Here is the example:
$(3)$ In general, $$ \mathbb Z_m \otimes_Z \mathbb Z_n \cong \mathbb Z/d\mathbb Z,$$where $d$ is the $g.c.d$ of the integers $m$ and $n.$ To see this, observe first that $$a \otimes b = a \otimes (b.1) = (ab) \otimes 1 = ab (1 \otimes 1),$$ from which it follows that $ \mathbb Z_m \otimes_Z \mathbb Z_n$ is a cyclic group with $1 \otimes 1$ as generator. Since $m (1 \otimes 1) = m \otimes 1 = 0 \otimes 1= 0 $ and similarly $n (1 \otimes 1) = 1 \otimes n = 0,$ we have $d (1 \otimes 1) = 0,$ so the cyclic group has order dividing $d.$ The map \varphi: $ \mathbb Z_m \times \mathbb Z_n \rightarrow \mathbb Z/d\mathbb Z$ defined by $\varphi(a \operatorname{mod}m, b \operatorname{mod}n) = ab \operatorname{mod} d$ is well defined since $d$ divides both $m$ and $n.$ It is clearly $\Bbb Z$-bilinear. The induced map $\Phi: \mathbb Z_m \otimes_Z \mathbb Z_n \rightarrow \mathbb Z/d\mathbb Z$ from corollary 12 maps $1 \otimes 1$ to the element $1 \in \mathbb Z/d\mathbb Z,$ which is an element of order $d.$ In particular $\mathbb Z_m \otimes_Z \mathbb Z_n$ has order at least $d.$ Hence $1 \otimes 1$ is an element of order $d$ and $\Phi$ gives an isomorphism $$ \mathbb Z_m \otimes_Z \mathbb Z_n \cong \mathbb Z/d\mathbb Z.$$
My questions are:
1- How can I prove that $\Phi$ is an isomorphism? what is the definition of it? what is a good candidate for the definition of its inverse?
2- In the fourth line, why we can move the $b$ between the first and second coordinate of the tensor product? I know that this is correct for elements of ring $R$ only i.e. $(rm) \otimes n = m \otimes (rn)$ but I do not know that we can also move the $m,n$'s.
3- Why $d(1 \otimes 1) = 0$?
4-why the cyclic group has order $d$ ? Is not the order of the group is its cardinality? why our group has $d$ elements? I can not see any proof to that.
5- Why $d$ divides both $m$ and $n$ leads to that the map $\varphi$ is well defined?
Answers for those queries will be greatly appreciated.
1.) Dummit and Foote give the definition of $\Phi$ in the same sentence as it first appears. Explicitly, they define $\Phi : (\mathbb Z / m \mathbb Z) \otimes_{\mathbb Z} (\mathbb Z / n \mathbb Z) \to \mathbb Z / d \mathbb Z$ by $\Phi[(1 + m \mathbb Z) \otimes_{\mathbb Z} (1 + n \mathbb Z)] = 1 + d \mathbb Z.$
2.) Each of the quotient rings is an abelian group (i.e., a $\mathbb Z$-module), and the tensor product is considered with respect to $\mathbb Z,$ so it is $\mathbb Z$-bilinear. Check the definition of $- \otimes_{\mathbb Z} -.$
3.) By definition, $d$ is the greatest common divisor of $m$ and $n,$ hence in particular, there exist integers $k$ and $\ell$ such that $mk + n \ell = d.$ Consequently, we have that $d + m \mathbb Z = n \ell + m \mathbb Z$ so that $$d[(1 + m \mathbb Z) \otimes_{\mathbb Z} (1 + n \mathbb Z)] = (d + m \mathbb Z) \otimes_{\mathbb Z} (1 + n \mathbb Z) = (n \ell + m \mathbb Z) \otimes_{\mathbb Z} (1 + n \mathbb Z).$$ Can you see why this is equal to $0?$
4.) Once you have shown that $(1 + m \mathbb Z) \otimes_{\mathbb Z} (1 + n \mathbb Z)$ generates $(\mathbb Z / m \mathbb Z) \otimes_{\mathbb Z} (\mathbb Z / n \mathbb Z)$ as an abelian group, it suffices to find the order of $(1 + m \mathbb Z) \otimes_{\mathbb Z} (1 + n \mathbb Z).$ (Why?) By (3.) above, it is $d.$
5.) Consider the map $\varphi : (\mathbb Z / m \mathbb Z) \times (\mathbb Z / n \mathbb Z) \to \mathbb Z / d \mathbb Z$ defined by $\varphi(a + m \mathbb Z, b + n \mathbb Z) = ab + d \mathbb Z.$ Given that $a + m \mathbb Z = a' + m \mathbb Z$ and $b + n \mathbb Z = b' + n \mathbb Z,$ we must show that $ab + d \mathbb Z = a' b' + d \mathbb Z.$ By hypothesis, we have that $m \,|\, (a - a')$ and $n \,|\, (b - b')$ so that $a - a' = mk$ and $b - b' = n \ell$ for some integers $k$ and $\ell.$ Using these two identities, can you verify that $ab + d \mathbb Z = a' b' + d \mathbb Z$?
One other way to see this is to prove the following more general fact.
Fact. If $R$ is a commutative unital ring with ideals $I$ and $J,$ then $$\frac R I \otimes_R \frac R J \cong \frac R {I + J}.$$
To prove this, you might show that for any $R$-module $M,$ we have that $(R / I) \otimes_R M \cong M / IM.$ I believe that you have already proved this in a separate exercise. From this, it follows that $(R / I) \otimes_R (R / J) \cong (R / J) / [I(R / J)].$ It suffices to prove that $I(R / J) = (I + J)/J.$
For our particular case, we have that $R = \mathbb Z,$ $I = m \mathbb Z,$ and $J = n \mathbb Z.$ In a Euclidean domain $R,$ we have that $aR + bR = \gcd(a, b)R.$ Considering that $\mathbb Z$ is a Euclidean domain, it follows that $m \mathbb Z + n \mathbb Z = d \mathbb Z.$ Putting this all together gives the desired result.