2
$\begingroup$

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.)

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
10
$\begingroup$

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).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.