3
$\begingroup$

I'm currently studying Atiyah-MacDonalds "Commutative Algebra". In Chapter 10 they introduce the $\mathfrak{a}$-adic topology on a ring $A$. The $p$-adic integers serve as an example.

My question now is: What is the reason for picking a prime $p$? Does this lead to more interesting objects than considering $\mathbb{Z}_{51}$?

Thanks!

$\endgroup$

3 Answers 3

5
$\begingroup$

The question needs precision on what you mean by $\mathbb{Z}_{51}$. Since you are studying Atiyah-Macdonald I assume what you meant to ask about is the completion of $\mathbb{Z}$ with respect to the ideal $51\cdot\mathbb{Z} \subset \mathbb{Z}$. If you mean with respect to some "metric" then you have other isues as quid pointed out.

But, using the definition of completion given in AM, Chapter 10, you get nothing more serious by considering composite numbers. Indeed, if $m$ and $n$ are coprime then $\mathbb{Z}/(mn)^r\mathbb{Z} \simeq \mathbb{Z}/m^r\mathbb{Z} \oplus \mathbb{Z}/n^r\mathbb{Z}$ in a canonical way (i.e. compatible with changing the $r$) and so the completion of $\mathbb{Z}$ with respect to the ideal $mn\cdot \mathbb{Z}$, which maybe I will call $\mathbb{Z}_{mn}^{\wedge}$ for now, is naturally isomorphic to the product ring $\mathbb{Z}_m^{\wedge} \times \mathbb{Z}_n^{\wedge}$.

So, for example, the 51-adic completion of $\mathbb{Z}$ is just the product of the $3$-adic integers $\mathbb{Z}_3$ and the $17$-adic integers $\mathbb{Z}_{17}$.

I should say things like this show up all the time in real life, e.g. the profinite completion $\hat{\mathbb{Z}}$ of $\mathbb{Z}$ is a natural enough object but it is really just the product of the $p$-adic integers over all primes $p$.

$\endgroup$
4
$\begingroup$

Considering a composite number $n$ in the construction, one loses $|ab|_n = |a|_n \ |b|_n$ or differently $v_n(ab)= v_n(a) + v_n(b)$.

This property is quite important in that context.

$\endgroup$
1
$\begingroup$

Let $R$ be a ring and for an ideal $I\subseteq R$, let $\hat{R}_I:= \lim\; R/I^n$ be the completion of $R$ with respect the $I$-adic topology.

Note that we have $\hat{R}_{I}\cong \hat{R}_{I^n}$ for any $n$ (algebraically but not topologically) by cofinality of the both cobsidered sequences and $\hat{R}_I\cong \hat{R}_{J_1}\times \hat{R}_{J_2} $ for $J_1$ and $J_2$ comaximal ideals (i.e. $J_1+J_2=R$) with $J_1J_2=I$ by the Chinese remainder theorem. Hence the general case over the integers can be reduced to the $p$- adic case (at least algebraically).

In your particular case, we will just have $\mathbb{Z}_3\times \mathbb{Z}_17$ algebraically and topologically.

$\endgroup$

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.