Denote $p$ a prime number and $\mathbb Z _p$ the ring of $p$-adic integers. We have a canonical injective ring homomorphism $:\mathbb Z \rightarrow \mathbb Z_p$ for all $p$. But $\mathbb Z$ is not the largest ring that maps into all $\mathbb Z_p$. Consider for instance the ring of all formal series $$ F := \left\{ \sum_{n=0}^\infty a_n \cdot n! \; ,\; 0\leq a_n\leq n \right\}. $$ Since $(n!)_n$ converges $p$-adically to zero for every $p$, this gives a canonical morphism $\phi_p: F\rightarrow \mathbb Z_p$.

I must admit that $\phi_p$ is not injective, but the intersection $\bigcap\limits_p \ker \phi_p$ is zero. So, what is known about rings $R$ containing $\mathbb{Z}$ and having "canonical" morphisms $\phi_p:R\rightarrow \mathbb Z_p$ with $\bigcap\limits_p \ker \phi_p=0$?

To start with, if $x\in R$ is an element that satisfies a polynomial equation with coefficients in $\mathbb Z$, then $x\in \mathbb Z$ (by Chinese remainder theorem). So $R$ must be a transcendental extension of $\mathbb Z$.

Is there a maximal ring resp. a good notion of "maximal element" of such rings? (For example, a ring $R'$ with an injection $:R\rightarrow R'$ making all $\phi_p$-diagrams commutative.)

  • $\begingroup$ No, $\frac{1}{p}$ is not in $\mathbb Z_p$. $\endgroup$ – Nicolas Malebranche Jan 1 '16 at 13:03
  • $\begingroup$ Ah, right, I'm thinking of $\Bbb Q_p$. My bad. (btw, it's \not\in, not \ni) $\endgroup$ – Wojowu Jan 1 '16 at 13:04
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    $\begingroup$ I think your conditions on $R$ just say that it is a subring of $\hat{\mathbb Z} \cong \prod_p \mathbb{Z}_p$. $\endgroup$ – tj_ Jan 1 '16 at 13:49
  • $\begingroup$ @tj_ I think you're right. So should we look at maximal subrings of $\hat{\mathbb Z}$ without zero divisors? $\endgroup$ – Nicolas Malebranche Jan 1 '16 at 14:03

The ring you are looking for is $\widehat{\mathbb{Z}}={\displaystyle\lim_{\leftarrow}\mathbb{Z}/N\mathbb{Z}}$.

This has canonical maps to $\mathbb{Z}_p={\displaystyle\lim_{\leftarrow}\mathbb{Z}/p^k\mathbb{Z}}$ induced by taking $\mathbb{Z}/N\mathbb{Z}\twoheadrightarrow \mathbb{Z}/p^k\mathbb{Z}$ (where $p^k$ is the biggest power of $p$ dividing $N$).

Your condition asks that the product map $\widehat{\mathbb{Z}}\to \prod_p \mathbb{Z}_p$ should be injective; in fact, by the Chinese Remainder Theorem, this map is an isomorphism: $\widehat{\mathbb{Z}}\cong \prod_p \mathbb{Z}_p$. So this is the universal such ring.

A broader perspective: passing from $\mathbb{Z}$ to $\widehat{\mathbb{Z}}$ amounts to taking the completion of $\mathbb{Z}$ w.r.t the topology induced by the inclusion $\mathbb{Z}\hookrightarrow \prod_p \mathbb{Z}_p$, so the ring $\widehat{\mathbb{Z}}$ we obtain is the closure of $\mathbb{Z}$ inside $\prod_p \mathbb{Z}_p$. The fact that this is all of $\prod_p \mathbb{Z}_p$ means that $\mathbb{Z}$ is dense in $\prod_p \mathbb{Z}_p$. This density is a weak form of "approximation" (it's the Chinese Remainder Theorem), which is significantly generalized by the classical theorems of weak and strong approximation (and even superstrong approximation).


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