Show that $\mathbb{Z}/(m_1) \otimes_{\mathbb{Z}/(n)} \mathbb{Z}/(m_2) \cong \mathbb{Z}/(m_1,m_2)$ 
Let $m_1,m_2\in \mathbb{N}$ s.t. $m_1\mid n$ and $m_2\mid n$. By that,
both $\mathbb{Z}/(m_1)$ and $\mathbb{Z}/(m_2)$ are
$\mathbb{Z}/(n)$-Algebras. Show that $$\mathbb{Z}/(m_1) \otimes_{\mathbb{Z}/(n)} \mathbb{Z}/(m_2) \cong \mathbb{Z}/(m_1,m_2)$$

Please see my edit below
Hi, i was working on this problem and would like to have some feedback and maybe some general advice regarding my approach (since i didn't manage to solve it).
My attempt:
I tried to invoke the universal property of the tensor product by first defining a map
\begin{align*} \varphi \colon \mathbb{Z}/(m_1) \times \mathbb{Z}/(m_2) &\to \mathbb{Z}/(m_1,m_2) \\ (z_1+(m_1),z_2+(m_2)) &\mapsto z_1z_2 + (m_1,m_2) \end{align*}
This map is bilinear, thus it induces a unique map
\begin{align*} \widetilde{\varphi} \colon \mathbb{Z}/(m_1) \otimes \mathbb{Z}/(m_2) &\to \mathbb{Z}/(m_1,m_2) \end{align*}
which is indeed well defined (by the universal property of the tensor product).
I was then trying to somehow naively guess an inverse and my attempt was trying something like
\begin{align*} \psi \colon \mathbb{Z}/(m_1,m_2) &\to \mathbb{Z}/(m_1) \otimes \mathbb{Z}/(m_2) \\ z + (m_1,m_2) &\mapsto z+(m_1)\otimes z+(m_2)\end{align*}
where $\overline{z_i} = z_i + (m_i)$ are the respective cosets.
However, i didnt succeed in proving well definedness and since the solution we were given to this exercise was in fact quite different to what i've tried, i wanted to ask whether my attempt is actually reasonable of if i went completely the wrong way.
Question 1: Is this attempt fine so far? Or should i have thought about something completely different while appempting to solve this problem?
Question 2: If my attempt was not reasonable, would you mind elaborating how i should have approached this problem instead?
Question 3: What role does the property of being an $\mathbb{Z}/(n)$-Algebra play here actually? I've treated everything as if i'd be working with modules, is this ok?
Thanks for any help!

Edit
I figured the following maps would actually do the trick:
\begin{align*} \varphi \colon \mathbb{Z}/(m_1) \otimes \mathbb{Z}/(m_2) &\to \mathbb{Z}/(m_1,m_2)\\ \left(z + (m_1)\right)\otimes \left(1 + (m_2) \right)&\mapsto z\cdot 1 + (m_1,m_2)\end{align*}
with the inverse map
\begin{align*} \psi \colon \mathbb{Z}/(m_1,m_2) &\to \mathbb{Z}/(m_1) \otimes \mathbb{Z}/(m_2) \\ z + (m_1,m_2) &\mapsto \left(z+(m_1)\right)\otimes \left(1+(m_2)\right)\end{align*}
It obviously holds that $$\varphi\circ\psi = \operatorname{id},\ \psi\circ\varphi = \operatorname{id}$$
Is my solution correct?
 A: Question 1: "Is this attempt fine so far? Or should i have thought about something completely different while appempting to solve this problem?"
Question 2: "If my attempt was not reasonable, would you mind elaborating how i should have approached this problem instead?"
Question 3: What role does the property of being an $Z/(n)$
-Algebra play here actually? I've treated everything as if i'd be working with modules, is this ok?
Answer Q3: Here is another approach using a formula for the tensor product: Since $n=k_1m_1$ it follows $\mathbb{Z}/(n)\cong \mathbb{Z}/(n)/(\overline{m_1})$ where $\overline{m_1}\subseteq \mathbb{Z}/(n)$ is the equivalence class of $m_1$. There is a general formula saying
$$FT.\text{  }A/I\otimes_A M \cong M/IM$$
for any ideal $I \subseteq A$ and any left $A$-module $M$. You may describe $\mathbb{Z}/(m_1) \cong (\mathbb{Z}/(n))/(\overline{m_1})$ since $n=k_1m_1$ and use $FT$.
From this you get isomorphisms
$$\mathbb{Z}(m_1)\otimes_{\mathbb{Z}/(n)} \mathbb{Z}/(m_2) \cong (\mathbb{Z}/(n))/(\overline{m_1})) \otimes_{\mathbb{Z}/(n)} \mathbb{Z}/(m_2) \cong $$
$$(\mathbb{Z}/(m_2))/\overline{m_1}\mathbb{Z}/(m_2) \cong \mathbb{Z}/(m_1,m_2).$$
Since there is an inclusion of ideals $(n) \subseteq (m_1)$, there is an exact sequence
$$0 \rightarrow (m_1)/(n):=(\overline{m_1}) \rightarrow \mathbb{Z}/(n) \rightarrow \mathbb{Z}/(m_1) \rightarrow 0.$$
A: Combining your two approaches, let's take \begin{align*}\Phi: \mathbb Z/(m_1)\otimes_{\mathbb Z/(n)}\mathbb Z/(m_2)&\to\mathbb Z/(m_1,m_2)\\(z_1+(m_1))\otimes_{\mathbb Z/(n)}(z_2+(m_2))&\mapsto z_1z_2+(m_1,m_2) \end{align*} and \begin{align*}\Psi:\mathbb Z/(m_1,m_2)&\to \mathbb Z/(m_1)\otimes_{\mathbb Z/(n)}\mathbb Z/(m_2)\\z+(m_1,m_2)&\mapsto(z+(m_1))\otimes_{\mathbb Z/(n)}(1+(m_2))\end{align*}
(I am using capital letters for the functions to avoid possible confusion with the functions in the question. Also note that I will write $\overline z$ for the residue class of $z$ when the context is clear, and also drop the indeces from tensor products.)

It is straightforward to show that $\Phi$ is a well-defined homomorphism using the universal property of tensor products. To show that $\Psi$ is well-defined, take $z,z'\in\mathbb Z$ such that $z-z'\in(m_1,m_2)$ and show that $\Psi(\overline z)=\Psi(\overline{z'}).$ It should be clear that $\Psi$ is a homomorphism.

Now, all that is left to do is to show that these two functions are inverses of each other. One direction $\Phi\circ\Psi=\mathrm{id}$ is obvious, the other one needs a little more work. Take a general element of $\mathbb Z/(m_1)\otimes\mathbb Z/(m_2)$, it has the form $z=\sum_{i=1}^n\overline x_i\otimes\overline y_i$. Then calculate \begin{align*}\Psi(\Phi(z))&=\sum_{i=1}^n\Psi(\overline{x_iy_i})\\&=\sum_{i=1}^n(x_iy_i+(m_1))\otimes(1+(m_2))\\&=\sum_{i=1}^n(y_i+(n))\cdot((x_i+(m_1))\otimes(1+(m_2)))\\&=\sum_{i=1}^n(x_i+(m_1))\otimes(y_i\cdot1+(m_2))=z,\end{align*} as desired.
