I'm not sure if this might instead be MathOverflow material, but I'll give it a shot here first anyway.
Does any ring $R$ exist that satisfies the following properties?
- $R$ is a totally ordered, commutative ring.
- $R$ has $\mathbb{Z}$ as an ordered subring.
- The ordering on $R$, when restricted to $\mathbb{Z}$, is the usual ordering on $\mathbb{Z}$.
- There exists some prime number p such that the ring of p-adic integers $\mathbb{Z}_p$ can be constructed as a quotient ring of $R$.
I'm also very curious about rings which might satisfy #1-3, and any of the following variants of #4:
For ALL prime numbers p, the different rings of p-adic integers $\mathbb{Z}_p$ can be constructed as quotient rings of the same $R$.
Instead of p-adic integers, the profinite completion $\mathbb{\hat{Z}}$ of the integers can be constructed as a quotient ring of $R$.
I'm very curious if such a ring could possibly exist. And barring that, I'm also curious if the analogous question for groups holds, i.e. replace "ordered commutative ring" with "ordered abelian group," "subring" with "subgroup," and treat $\mathbb{Z}_p$ as a group where multiplication has been forgotten.
EDIT: to add, I'm especially interested in the case where $R$ is an integral domain, but I'm also curious about $R$ that have zero divisors too.
Any ring $R$ is a quotient of the ring $\Bbb Z[R]$ (with one indeterminate $[x]$ for each element $x \in R$) by the ideal generated by the elements $[x] + [y] - [x+y], [0], [x]+[-x], [x]*[y] - [xy]$. (this is needlessly large : if you want, you can use a generating set for $R$ instead of $R$ as indeterminates).
So we only need to show that for any set $S$, the ring $\Bbb Z[S]$ can be ordered. In order to do this, pick any total order on $S$. This induces a lexicographic ordering on the monomials (in fact any order on monomials such that $x \le y \Leftrightarrow xz \le yz$ is fine), and an order on $R$ by looking at the sign of the coefficient of the monomial of highest weight.