# What is the isomorphism in: $\mathbb{Z}_2\oplus\mathbb{Z}_3\cong\mathbb{Z}_6$

Disclaimer..
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)?

• 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. – Qiaochu Yuan Mar 7 '17 at 19:47
• Oh ok so this is precisely the chinese remainder theorem. – 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}
• 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. – Bernard Mar 7 '17 at 22:13