Ring structure of $m\mathbb Z/n\mathbb Z$ I am interested in the ring structure of $I=m\mathbb Z/n\mathbb Z$ where $m\mid n$.
I have looked at those for various $m$ and $n$ and it seems that $I\cong\mathbb Z_\frac{n}{m}$. I have tried to prove this with the homomorphism theorem by composing the embedding $\phi :m\mathbb Z\hookrightarrow \mathbb Z$ and the projection $\pi :\mathbb Z\to\mathbb Z_\frac{n}{m}$. But I'm not sure how to verify that $\ker(\pi\circ\phi)=n\mathbb Z$. 
Obviously $\phi$ is injective so $\ker(\pi\circ\phi)=\{mz\in m\mathbb Z\mid \phi(mz)\in \frac{n}{m}\mathbb Z\}$. How is this $n\mathbb Z$ though?
 A: Note If $n=4, m=2$ then $\frac{2\mathbb Z}{4 \mathbb Z}$ is a non unitary ring where all elements are nilpotent.
In general I think that $\frac{m\mathbb Z}{n \mathbb Z}$ is not unitary, unless $m=1$. In fact all its elements are zero divisors.
I think that they are isomorphic as abelian groups but not as rings.
A: In reality $\ker(\pi \circ \phi) = m\mathbb{Z} \cap \frac{n}{m}\mathbb{Z}=\text{lcm}(m,\frac{n}{m})\mathbb{Z}$, which will not be $n\mathbb{Z}$ if there is a prime $p$ dividing both $m$ and $n$ but dividing $n$ with higher exponent than $m$ (for example $n = p^2$ and $m = p$, in which case $\text{lcm}(m, \frac{n}{m}) = \text{lcm}(p, p) = p$).
For this example, you'd have the maps $$p\mathbb{Z} \xrightarrow{\phi} \mathbb{Z} \xrightarrow{\pi} \mathbb{Z}_p$$ with $\text{im}(\pi \circ \phi) = \{0\}$, so the homomorphism theorem concludes $p\mathbb{Z}/p\mathbb{Z} \cong \{0\}$ which is true, but not what you wanted.
To prove the original claim (as abelian group isomorphisms) note that $m\mathbb{Z}/n\mathbb{Z}$ is a subgroup of $\mathbb{Z}/n\mathbb{Z} = \mathbb{Z}_n$ and subgroups of cyclic groups are cyclic.
By the third isomorphism theorem we have $$\frac{\mathbb{Z}/n\mathbb{Z}}{m\mathbb{Z}/n\mathbb{Z}} \cong \mathbb{Z}/m\mathbb{Z}$$ and hence by Lagrange's theorem we find $|m\mathbb{Z}/n\mathbb{Z}| = \frac{n}{m}$.
I'd like to see a simpler proof, I am not convinced the 3rd isomorphism theorem is really necessary...
A: This approach does not quite work, since it is possible that $z$ to be a multiple of both $n/m$ and $m$ without being a multiple of $n$ - which is exactly what you're trying to show. For instance, if $n=m^2$, then being a multiple of $n/m$ and $m$ is the same as being a multiple of $m$ alone - which obviously does not imply that a number is a multiple of $m^2$.
The correct thing to do would be to compose the embedding $\phi:m\mathbb Z\rightarrow \mathbb Z$ withe the reduction $\pi:\mathbb Z\rightarrow \mathbb Z_n$. Obviously, since $\pi$ is injective, the kernel of the composition is just the kernel of $\pi$ - which is $n\mathbb Z$ - intersected with $m\mathbb Z$, which contains $n\mathbb Z$. However, you then have the issue that $\pi\circ \phi$ is not injective; its image is the elements of $\mathbb Z_n$ which are of the form $m\cdot x$ for some $x\in \mathbb Z_n$. That is, it is the ideal $(m)$ in $\mathbb Z_n$.
