# What is the topology on $G=\prod\limits_{n=1}^{\infty}\mathbb Z/(p^n)$?

Let $$p$$ be any prime number and $$G=\prod\limits_{n=1}^{\infty}\mathbb Z/(p^n)$$ be the direct product of the finite cyclic groups $$\mathbb Z/(p^n)$$. Let $$H=\mathbb Z_p$$ be the group of $$p$$-adic integers with its usual norm $$|.|_p$$. Then the function $$f_n:G\to H$$ sending a sequence $$(x_k)_{k=1}^{\infty} \in G$$ to its $$n$$-th term $$x_n \in \{0, 1, \dots , p^n − 1\} \subset \mathbb Z_p$$ is a continuous $$p^{−n}$$–homomorphism.

As there is a term "continuity" of the map $$f_n:G\to H$$, there should be topologies on both $$G$$ and $$H$$, with respect to which $$f_n$$ is continuous.

Here, I've two questions.

Question 1 : It seems that the topology on $$H=\mathbb Z_p$$ is generated by the metric induced by the norm $$|.|_p$$. Is it true?

Question 2 : What is the topology on $$G=\prod\limits_{n=1}^{\infty}\mathbb Z/(p^n)$$?

• A1: Yes. A2: Most likely the product topology of discrete topologies on the factors. I don't know of other "natural" topologies for that space. – Jyrki Lahtonen Aug 20 at 7:05
• I agree. If you construct the $p$-adic integers more algebraically (not via completions and metrics, but via inverse limits) one will also take $\mathbb{Z}/p^n\mathbb{Z}$ as a discrete space, take the product over $n \geq 1$ and realize the $p$-adic integers as a subspace (which is just the concrete construction of the inverse limit in the category of groups). – ThorWittich Aug 20 at 7:11
• @JyrkiLahtonen What is happened if we take other topology instead of discrete topology on the factors of $G$? – BijanDatta Aug 20 at 7:56
• Connection with adeles and ideles (mathoverflow.net/q/41253/88984) ? – Jean Marie Aug 20 at 8:20
• There is no homomorphism $\Bbb{Z}/(p^n) \to \Bbb{Z}_p$ except the one sending everything to $0$. – reuns Aug 20 at 10:02