If a is a multiple of b, then prove $\mathbf{Z}_a /\langle b \rangle = \mathbf{Z}_b$
I think I get the intuition behind this. When I play with toy examples this makes sense $\mathbf{Z}_4 /\langle 2 \rangle = \mathbf{Z}_2$ or even $\mathbf{Z}_{18} /\langle 2 \rangle = \mathbf{Z}_2$ makes sense.
But I am pretty lost at proving this.
I see that I am suppose to use Ring Homomorphism Properties, specifically:
$Z/\ker \phi \cong S$
So I guess the game becomes how to to apply this theorem? Which would mean we would have to find a ring morphism of $\mathbf{Z}_a \rightarrow \mathbf{Z}_b$ with a $\ker \phi = \langle b \rangle$
I guess since there is a Theorem that states "Let R be a ring with unity 1. If 1 has order n under addition then the characteristic of R is n." then $\ker \phi = \langle b \rangle$.
So then we would have:
$Z/\ker \phi \cong S \rightarrow Z/\langle b \rangle = \mathbf{Z}_b$
Not really sure what I should do from here.
I don't know if this is really even a correct way to prove this.
I have a feeling that my intuition of why this works VS the correct to prove this may not be aligned.
Just consider $\phi(n) = n \bmod b$. Prove that it's well defined, is an epimorphism and its kernel is $\langle b\rangle$.