Are we allowed to interchange product and inverse limits?

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Currently, I am trying to show that the profinite completion $hatmathbbZ$ of $mathbbZ$ is isomorphic to $prod_p mathbbZ_p$ (as topological groups) where $p$ runs through all prime numbers and $mathbbZ_p$ denotes the ring of $p$-adic integers.



By definition, we know that $hatmathbbZ$ is the inverse limit of $mathbbZ/nmathbbZ$ and $mathbbZ_p$ is the inverse limit of $mathbbZ/p^nmathbbZ$. Furthermore, the Chinese Remainder Theorem tells us that if $n = prod_p p^e_p(n)$ denotes the prime factorization of $n$, then we have $$ mathbbZ/nmathbbZ simeq prod_p mathbbZ/p^e_p(n) mathbbZ.$$



So I would like to show that
$$
hatmathbbZ = varprojlim_n mathbbZ/nmathbbZ simeq varprojlim_n prod_p mathbbZ/p^e_p(n) mathbbZ = prod_p varprojlim_n mathbbZ/p^e_p(n) mathbbZ = prod_p varprojlim_n mathbbZ/p^n mathbbZ = prod_p mathbbZ_p,
$$



but I am not sure about these steps, especially with the third and fourth step. Could you please give me an easy explanation whether this works or not? Thank you!







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  • 2




    Yes, you can. A product is a particular case of an inverse limit and you can interchange inverse limits.
    – d.t.
    Aug 24 at 9:46










  • More generally you can interchange any two limits (in the categorical sense; in particular "direct limits" are not limits)
    – Max
    Aug 24 at 9:51






  • 1




    Okay, that is good to know! Could you maybe explain the second last step as well? I do not really see why this is true.
    – Diglett
    Aug 24 at 9:55














up vote
3
down vote

favorite












Currently, I am trying to show that the profinite completion $hatmathbbZ$ of $mathbbZ$ is isomorphic to $prod_p mathbbZ_p$ (as topological groups) where $p$ runs through all prime numbers and $mathbbZ_p$ denotes the ring of $p$-adic integers.



By definition, we know that $hatmathbbZ$ is the inverse limit of $mathbbZ/nmathbbZ$ and $mathbbZ_p$ is the inverse limit of $mathbbZ/p^nmathbbZ$. Furthermore, the Chinese Remainder Theorem tells us that if $n = prod_p p^e_p(n)$ denotes the prime factorization of $n$, then we have $$ mathbbZ/nmathbbZ simeq prod_p mathbbZ/p^e_p(n) mathbbZ.$$



So I would like to show that
$$
hatmathbbZ = varprojlim_n mathbbZ/nmathbbZ simeq varprojlim_n prod_p mathbbZ/p^e_p(n) mathbbZ = prod_p varprojlim_n mathbbZ/p^e_p(n) mathbbZ = prod_p varprojlim_n mathbbZ/p^n mathbbZ = prod_p mathbbZ_p,
$$



but I am not sure about these steps, especially with the third and fourth step. Could you please give me an easy explanation whether this works or not? Thank you!







share|cite|improve this question
















  • 2




    Yes, you can. A product is a particular case of an inverse limit and you can interchange inverse limits.
    – d.t.
    Aug 24 at 9:46










  • More generally you can interchange any two limits (in the categorical sense; in particular "direct limits" are not limits)
    – Max
    Aug 24 at 9:51






  • 1




    Okay, that is good to know! Could you maybe explain the second last step as well? I do not really see why this is true.
    – Diglett
    Aug 24 at 9:55












up vote
3
down vote

favorite









up vote
3
down vote

favorite











Currently, I am trying to show that the profinite completion $hatmathbbZ$ of $mathbbZ$ is isomorphic to $prod_p mathbbZ_p$ (as topological groups) where $p$ runs through all prime numbers and $mathbbZ_p$ denotes the ring of $p$-adic integers.



