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I have seen people mentioning that this equation is a consequence of the Chinese Remainder Theorem in some old posts, but I did not quite understand how. Can anyone elaborate on that?

Thanks in advance.

Kevin
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  • Could you please link a post where it referes to chinese remainder theorem.. – AgentS Nov 17 '19 at 05:02
  • It comes from the constructive proof of the theorem. Look at this page. – David Nov 17 '19 at 05:14
  • Where do you see the gcd * lcm law there? For ideals in a domain it is equivalent to one form of CRT, see $(27)$ and $(14)$ in these characterizations of Prufer domains. – Bill Dubuque Nov 17 '19 at 16:09
  • The use of Chinese Remainder Theorem is mentioned here: https://math.stackexchange.com/questions/470807/prove-that-gcdm-n-times-mboxlcmm-n-m-times-n/470812 – Kevin Nov 17 '19 at 16:32
  • But no one seem to have elaborated on that. – Kevin Nov 17 '19 at 16:33
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    See also https://math.stackexchange.com/questions/2205618/proving-that-mathbbz-m-oplus-mathbbz-n-cong-mathbbz-d-oplus-mathbbz – lhf Nov 17 '19 at 18:27
  • See also this ancient question – Bill Dubuque Nov 18 '19 at 16:34
  • Even though the answer by Lukas Heger in the link given by Ihf proves $\operatorname{gcd}(m,n)\cdot\operatorname{lcm}(m,n)=m\cdot n$ by mentioning the Chinese Remainder Theorem, the CRT can be avoided. Namely, the answer proves the stronger result $\mathbb{Z}_m\oplus \mathbb{Z}_n \cong \mathbb{Z}_d\oplus \mathbb{Z}_l $ as groups, where $l=\mathrm{lcm}(m,n)$ and $d=\gcd(m,n)$. The proof uses the CRT and the identity $\operatorname{max}(m,n)+\operatorname{min}(m,n)=m+n$. Now, if you are only interested in proving $\operatorname{gcd}(m,n)\operatorname{lcm}(m,n)=m\cdot n$, … – Max Demirdilek Apr 15 '23 at 06:10
  • … you only need the identity $\operatorname{max}(m,n)+\operatorname{min}(m,n)=m+n$. Employing the CRT is an unnecessary detour. – Max Demirdilek Apr 15 '23 at 06:11

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