Let $R$ be a ring with an injective homomorphism $\varphi : \mathbb{Z} \to R$ such that $\varphi (n)$ is a unit if $n \not \in 13\mathbb{Z}$. Determine if $R$ is a field or an integral domain or neither of those.
I eliminated the possibility that $R$ is a field. Suppose that $R$ were a field. Since $\varphi$ is injective, $\ker \varphi = 0$. Thus, $\varphi (\mathbb{Z})$ is an ideal and it contains $1_{R}$ and hence $\varphi (\mathbb{Z}) = R$. But then $R$ and $\mathbb{Z}$ are isomorphic, which is not possible because $\mathbb{Z}$ is not a field. Does this work?
Now, is it true that $R$ has to be integral domain? I know only one example which satisfies the hypothesis and hence was unable to disprove it.
I'll appreciate it if someone drops some hints for me. Thanks in advance!
Update: My proof that $R$ is not a field is flawed.
Here's a counterexample that $R$ doesn't necessarily have to be a field. Let $S=\mathbb{Z} \setminus 13\mathbb{Z}$. Then $S$ is multiplicative set of the ring $\mathbb{Z}$.
Then $S^{-1}\mathbb{Z}=\{ \frac{n}{s} : n\in \mathbb{Z} , s \in S \}$ is a ring and there is a injective ring homomorphism from $\mathbb{R}$ to $S^{-1} \mathbb{Z}$ given by $a \mapsto \frac{a}{1}$ such that for all $s \in S$, image of $s$ is a unit in $S^{-1}\mathbb{Z}$.
Now, $\frac{13}{12} \in S^{-1}\mathbb{Z}$ but its inverse is not. Hence we are done.