By definition, we know that $hatmathbbZ$ is the inverse limit of $mathbbZ/nmathbbZ$ and $mathbbZ_p$ is the inverse limit of $mathbbZ/p^nmathbbZ$. Furthermore, the Chinese Remainder Theorem tells us that if $n = prod_p p^e_p(n)$ denotes the prime factorization of $n$, then we have $$ mathbbZ/nmathbbZ simeq prod_p mathbbZ/p^e_p(n) mathbbZ.$$



So I would like to show that
$$
hatmathbbZ = varprojlim_n mathbbZ/nmathbbZ simeq varprojlim_n prod_p mathbbZ/p^e_p(n) mathbbZ = prod_p varprojlim_n mathbbZ/p^e_p(n) mathbbZ = prod_p varprojlim_n mathbbZ/p^n mathbbZ = prod_p mathbbZ_p,
$$



but I am not sure about these steps, especially with the third and fourth step. Could you please give me an easy explanation whether this works or not? Thank you!







share|cite|improve this question












Currently, I am trying to show that the profinite completion $hatmathbbZ$ of $mathbbZ$ is isomorphic to $prod_p mathbbZ_p$ (as topological groups) where $p$ runs through all prime numbers and $mathbbZ_p$ denotes the ring of $p$-adic integers.



By definition, we know that $hatmathbbZ$ is the inverse limit of $mathbbZ/nmathbbZ$ and $mathbbZ_p$ is the inverse limit of $mathbbZ/p^nmathbbZ$. Furthermore, the Chinese Remainder Theorem tells us that if $n = prod_p p^e_p(n)$ denotes the prime factorization of $n$, then we have $$ mathbbZ/nmathbbZ simeq prod_p mathbbZ/p^e_p(n) mathbbZ.$$



So I would like to show that
$$
hatmathbbZ = varprojlim_n mathbbZ/nmathbbZ simeq varprojlim_n prod_p mathbbZ/p^e_p(n) mathbbZ = prod_p varprojlim_n mathbbZ/p^e_p(n) mathbbZ = prod_p varprojlim_n mathbbZ/p^n mathbbZ = prod_p mathbbZ_p,
$$



but I am not sure about these steps, especially with the third and fourth step. Could you please give me an easy explanation whether this works or not? Thank you!









share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Aug 24 at 9:36









Diglett

579311




579311







  • 2




    Yes, you can. A product is a particular case of an inverse limit and you can interchange inverse limits.
    – d.t.
    Aug 24 at 9:46










  • More generally you can interchange any two limits (in the categorical sense; in particular "direct limits" are not limits)
    – Max
    Aug 24 at 9:51






  • 1




    Okay, that is good to know! Could you maybe explain the second last step as well? I do not really see why this is true.
    – Diglett
    Aug 24 at 9:55












  • 2




    Yes, you can. A product is a particular case of an inverse limit and you can interchange inverse limits.
    – d.t.
    Aug 24 at 9:46










  • More generally you can interchange any two limits (in the categorical sense; in particular "direct limits" are not limits)
    – Max
    Aug 24 at 9:51






  • 1




    Okay, that is good to know! Could you maybe explain the second last step as well? I do not really see why this is true.
    – Diglett
    Aug 24 at 9:55







2




2




Yes, you can. A product is a particular case of an inverse limit and you can interchange inverse limits.
– d.t.
Aug 24 at 9:46




Yes, you can. A product is a particular case of an inverse limit and you can interchange inverse limits.
– d.t.
Aug 24 at 9:46












More generally you can interchange any two limits (in the categorical sense; in particular "direct limits" are not limits)
– Max
Aug 24 at 9:51




More generally you can interchange any two limits (in the categorical sense; in particular "direct limits" are not limits)
– Max
Aug 24 at 9:51




1




1




Okay, that is good to know! Could you maybe explain the second last step as well? I do not really see why this is true.
– Diglett
Aug 24 at 9:55




Okay, that is good to know! Could you maybe explain the second last step as well? I do not really see why this is true.
– Diglett
Aug 24 at 9:55















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