Is division allowed in rings and fields?

Question

Is division allowed in ring and field?

The definition of ring I am using here does not require the presence of multiplicative inverse.

I think in general, division is not a well-defined operation in rings because division can only occur if multiplicative inverse exists (not sure about this though). For field, since multiplicative inverse exists for all elements in the field, if we want to perform the operation $x/y$ whereby $x$ and $y$ belongs to the field, we multiply $x$ by $1/y$.

Is my understanding correct?

This is my first abstract algebra class, so I am still quite confused.

Answer 1

The simplest example of a ring is that of the integers $\mathbb Z$, and this is in some sense the motivating example of a ring.

Division in $\mathbb Z$ is highly restrictive - we can only divide $m$ by $n$ if $m$ is a multiple of $n$. As such, division is not defined as a binary operation - there is no function that gives a multiplicative inverse.

The multiplicative inverse is not defined in a general ring $R$.

The definition of a field is a ring such that every non-zero element has a multiplicative inverse. Be careful though - we still cannot define a multiplicative inverse on the whole field, since $0$ has no inverse!

However, if $\mathbb F$ is a field, then $\mathbb F^*=\mathbb F\setminus \{0\}$ is a group under multiplication, so does have a multiplicative inverse. As such, if $y$ is a non-zero element of a field, it has an inverse which is usually denoted $y^{-1}$.

Answer 2

You are correct.

The most basic ring is $\mathbb Z$, the ring of integers. Can we divide in the ring of integers, and always get an integer? What if we divide 3 by 2?

Some rings have zero divisors. That is, nonzero $a$ and $b$ so that $ab=0$. So $a\cdot b=0=a\cdot 0$. Can we divide $0$ by $a$ and get $b$?

Answer 3

What we always have in a ring (or field) is addition, subtraction, multiplication. Division $a/b$, that is the existence and uniqueness of a solution to $bx-a=0$ is different. Even with a field there is not always a soltution (namly if $b=0$ and $a\ne 0$), or it may not be unique (namely if $a=b=0$), so even in a field we only have division if what we divide by is not $0$.

In a ring, additional obstacles may occur. First of all the non-existence, e.g. in $\mathbb Z$ there is no solution to $3x-2=0$. On the other hand $3x-6=0$ does have a solution - and it is unique. If in a ring $R$ with $1$ the equation $bx-1=0$ has a solution, then for all $a$ the equation $bx-a=0$ has a solution. In such a case, $b$ is called a unit of $R$. For example, the units of $\mathbb Z$ are $1$ and $-1$. If all elements $\ne0$ are units, then in fact $R$ is a field.

Another obstacle is the lack of uniqueness. It may happen that two solution exist, i.e. $x_1\ne x_2$ with $bx_1-a=bx_2-a$. This leads to $b(x_1-x_2) = 0$ and in a general ring we can not conclude from this that $b=0$. In such a case, $b$ is a divisor of zero. For example, in $\mathbb Z/12\mathbb Z$, we have $2\cdot 3 = 2\cdot 9 = 6$, so there is no unique meaning we could assign to $6/2$. ($2$ is a divisor of zero because $2\cdot 6=0$ in this ring).

So in a ring the following cases are possible:

• If $b$ is a zero divisor (including $b=0$ itself), a solution of $bx-a=0$ may exist, but even if it exists, it is not unique. Hence divsion by $b$ is not defined.
• If $b$ is a unit, everything is fine: $bx-a=0$ always has one and only one solution. This solution can be denoted as $\frac ab$
• For other $b$ (i.e. neither a unit nor a divisor of $0$), the existence may depend on $a$, but if a solution exists, it is unique. It does make sense to write $\frac ab$ to denote this solution.