Let $m,n \in \mathbb{Z} \setminus \lbrace 0 \rbrace $, consider $$\varphi: \begin{cases} \mathbb{Z}_{/<m \cdot n >} &\longrightarrow \mathbb{Z}_{/<m>} \times \mathbb{Z}_{/<n>} \\ [a]_{mn} &\longmapsto ([a]_m, [a]_n) \end{cases} $$ find the kernel of $\varphi$
My approach: I managed to show that $\varphi$ is well-defined, that is the mapping is independent of the choice of representatives. Also I managed to show that the mapping defines a ring morphism. Thus the question about the kernel becomes relevant.
However I am utterly lost when dealing with the kernel of this mapping. In fact I don't even know if there is a general approach to this kind of problem since it's my first time dealing with ring morphisms.
Do I just make an assumption and than show that I was right? Meaning verify both inclusions $\subset$ and $\supset$.
Since $[a]_{mn} := \lbrace b \in \mathbb{Z} \mid \exists k \in \mathbb{Z}: b=a+k\cdot(mn) \rbrace $ my first guess for $\ker$ would have been that $\ker \varphi = [e]$ but that is more out of the blue then anything else. Is there any method or way I could think about this problem?
I'd appreciate some hints to get me started.
Update: After plugging in some numbers for $n,m$ and playing with coprime and non coprime integers I came up with the following statement $$ \ker \varphi = \lbrace k \cdot \text{lcm}(m,n) \mid k \in \mathbb{Z} \text{ and } m,n \in \mathbb{Z} \setminus \lbrace0 \rbrace \rbrace=:A $$
Now my problem lies in verifying the above.
Attempt: " $\supset$" Let $x \in A \implies \exists k \in \mathbb{Z}$ and $m,n \in \mathbb{Z} \setminus \lbrace 0 \rbrace$ such that $x=k \cdot \text{lcm}(m,n)$. So that means that $m$ divides $x/k$ and also $n$ divides $x/k$ thus: $$\exists h,q \in \mathbb{Z}: hm =\frac{x}{k}, qn = \frac{x}{k} $$ I notice that the $k$ seems to be dangerous here, need to make sure that it is not equal to zero. However this means by definition that $$x \in [0]_n \text{ and } x \in [0]_m \overset{?}\implies x \in [0]_{nm} \implies x \in \ker \varphi $$
For "$\subset$" let $x \in \ker \varphi$ that is $\varphi x = ([0]_m, [0]_n) \in \mathbb{Z}_{/<m>} \times \mathbb{Z}_{/<n>} $ but I know from a lemma that morphisms send neutral elements to neutral elements, hence $x=[0]_{nm} \implies \exists k \in \mathbb{Z}: x = kmn$ so $m$ divides $x$ and also $n$ divides $x \implies x \in A$.
The $"\subset"$ part seems right to me, although we never even introduced the lcm in class, hence it feels alien to me to work with it. The $"\supset"$ part can't be right I'm afraid, especially the implication indicated by a ?.