I was trying to use the following result which says: if $I$ is a ideal of $R$ such that $R\setminus I$ is the set of all units of $R$ then $I$ is the only maximal ideal of $R$.
I showed that the units of $R$ are $\overline{ aX+1 }$ where $a \in \mathbb{Z}$ but it seems that the complement of the preceding set isn't an ideal. So, it looks to me as if $R$ is not a local ring.
I wasn't able to prove my claim. Can someone drop some hints? Thanks in advance.
For any prime $p$, you can readily find a homomorphism $R\to \Bbb F_p$