# Are fractions with zero divisors in the denominator never well defined?

Are fractions with zero divisors in the denominator never well defined?

I know that for a fraction in modular arithmetic to be well defined, the denominator must not be a zero divisor, e.g:

$$x \equiv \frac a b \pmod p$$

Then $b\neq k \cdot p, k\in \Bbb{Z}$, but what about fractions in other rings?

It is common in commutative algebra to define for $R$ a commutative ring, and $S$ a multiplicatively closed set (a set containing $1_R$ and $st$ for all $s,t \in S$), a ring of fractions $S^{-1}R$, sometimes also called a localization.

This construction can be done exactly as one constructs rational numbers from the integers when there are no zero-divisors present, but in the presence of zero-divisors one needs a twist and the resulting structures behave somewhat differently.

Let us recall as a warm-up how one can construct the rationals from the integers.

When are two fractions $\frac{n_1}{d_1}$ and $\frac{n_2}{d_2}$ equal? When $n_1d_2= n_2d_1$.

Turning things around, a common way to construct the rational numbers from the integers is by defining a relation on $\mathbb{Z} \times (\mathbb{Z} \setminus \{0\})$ by $(n_1, d_1) \sim (n_2, d_2)$ if $n_1d_2= n_2d_1$, which can be shown to be an equivalence relation.
And then one denotes the equivalence class of some $(n,d)$ by $\frac{n}{d}$, extends the operations and so on.

More generally one can do the very same thing for some multiplicatively closed subset $S \subset \mathbb{Z} \setminus \{0\}$ (for example all powers of $2$). Then one gets a subset, even a subring, of the rational numbers, those who can be written with a denominator from $S$, one might denote this $S^{-1}\mathbb{Z}$.

Still more generally, one can do the very same thing for any domain $R$ (a ring without zero-divisors) and multiplicatively closed subset $S \subset R\setminus \{0\}$. And denotes the set of equivalence classes $\frac{n}{d}$ with $n \in R$ and $d \in S$ by $S^{-1}R$. This set can be given the structure of a ring in a natural way (just do addition and multiplication as you are used to doing it from the rationals) and is called a ring of fractions or a ring of quotients or a localization of $R$ by $S$.
(If one takes $S= R \setminus \{0\}$ one even gets a field, the so-called fraction field, or field of quotients, of $R$.)

So far so good, but here we always avoided zero-divisors as denominators.

## What now with the zero-divisors?

One can give a slight twist to the construction, namely in this way:

We define the relation by $(n_1, d_1) \sim (n_2, d_2)$ if $d(n_1d_2)=d( n_2d_1)$ for some $d \in S$, which can again be shown to be an equivalence relation.
So we do not require $n_1d_2=n_2d_1$, but only require that the two are equal upon multiplication by some $d \in S$. If $S$ does not contain any zero-divisors this does not change a thing as we could cancel the $d$, but if there are zero-divisors we cannot.

The reason we do this is that if we do not do it we cannot show it is an equivalence relation.

Now, again one denotes the equivalence class of $(n,d)$ by $\frac{n}{d}$ and the set of all equivalence classes is a ring denoted $S^{-1}R$.

An important point to note is that one often will want that $r \mapsto \frac{r}{1}$ from $R \to S^{-1}R$ is injective so that one can consider $R$ as a subset of $S^{-1}R$.

This is true when $S$ contains no zero-divisors, but otherwise it might not be true. There are also some related issues where one needs to be more careful when dealing with these objects than without zero-divisors and one best always makes the ring $S^{-1}R$ clear and does not just write a fraction in that case. (Note that equivalence does not only depend on the elements in the couples under consideration but also the set $S$, so it needs to be clear.)

Thus, often, one does not allow zero-divisors as denominators in fractions.

However, there is a common and useful construction in commutative algebra that allows to consider them. Some care is however needed in dealing with them for example as in a ring $S^{-1}R$ it can be true that $\frac{r_1}{1}= \frac{r_2}{1}$ for $r_1, r_2 \in R$ distinct.

For further details on this construction see for example the Wikipedia page on localization.

Summary: In brief and independent of understanding of the above construction in detail: fractions with zero-divisors in the denominator can be defined but they behave in somewhat different ways and often one restricts to only non-zero divisors in the denominator.

• But rings of fractions generally are not localizations, e.g. see the Math Reviews here (this is one of the most common errors in commutative algebra). – Bill Dubuque May 25 '15 at 18:35
• What do you mean by they are "not localizations"? I said that $T^{-1}R$ is sometimes called a localization (of $R$ by $T$); this is done, e.g., on the page I link to. That being said, I would not usually call them localizations except for more restricted types of sets $T$. The reason I mentioned the word "localization" at all is that the relevant wiki page has this as title. – quid May 25 '15 at 18:43
• So what? I feel we are taking sideways here. By $T^{-1}R$ I mean precisely what is constructed on the page I link to. I would call this a ring of fractions, or a localization. That's all. That there are other structures that might also be called a ring of fractions seems tangential to me. Especially as I did not say something like "the rings of fractions are." – quid May 25 '15 at 19:07