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I'm reading Jean-Pierre Serre's 1970 "Cours d'arithmetique". I'm having trouble reading the beginning of his chapter 2 devoted for example to $\mathbb Z_3$, $3$-adic numbers that I discover and that interest me because of questions I have about the ring sequence $$(\mathbb Z/2\mathbb Z,\mathbb Z/6\mathbb Z,\mathbb Z/30\mathbb Z,\dotsc,\mathbb Z/p\#\mathbb Z,\dotsc)$$ He writes

Let $n\geq 1, A_n:=\mathbb Z/3^n\mathbb Z$. An element of $A_n$ clearly defines an element of $A_{n-1}$

What I understand here is that, if $\textbf{a}\in A_n$ then $\textbf{a}=a+3^n\mathbb Z$. And if we do the Euclidean division of $a$ by $3^{n-1}$ $$a=3^{n-1}q+r$$with $0\leq r\leq 3^{n-1}$. Then $$a+3^n\mathbb Z=3^{n-1}q+r+3^n\mathbb Z=r+3^{n-1}(q+3\mathbb Z)\subset r+3^{n-1}\mathbb Z$$ For example, $\mathbb Z/9\mathbb Z\to \mathbb Z/3\mathbb Z$ $$0\mapsto0$$ $$1\mapsto 1$$ $$2\mapsto 2$$ $$3\mapsto 0$$ $$4\mapsto 1$$ $$5\mapsto 2$$ $$...$$ Then he writes

This results in a homomorphism $$\varphi_n:A_n\to A_{n-1}$$ which is surjective, and of kernel $3^{n-1}A_n$

No problem here. Then he writes

[...] By definition, an element of $\mathbb Z_3$ is $x=(...,x_n,...,x_1)$, with $$x_n\in A_n\land \varphi_n(x_n)=x_{n-1} \text{ if }n\geq 2$$

  1. Is my comprehension of what is presented by Serre as "clear", correct ? Probably you would have to complete to take another representative of $\textbf{a}$ and make sure that you get the same $r$ and talk about uniqueness.

  1. I wonder if we can do the same with the sequel $$(\mathbb Z/2\mathbb Z,\mathbb Z/6\mathbb Z,\mathbb Z/30\mathbb Z,...,\mathbb Z/p\#\mathbb Z,...)$$

    For example, $\varphi : \mathbb Z/6\mathbb Z\to \mathbb Z/2\mathbb Z$ defined by $$0\mapsto 0$$ $$1\mapsto 1$$ $$2\mapsto 0$$ $$3\mapsto 1$$ $$4\mapsto 0$$ $$5\mapsto 1$$

    And if so, what would be the "numbers" we would get then?

Stéphane Jaouen
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1 Answers1

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Your comprehension for 1. is correct. In fact I am somewhat surprised that Serre doesn't do it the other way around: There is a rather obvious injective group homomorphism $\mathbb{Z} / p\hookrightarrow p^{n - 1} \mathbb{Z} / p^n \subset \mathbb{Z} / p^n$ given by $a \mapsto p^{n - 1}a$ which gives rise to a quotient map (of rings!) $\mathbb{Z}/p^n \twoheadrightarrow (\mathbb{Z}/p^n) / (p^{n - 1} \mathbb{Z} / p^n) \cong \mathbb{Z}/p^{n - 1}$. This map has the form $a \to a \pmod{p^{n - 1}}$ which agrees with what you computed using division.

For 2., there is no problem generalizing this: Given $\mathbb{Z} / n$ and $\mathbb{Z} / m$ with $m \mid n$, you analogously get a group homomorphism $\mathbb{Z} / (n / m) \hookrightarrow m \mathbb{Z} / n \subset \mathbb{Z} / n$ and a quotient map $\mathbb{Z} / n \twoheadrightarrow (\mathbb{Z} / n) / (m \mathbb{Z} / n) \cong \mathbb{Z} / m$. Specializing to your example and applying the CRT, we see that our system is in fact of the form $(\mathbb{Z} / 2 \times \ldots \times \mathbb{Z} / p_n, \varphi_n)_{n \geq 1}$ where $\varphi_n\colon \mathbb{Z} / 2 \times \ldots \times \mathbb{Z} / p_n \twoheadrightarrow \mathbb{Z} / 2 \times \ldots \times \mathbb{Z} / p_{n - 1}$ is the projection onto the first $n - 1$ coordinates.

Define now, analogous to what Serre does for $\mathbb{Z}_3$, $$ P = \{(\ldots, x_n, \ldots, x_1) \mid x_n \in \mathbb{Z} / 2 \times \ldots \times \mathbb{Z} / p_n,\ \varphi_n(x_n) = x_{n - 1} \text{ if } n \geq 2\} $$ (more formally, one notes that $(\mathbb{Z} / 2 \times \ldots \times \mathbb{Z} / p_n, \varphi_n)_{n \geq 1}$ forms an inverse system and then takes the inverse limit $P = \varprojlim (\mathbb{Z} / 2 \times \ldots \times \mathbb{Z} / p_n, \varphi_n)$). Then the map $\prod_p \mathbb{Z} / p \to P$, $(x_n)_n \mapsto ((x_1), (x_1, x_2), (x_1, x_2, x_3), \ldots) \in P$ is an isomorphism: Clearly it is bijective and a ring map, so your construction just recovers a product.

Ben Steffan
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  • For example, $(...,21,21,21,21,3,1)\in P$. What is this "number" ? I would like to say : it is $21$. – Stéphane Jaouen Apr 02 '24 at 14:29
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    @StéphaneJaouen I guess you could say that, sure. You do get an injective map $\mathbb{Z} \to \prod_p \mathbb{Z} / p$ with respect to which your element corresponds to $21 \in \mathbb{Z}$. But this map is clearly not surjective: For instance $(\ldots, p_n \cdot p_{n - 1} \cdot \ldots \cdot 3, \ldots, 7 \cdot 3, 3, 1)$ is not in the image of that map. (This is nothing special about $P$: it also holds of $\mathbb{Z}_3$, and $\mathbb{Z}_p$ more generally). – Ben Steffan Apr 02 '24 at 14:55
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    To put a different spin on it: The elements of $\mathbb{Z}_p$ we call numbers due to their number-theoretic significance, even though they need not correspond to elements of $\mathbb{Z}$, or $\mathbb{Q}$, or a number field, etc. I would not call the elements of $P$ numbers, at least not until you have a good justification for that terminology. – Ben Steffan Apr 02 '24 at 15:00
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    @StéphaneJaouen I'm not sure that's a great idea: $\mathbb{Z}_p$ gets to be called "$p$-adic integers" since it's the $I$-adic completion of $\mathbb{Z}$ where $I$ is the ideal $I = (p)$ (this is probably anachronistic but the point still stands). $P$ does not arise in this way, so you'll make people confused and/or upset at you if you do that :) – Ben Steffan Apr 03 '24 at 11:54
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    I found other examples to illustrate the fact that $f:\mathbb{Z} \to \prod_p \mathbb{Z} / p$ is not surjective. Let $n$ be an integer. Then $n$ has only a finite number of divisors. So, for example $(1,0,0,0,0,0,0,0,0,0,...)$ is not in the range of $f$. – Stéphane Jaouen Apr 16 '24 at 05:44
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    @StéphaneJaouen ...that's the idea behind my counterexample as well: if that particular element was in the image, then it would have to be divisible by every prime except 2. – Ben Steffan Apr 16 '24 at 11:00