# 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.

- 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.
- 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 \(\frac{2}{3}\).
- 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 \(k\), \((2,k)\cdot y = x\).
- 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:

- 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.
- 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)