F\setminus \{0\}$$ is a group! Then another way to define a field is as a ring that is a commutative group under addition, and where $$F\setminus \{0\}$$ is a commutative group as well.

For example, we've already seen the group $$\mathbb{Q}^{\times}$$, which is just $$\mathbb{Q}$$ with the $$0$$ removed. Since this is a commutative group, $$\mathbb{Q}$$ is a field.

Example 8.0.2

For which values of $$n$$ is $$\mathbb{Z}_n$$ a field?

All of this works fine in $$\mathbb{R}$$, $$\mathbb{Q}$$ and $$\mathbb{C}$$: in these rings, for every $$x$$ and $$y$$, we can find a unique number $$z$$ such that $$zy=x$$. In other rings, though, things can go wrong in a number of different ways.

1. The first problem that could arise is that $$y$$ has no multiplicative inverse. For example, in $$\mathbb{Z}_6$$, there is no number $$z$$ such that $$2\cdot z=1$$. Likewise, almost no element of $$\mathbb{Z}$$ has a multiplicative inverse.
2. It could be that for a given $$x$$ and $$y$$, there is no quotient $$z=\frac{x}{y}$$. An example of this occurs in $$\mathbb{Z}$$, where (for example) there is no number $$\frac{2}{3}$$.
3. It could happen that the quotient $$z$$ exists but is not unique. For example, consider the product ring $$\mathbb{Z}\times \mathbb{Z}$$. Let $$x=(4,0)$$ and $$y=(2,0)$$. Then for any integer $$k$$, $$(2,k)\cdot y = x$$.
4. There's also a problem if the ring $$R$$ is not commutative. It could occur that $$yz=x$$ but $$zy\neq x$$. Which 'side' of $$x$$ is our division happening on?

We'll see that the different ways of resolving these questions give rise to definitions of different kinds of rings.

#### Zero-divisors

We'll first consider the question of multiplicative inverses. For a start, in any non-zero ring, $$0$$ does not have a multiplicative inverse: For any $$x$$ we have have $$x\cdot 0=0$$, so it can't be the case that $$x\cdot 0=1$$. This situation is familiar from working with the rational and real numbers. But there can be other elements without a multiplicative inverse.

For example, consider $$\mathbb{Z}_6$$. The elements $$1$$ and $$5$$ have multiplicative inverses: $$1\cdot 1=1$$ and $$5\cdot 5=1$$. But none of the other elements have a multiplicative inverse! For example, if we multiply $$2$$ times each element of $$\mathbb{Z}_6$$, we get the list $$[0,2,0,2,0,2]$$. Since $$1$$ isn't in the list, $$2$$ has no multiplicative inverse. Something interesting is happening in that list of multiples of $$2$$, though: there are many zeroes!

Definition 8.0.3

Let $$x\in R$$. Then $$x$$ is a zero-divisor if there exists $$y$$ such that $$x\cdot y=0$$.

Exercise 8.0.4

Describe all of the zero-divisors in the ring $$\mathbb{Z}\times \mathbb{Z}$$.

We have the following immediate result.

Proposition 8.0.5

For $$x\in R$$, $$x$$ cannot be both invertible and a zero-divisor.

Proof 8.0.6

Suppose that $$x$$ is invertible and a zero-divisor, and let $$y\neq 0$$ with $$xy=0$$. Then $$y=(x^{-1}x)y=x^{-1}(xy)=x^{-1}0=0$$, a contradiction.

As a result, the presence of zero-divisors means that there are non-invertible elements in the ring, and thus throws our division project into jeopardy. Furthermore, zero divisors also contribute to non-uniqueness of division: if $$ry=0$$ and $$x=zy$$, then we also have $$x=(z+r)y$$, so that both $$z$$ and $$z+r$$ can be considered as solutions to $$\frac{x}{y}$$.

To give a concrete example of this phenomenon, consider again $$\mathbb{Z}_6$$. What is the quotient $$\frac{4}{2}$$? Obviously, $$2$$ is an answer, since $$2\cdot 2=4$$. But we also have $$2\cdot 5=4$$, so we could also write $$\frac{4}{2}=5$$ just as easily. Notice that $$(2+3)\cdot 2=4+0=4$$; this is exactly the case described above.

Interestingly, for elements which are neither invertible nor zero-divisors, we still have a cancelation law:

Corollary 8.0.7

Suppose that $$r\neq 0$$ is not a zero-divisor and $$rx=ry$$. Then $$x=y$$.

Proof 8.0.8

We have $$rx-ry=r(x-y)=0$$. Then since $$r$$ is not a zero-divisor, we must have $$x-y=0$$, so that $$x=y$$.

One can use this result directly to prove the following:

Corollary 8.0.9

If $$r$$ is not a zero divisor, then the quotient $$\frac{x}{r}$$ is unique if it exists.

Then we see that the presence of zero-divisors is a major impediment to doing division in rings. Rings without zero-divisors will then be nice to work with!

Definition 8.0.10: Integral Domain

A commutative ring with no zero-divisors is called an integral domain.

Every field is an integral domain, since every non-zero element of a field is invertible. The primary example of an integral domain that is not a field is the integers: There are no non-zero integers where $$xy=0$$, but most integers don't have multiplicative inverses, so $$\mathbb{Z}$$ is not a field.

Then we seem to have an answer to the problem of division for commutative rings:

1. The best-case scenario is when every element has an inverse. Such rings are called division rings, or (if the ring is also commutative) fields.
2. The next-best case is when there are no zero divisors. These are the integral domains.

In the next two sections, we'll look at two different ways to 'solve' a division problem in an integral domain. The first way is to introduce fractions, which allow us to find inverses for any element of the ring. The second - available only in some rings - allows us to do division with a remainder.

Exercise 8.0.11

Show that $$M_{n,n}(\mathbb{Q})$$, the ring of $$n\times n$$ matrices, is not an integral domain.

### Contributors

• Tom Denton (Fields Institute/York University in Toronto)