Division notation in ring Let $a,b\in R$ where $R$ is a ring.
Is there any problem/ambiguity with this notation $\frac{ab}{a}=b$, where $a\neq 0$?
What are the minimal conditions (e.g. commutative, UFD, etc) that we need such that the above makes sense?
Thanks for any help.
 A: First of all, for the non-commutative case I'd say that this is a problem. It is not clear if you mean something like $aba^{-1}$ or $a^{-1}ab$. So I think this should be reserved for the commutative case.
Let's now look at the commutative case.
If $a$ is a zero divisor and $ab = 0, b \neq 0$, the equality you wrote, i.e. $\frac{ab}{a} = b$ doesn't really make sense.
So let's assume for a second that $R$ is an integral domain. Then, analogous to the construction of $\mathbb{Q}$ from $\mathbb{Z}$ we can construct a field in which we can embed $R$. This field is called field of fractions and usually denoted by $Q(R)$ or $\text{Quot}(R)$. I shortly explain the construction: On the set $R \times R \setminus \{0\}$, we defined the equivalence relation $(a,b) = (a',b')$ iff $ab' = a'b$. Let's write $(a,b) = \frac{a}{b}$, then we see that this just means that $\frac{a}{b} \sim \frac{a'}{b'}$ iff $ab' = a'b$ and define $Q(R) = (R \times R \setminus \{0 \}) / \sim$. Multiplication and addition is defined as in $\mathbb{Q}$:
$$\frac{a}{b} + \frac{a'}{b'} = \frac{ab' + a'b}{bb'}, \quad \frac{a}{b} \cdot \frac{a'}{b'} = \frac{aa'}{bb'}.$$
In fact, this construction can be made a bit more general. Let $R$ be a commutative ring and $S \subseteq R$ a multiplicative closed set, i.e. for $s,t \in S$ we have $s \cdot t \in S$ (and let's assume $1 \in S$). Then we define the equivalence relation on $R \times S$ to be $(r,s) \sim (r',s')$ iff there exists $t \in S$ with $t(rs' - r's) = 0$. The quotient is denoted by $S^{-1}R$, multiplication and addition is defined as above and one usually also writes $\frac{r}{s}$ for the elements $(r,s) \in S^{-1}R$. Contrary to the field of fractions we don't have always an injective homomorphism given by $r \mapsto \frac{r}{1}$, the problem being the zero divisors. That was the reason why we had to introduce the element $t$ in the definition of the equivalence relation (you should show that this is in fact an equivalence relation and that you need $t$ if $R$ is not an integral domain).
A: In domains (for instance polynomial rings over domains) this is used quite frequently: one often finds an expression such as
$$
\frac{X^n-1}{X-1}
$$
to denote $X^{n-1}+X^{n-2}+\dots+X+1$ or, more generally,
$$
\frac{f(X)}{X-a}
$$
where $f(X)$ is a polynomial and $a$ is a root thereof, so this denotes a well defined polynomial.
This makes sense in the field of quotients, so it's just a slight abuse of language. With this convention, the equality
$$
\frac{ab}{a}
$$
always holds in a domain, provided $a\ne0$. Other examples are found when dealing with a greatest common divisor $d$ of $a$ and $b$: writing
$$
\frac{a}{d}
$$
in such cases is quite frequently found. Again, this stands for the (unique) element $c$ such that $cd=a$ and we can perform the operation in the field of quotients.
When the ring is not a domain, in particular if it is not commutative, such a notation is to be carefully avoided: we might have $ab=0$ with $b\ne0$, so $ab/a=b$ would even be false. Even if $a$ is invertible, the notation might be interpreted both as $a^{-1}ab=b$, but also as $aba^{-1}$ which may well be different from $b$.
When talking about rings of quotients, say $S^{-1}R$, where $R$ is a commutative ring and $S$ is a multiplicative set, the notation $a/s$ ($a\in R$ and $s\in S$) has a special meaning and $ab/a=b$ should be avoided as well, even if $a\in S$, unless the canonical map $R\to S^{-1}R$ is injective. This is the case when $R$ is a domain, but then we are essentially back to the situation at the beginning.
A: The notation "$/a$" is a shorthand for the inverse of $a$, so one necessary feature is that $a$ is a unit in $R$, that is, there exists a $c$ such that $ac=ca=1$ and then you could write $a^{-1}=\frac{1}{a}=c$.
The second thing is that it is necessary to consider commutativity of the ring with this notation. There is no problem when the ring is commutative. But when the ring is noncommutative, $\frac{ab}{a}$ is ambiguous, because it does not tell us what order $c$ is being multiplied with $ab$. It could potentially mean $a^{-1}ab$ or it could be $aba^{-1}$. The first one reduces to just $b$ automatically, but there is no reason for $b=aba^{-1}$.
For example, in the quaternions, $iji^{-1}=-j\neq j$.
