Proof of $(\mathbb{Z}/m\mathbb{Z}) \otimes_\mathbb{Z} (\mathbb{Z} / n \mathbb{Z}) \cong \mathbb{Z}/ \gcd(m,n)\mathbb{Z}$

The part that is missing is pretty much the essence of the following (incomplete) alternative proof.

Determine the kernel of $$ \begin{array}{rlrl} g: & \mathbb{Z} & \rightarrow & (\mathbb{Z}/m\mathbb{Z}) \otimes_\mathbb{Z} (\mathbb{Z} / n \mathbb{Z}) \\ & z & \to & z (1 \otimes 1) \end{array} $$ That is: When is it true that $z (1 \otimes 1)$ is null? Since it is true for $z \in m\mathbb{Z} \cup n\mathbb{Z}$, then you know that it is true for the ideal generated by it: $\langle \mathrm{gcd}(m,n)\rangle \subset \mathrm{ker}(g)$.

You know that the map is surjective because $1 \otimes 1$ is a generator. If you show that $\mathrm{ker}(g) \subset \langle \mathrm{gcd}(m,n) \rangle$, you will have the isomorphism you claim. This is the part you are missing. It is equivalent to showing that your $f$ is well-defined.

So, the conclusion is that it is not right.

The better way to define a homomorphism from $\mathbb{Z}/m\mathbb{Z}\otimes \mathbb{Z}/n\mathbb{Z}$ to $\mathbb{Z}/\gcd(m,n)\mathbb{Z}$ is via the universal property.

Note that the map $\mathbb{Z}/m\mathbb{Z}\times \mathbb{Z}/n\mathbb{Z}\to \mathbb{Z}/\gcd(m,n)\mathbb{Z}$ defined by $$(x+m\mathbb{Z},y+n\mathbb{Z})\mapsto xy+\gcd(m,n)\mathbb{Z}$$ is well-defined and also bi-linear, thus by the universal property of tensor product, there is a linear map $f:\mathbb{Z}/m\mathbb{Z}\otimes \mathbb{Z}/n\mathbb{Z}\to \mathbb{Z}/\gcd(m,n)\mathbb{Z}$ such that $$f(x+m\mathbb{Z}\otimes y+n\mathbb{Z})=xy+\gcd(m,n)\mathbb{Z}.$$ Verify that the linear map $g:\mathbb{Z}/\gcd(m,n)\mathbb{Z}\to\mathbb{Z}/m\mathbb{Z}\otimes\mathbb{Z}/n\mathbb{Z}$ defined by $$g(z+\gcd(m,n)\mathbb{Z})=(z+m\mathbb{Z})\otimes (1+n\mathbb{Z})$$ is well-defined, and we also have $g\circ f=1, f\circ g=1$, thus $f$ is isomprhism. To see $g$ is well-defined, you may use the equality that $\gcd(m,n)=am+bn$ for some integers $a,b\in\mathbb{Z}$.

Let $M$ be a $R$-module and let $I$ be an ideal of $R$. Then we have $\frac RI\otimes_RM\cong\frac M{IM}$.

First proof: consider the exact sequence $0\to I\to R\to\frac RI\to0$. Tensoring with $M$ and using right-exactness of the tensor product we get $I\otimes_RM\to R\otimes_RM\to\frac RI\otimes_RM\to0$. We know that $R\otimes_RM\cong M$ via $r\otimes m\mapsto rm$; under this isomorphism the image of $I\otimes_RM\to R\otimes_RM$ is precisely $IM$, and since this image is precisely the kernel of $R\otimes_RM\to\frac RI\otimes_RM$, the claim follows from the isomorphism theorem.

Second proof: the mapping $\frac RI\times M\to\frac M{IM}$ given by $(r+I,m)\mapsto rm+IM$ is well-defined and $R$-bilinear, so it induces a $R$-linear mapping $\frac RI\otimes_RM\to\frac M{IM}$, which is obviously surjective, and the inverse mapping is given by $m+IM\mapsto(1+I)\otimes m$; this mapping is well-defined because if $m\in IM$, say $m=\sum_k i_km_k$, with $i_k\in I, m_k\in M$, then by $R$-bilinearity of $\otimes$ we have $(1+I)\otimes m=\sum_k\bigl[i_k(1+I)\bigr]\otimes m_k=\sum_k0_{R/I}\otimes m_k=0$.

In general, if $N$ is a $R$-submodule of a module $M$, then

$$I\frac MN=\frac{IM+N}N\,,$$

so

$$m\mathbb Z\,\frac{\mathbb Z}{n\mathbb Z}=\frac{(m\mathbb Z)\mathbb Z+n\mathbb Z}{n\mathbb Z}=\frac{m\mathbb Z+n\mathbb Z}{n\mathbb Z}=\frac{\gcd(m,n)\mathbb Z}{n\mathbb Z}\,.$$

Finally, applying the first result we get

$$\frac{\mathbb Z}{m\mathbb Z}\otimes_{\mathbb Z}\frac{\mathbb Z}{n\mathbb Z}\cong\frac{\frac{\mathbb Z}{n\mathbb Z}}{m\mathbb Z\,\frac{\mathbb Z}{n\mathbb Z}}=\frac{\frac{\mathbb Z}{n\mathbb Z}}{\,\frac{\gcd(m,n)\mathbb Z}{n\mathbb Z}\,}\cong\frac{\mathbb Z}{\gcd(m,n)\mathbb Z}\,.$$

ADDENDUM

The same reasoning shows that for any ideals $I,J$ in $R$ we have $\frac RI\otimes_R\frac RJ\cong\frac R{I+J}$.