What is the equivalence class $a/0$ equal to?

Question

Let $D$ be a domain. Let $D^{*}$ be the set of non-zero elements of $D$. We define a relation $\sim$ on $D\times D^{*}$ as follows: $(a, b)\sim(c, d)$ if and only if $ad=bs$. It is easily checked that $\sim$ is an equivalence relation. We denote the equivalence classes with respect to this relation as $a/b$. Let $F$ denote the quotient set.

I wonder to know what is the equivalence class $a/0$ equal to? If $a/0=b/c$, then $ac=0$ and since $D$ is a domain, then $a=0$.

What about the equivalence class $0/0$?

May you help me, please? Thank you in advance.

Answer 1

$a/0$ doesn't have a class, since $(a,0)$ is not an element of $D \times D^*$. (unless $D$ is the zero ring)

For a variation on this construction where one does have $a/0$, see the projective line. If you want $0/0$ too, see the wheel of fractions

---

One can consider the localization of a ring with respect to more general sets than $D^*$. If you include $0$ in the set of elements you invert, then the resulting ring will be the zero ring.

Answer 2

There are no such equivalence classes. Only elements in $D\times D^*$ have equivalence classes, by definition. So no element whose second component is zero can have an equivalent class. It is further instructive to see what will happen if you attempt a similar construction starting with the set $D\times D$.