I am studying the ring of integers modulo $n$, $\mathbf{Z_n}$, which consists of the set of all cosets $\overline{a}$ and the latter is the set of all the elements $b\in{\mathbf{Z}}$ such that $n|a-b$.
The ring $\mathbf{Z_2}$ is equal to the set of cosets {${\overline{0},\overline{1}}$}; the set $\mathbf{Z_1}$ is equal to {$\overline{0}$}=$\mathbf{Z}$.
I am confused in finding $\mathbf{Z_0}$. In my opinion it is the empty set since the division by zero does not make sense.
But by the first isomorphism theorem it follows that $\mathbf{Z_0}$ is isomorphic to $\mathbf{Z/0Z}$=$\mathbf{Z}$ which means that $\emptyset$ is isomorphic to $\mathbf{Z}$?
What is the set $\mathbf{Z_0}$ equal to?
Would you help me please? Thank you in advance.
Say that $a \equiv_n b$ if $a - b = kn$ for some nonzero integer $k$. Then $\equiv_n$ is the usual equivalence relation modulo $n$ for $n\in\mathbb{Z}$. On the other hand, $\equiv_0$ can be made sense of, whereas "division by zero" doesn't really make sense.
Specifically, $a \equiv_0 b$ if $a-b = k0$ for some nonzero integer $k$. But this means that $a-b = 0$, which implies that $a=b$. Hence $\overline{a} = \{a\}$ for each $a\in\mathbb{Z}$. This implies that $\mathbb{Z}_0 = \{\overline{a} : a\in\mathbb{Z}\} \cong \mathbb{Z}$, where the isomorphism is the obvious choice.