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Fix a positive integer $p$, possibly prime.

For each natural number $n$, there is a ring $\mathbb{Z}/p^n \mathbb{Z}$ together with a distinguished ring homomorphism $$\pi_n:\mathbb{Z} \rightarrow \mathbb{Z}/p^n \mathbb{Z}.$$ For any given integer $k$, we can think of $\pi_n(k)$ as the remainder when $k$ is divided by $p^n$.

Now clearly, $\pi_n$ will never be injective. If we wanted to get around this, we could trying bundling up all the $\mathbb{Z}/p^n \mathbb{Z}$ together as a direct product. So define:

$$\mathbb{Z}/p^\mathbb{N} \mathbb{Z} = \prod_{n:\mathbb{N}} \mathbb{Z}/p^n\mathbb{Z}$$

There is, of course, a unique ring homomorphism $\pi : \mathbb{Z} \rightarrow \mathbb{Z}/p^\mathbb{N} \mathbb{Z}.$ In fact, it can be computed pointwise: $$(\pi(k))_{n} = \pi_n(k)$$

In some sense, we might say that $\pi(k)$ is the sequence of remainders obtained by attempting to divide $k$ by increasingly large powers of $p$.

The cool thing about this is that:

  • I'm pretty sure that $\pi$ is always injective; hence, we've included the integers in a bigger number system. Perhaps there are situations where we're working away in $\mathbb{Z}$ but we find that there aren't "enough" integers to do what we're trying to do; if so, one idea would be to look in $\mathbb{Z}/p^\mathbb{N} \mathbb{Z}$ for an appropriate choice of $p.$

  • Suppose $p$ is a prime number. Then to compute the number of copies of $p$ in the prime factorization of an integer $k$, we can simply count the number of $0$'s in the sequence $\pi(k)$. Furthermore, these $0$'s will always occur as an initial segment.

Question. Is $\mathbb{Z}/p^\mathbb{N} \mathbb{Z}$ widely studied, is it associated with a standard name or notation, and where can I learn more about it?

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  • $\begingroup$ These are related to the p-adic numbers. They are usually constructed using limits, and are maybe what you are looking for. $\endgroup$ – Michael Jun 10 '15 at 13:43
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    $\begingroup$ There are two constructions I know of that come close to yours. First there is the pro-finite completion $\hat{\mathbb{Z}}$ of $\mathbb{Z}$. It is defined as the inverse limit $\lim_{\leftarrow} \mathbb{Z}/n\mathbb{Z}$. Next there are the $p$-adic numbers $\mathbb{Z}_p$ defined in a similar way as the inverse limit $\lim_{\leftarrow} \mathbb{Z}/p^n\mathbb{Z}$. If you don't know inverse limits, check them out, they're great! :) $\endgroup$ – Zardo Jun 10 '15 at 13:44
  • $\begingroup$ @Zardo, interesting, however I'm not quite sure why $\mathrm{lim}_{n:\mathbb{N}} \mathbb{Z}/n\mathbb{Z}$ isn't just $\mathbb{Z}$ (up to isomorphism). Note in particular that for $n=0$ we get $\mathbb{Z}/n\mathbb{Z}=\mathbb{Z}/0\mathbb{Z} \cong \mathbb{Z}$ $\endgroup$ – goblin Jun 11 '15 at 13:34
  • $\begingroup$ Look here: math.stackexchange.com/questions/256732/… $\endgroup$ – Zardo Jun 11 '15 at 15:34
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    $\begingroup$ @Zardo, judging by the description at wikipedia, the trick is to omit the $n=0$ case. If so, then we can do this for any algebraic structure. Let $T$ denote a Lawvere theory and $X$ denote a $T$-algebra. Write $\mathrm{Cong}(X)$ for the poset of congruences of $X$. There is a subposet of finite congruences, by which I mean congruences with finitely many blocks. Denote this $\mathrm{FinCong}(X)$. There is a functor $X/- : \mathrm{FinCong}(X) \rightarrow T\mathbf{Alg}$. Define that $\hat{X}$ is the limit of this functor. $\endgroup$ – goblin Jun 11 '15 at 15:44
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$\newcommand\Z{\mathbf{Z}}$

Call the ring $R$. You will notice that the image of $\Z$ in $R$ under the map $\phi:\Z \rightarrow R$ you constructed has the property that, if you write is as:

$$\phi(a) = (a_1,a_2,a_3, \ldots )$$

then:

  1. $a_2 \in \Z/p^2$ has remainder $a_1$ when reduced modulo $p$,
  2. $a_3 \in \Z/p^3$ has remainder $a_2$ when reduced modulo $p^2$,
  3. $a_4 \in \Z/p^4$ has remainder $a_3$ when reduced modulo $p^3$,

and so on...

So you might consider replacing your ring $R$ by the subring $A \subset R$ consisting of (infinite) sequences which satisfy this property (namely, the ring consisting of sequences such that that each element in $\Z/p^{n+1}$ reduces modulo $p^n$ to the corresponding element in $\Z/p^n$ term). If you do that, then the corresponding ring $A$ is exactly the $p$-adic numbers $\Z_p$. The ring $A$ has all the nice properties you noted about $R$, but (in addition) has much nicer ring theoretic properties: it's torsion free, it's reduced and irreducible (so an integral domain), it's a complete local ring, it has a natural topology, and finally, the image of $\Z$ under the map $\phi$ is dense with respect to this topology. (Equivalently, for any element $a \in A$, there is a $x_n \in \Z$ so that $\phi(x_n)$ and $a$ agree to the first $n$ terms.)

Naturally, the ring of $p$-adic numbers $\mathbf{Z}_p$ is indeed widely studied and very useful. (If you insist on considering $R$ instead, then I don't imagine it has a name or is particularly studied.)

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