Let $D$ be an integral Domain. Show that $\langle r\rangle = \langle s\rangle$ then $s = ur$ for some unit $u \in D$ My basic question is about the notation- in regards to $\langle r\rangle$ and $\langle s\rangle$. What exactly are these things? I have seen that bracket notation in Group theory. Mainly in relation to cyclic groups and generators. 
Is it being used in the same sense here? And if so, how? Since a an Integral domain is a ring and it has two operations, which one is the generator of the set? 
 A: In this case, we have $\langle x\rangle = \{rx \mid r\in R\}$.  For (commutative) rings, this is more commonly written as $(x)$.
In general, $\langle S \rangle$ is the smallest ideal containing $S$.
A: I strongly suspect, based upon the context provided by the title as well by the body of the question, as well as my own familiarity with such matters, that the the notation $\langle r \rangle$ refers to the principal ideal generated by $r \in D$; that is,
$\langle r \rangle = \{ rd \mid d \in D \} = rD; \tag{1}$
$\langle r \rangle$ is multiplicatively generated since it is the set of products of $r$ with all elements of $D$.  Nevertheless, it is closed under addition and subtraction, viz.
$rd_1 + rd_2 = r(d_1 + d_2) \in rD, \tag{2}$
$rd_1 - rd_2 = r(d_1 - d_2) \in rD; \tag{3}$
it is also closed under multiplication by an arbitrary element of $D$: for $rd \in rD$ and $a \in D$ we have
$a(rd) = a(dr) = (ad)r \in Dr = \langle r \rangle. \tag{4}$
As far as the title question is concerned, if
$\langle r \rangle = \langle s \rangle, \tag{5}$
then since $D$ is an integral domain there is a multiplicative identity $1_D \in D$ such that $1_D a = a 1_D = a$ for all $a \in D$; thus
$r = r 1_D \in rD = \langle r \rangle = \langle s \rangle = sD; \tag{6}$
this further implies that there is some $u \in D$ with 
$r = su; \tag{7}$
reversing the roles of $r$ and $s$ shows that we also have
$s = rv \tag{8}$
for some $v \in D$; combining (7) and (8) yields
$r = us = u(rv) = (uv)r \tag{9}$
or
$(1_D - uv)r = 0; \tag{10}$
since $D$ is an integral domain, $r \ne 0$ yields
$uv = 1_D, \tag{11}$
showing $u$ and $v$ are both units in $D$.  If $r = 0$,
$\langle s \rangle = \langle r \rangle = \langle 0 \rangle = \{ 0 \}; \tag{12}$
thus since $s = 1_D s  \in \langle s \rangle$, $s = 0$ as well as $r$; we have
$r = su$ whether $u$ is a unit or not.
