Quotient of Ideals in the ring of integers My textbook defines the quotient $A:B$ of two ideals $A,B$ of a ring $R$ to be
$A:B=\{r\in R\mid rb\in A~ \forall b\in B\}$. This is equivalent to
$A:B= \{r\in R\mid rB\subseteq A\}. $
A previous exercise asked "In the ring $\mathbf{Z}$, what is $(m)(n)$?," and I proved that $(m)(n)=(mn)$. 

My current exercise asks "In the ring $\mathbf{Z}$, what is $(m):(n)$? I'm assuming this answer will be similar in form to the last one.

So far I've managed to get that if $m$ is prime then $(m)$ contains a unique element and so $(m):(n)=\{0,1\}$. I'm now stuck on how to further classify $(m):(n)$ for the integers. I'm also not really sure what kind of answer I'll be able to give.
Thanks for any help!
 A: Let us investigate the definition $A:B = \{r\mid rB\subseteq A\}$ for $A = (m), B = (n)$. 
It says that $r \in A: B$ iff $\forall x \exists y:rnx = my$. Obviously it is sufficient to find an $y_1$ for $x = 1$, since then $y_x = xy_1$.
Thus the condition to solve becomes $rn = my$, or, phrased differently, $m \mid rn$.
Now the final step is to take $n$ out of the equation by dividing by $\gcd(m,n)$: $$m \mid rn \iff \left.\frac{m}{\gcd(m,n)}~\middle\vert~ r \frac{n}{\gcd(m,n)}\right.$$
By definition of $\gcd(m,n)$, the two fractions are coprime and we are left with the defining condition: $$\frac m{\gcd(m,n)}\mid r$$
We conclude that $(m) : (n) = \left(\dfrac m{\gcd(m,n)}\right)$.
A: Let's consider a more general problem, with $A$ and $B$ arbitrary ideals of a commutative ring $R$. Then
$$
(A:B)=\{r\in R:rB\subseteq A\}
$$
Let's try our hand with
$$
(A:A+B)=\{r\in R:r(A+B)\subseteq A\}
$$
If $r\in (A:B)$, then $r(A+B)\subseteq rA+rB\subseteq A+A=A$, so $r\in(A:A+B)$. Conversely, if $r\in(A:A+B)$, then $rB\subseteq r(A+B)\subseteq A$, so $r\in(A:B)$.
Thus your problem is finding
$$\def\Z{\mathbb{Z}}
(m\Z:n\Z)=(m\Z:(m\Z+n\Z))=(m\Z:d\Z)
$$
where $d=\gcd(m,n)$.
The task is now much easier, because $m\Z\subseteq d\Z$. Saying that $r\in(m\Z:d\Z)$ is equivalent to saying that $rd\in m\Z$, that is,
$$
rd=mk
$$
for some $k$, which in turn is
$$
r=\frac{m}{d}k
$$
so $r\in(m/d)\Z$.
Therefore
$$
(m\Z:n\Z)=\frac{m}{\gcd(m,n)}\Z
$$
