I know it's a very common, very basic, baby problem but I really had no idea how to google it.
So I'm asking here, while apologizing, for a link, reference, or short explanation.

What is the isomorphism in: $$\mathbb{Z}_2\oplus\mathbb{Z}_3\cong\mathbb{Z}_6$$ (and is it a ring morphism)?

  • 3
    $\begingroup$ Yes, it's a ring isomorphism. This is a very special case of the Chinese remainder theorem. It's easier to define the map that goes the other way first. $\endgroup$ – Qiaochu Yuan Mar 7 '17 at 19:47
  • $\begingroup$ Oh ok so this is precisely the chinese remainder theorem. $\endgroup$ – C-Star-W-Star Mar 7 '17 at 19:52

The natural homomorphism \begin{align} \mathbf Z &\longrightarrow \mathbf Z/2\mathbf Z \times\mathbf Z/3\mathbf Z \\ n&\longmapsto(n\bmod2,n\bmod3) \end{align} has kernel $6\mathbf Z $, hence induces an isomorphism: \begin{align} \mathbf Z/6\mathbf Z &\longrightarrow \mathbf Z/2\mathbf Z \times\mathbf Z/32\mathbf Z \\ n\bmod6&\longmapsto(n\bmod2,n\bmod3) \end{align} The inverse isomorphism can be defined in terms of a Bézout's identity: if $2u+3v=1$ is such a relation, say $-2+3=1$, the inverse isomorphism is defined as follows: \begin{align} \mathbf Z/2\mathbf Z \times\mathbf Z/3\mathbf Z &\longrightarrow \mathbf Z/6\mathbf Z \\ (a\bmod2,b\bmod3)&\longmapsto 2ub+3va\bmod6 \enspace(=3a-2b\bmod6) \end{align}

  • $\begingroup$ "hence induces an isomorphism" because both sides have the same number of elements. $\endgroup$ – lhf Mar 7 '17 at 22:07
  • 2
    $\begingroup$ Rather because the kernel is $6\mathbf Z $. The isomorphism does not rely on the finiteness of the ring: it is valid for the abstract Chinese remainder theorem. $\endgroup$ – Bernard Mar 7 '17 at 22:13
  • $\begingroup$ It is not immediately clear that the natural homomorphism is surjective. $\endgroup$ – lhf Mar 7 '17 at 23:27
  • $\begingroup$ It uses Bézout's identity. Have you solved simultaneous congruences? Another argument is an inverse map is given. $\endgroup$ – Bernard Mar 7 '17 at 23:31
  • $\begingroup$ @Bernard: That is so simple with the kernel 6Z, thank you!!! :) $\endgroup$ – C-Star-W-Star Mar 8 '17 at 8:02

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.