In general rings, what do principal ideals look like? For some ring $R$ (no $1$ or commutativity necessary) and $a \in R$, we defined the principal ideal $(a)$ by
$$ (a) := \bigcap \{ I : a \in I \subseteq R, I \text{ is an ideal}\}.$$
Now as a homework question, we shall show that $(a)$ always is of the form
$$ (a) = \{ra + as + z\cdot a + \sum_{i=1}^m r_ias_i : r,a,r_i,s_i \in R, z\in \mathbb Z \},$$
where $z\cdot a$ is defined as repeated addition of $a$ with itself. Of course, any element of this form is contained in $(a)$, but I'm clueless as how to show the opposite inclusion. Could you give any hints on how to arrive at that any $x\in (a)$ can be written in this special sum form? 
 A: Clearly, every element in
$$ \{ra + as + z\cdot a + \sum_{i=1}^m r_ias_i : r,a,r_i,s_i \in R, z\in \mathbb Z \}$$
must lie in every ideal that contains $a$, since each of $ra$, $as$, $z\cdot a$, and each $r_ias_i$ will like in an ideal that contains $a$. Thus, this set is contained in the intersection.
In order to show that this set equals the intersection, we need to show that this set is in fact an ideal that contains $a$, because this will establish that it is one of the sets being intersected.
That the set contains $a$ follows by setting $r=s=m=0$ and $z=1$. 
Now let $ra+as+z\cdot a + \sum_{i=1}^m r_ias_i$ and $ta+au+w\cdot a + \sum_{j=1}^n t_jau_j$ be two elements of the set. Then their difference is
$$(r-t)a + a(s-w) + (z-w)\cdot a + \sum x_kay_k$$
where $k$ ranges from $1$ to $n+m$, $x_k=r_k$ and $y_k=s_k$ for $k=1,\ldots,m$, and $x_k=t_{k-m}$, $y_k = u_{k-m}$ for $k=m+1,\ldots,n$. Thus, the set is a subgroup.
Multiplying $ra + as + z\cdot a + \sum_{i=1}^m r_ias_i$ by $x$ on the left gives
$$(xr+z\cdot x)a + a0 + 0\cdot a + \sum_{i=0}^m t_iau_i$$
where $t_0=x$, $u_0 = s$, $t_i=xr_i$ for $i=1,\ldots,m$, and $u_i=s_i$ for $i=1,\ldots,m$. This is an element of the set.
Similarly, if we multiply by $x$ on the right we get
$$0a + a(sx+z\cdot x) + \sum_{i=0}^m v_iaw_i,$$
where $v_0 = r$, $w_0=x$, $v_i=r_i$ for $i=1,\ldots,m$, and $w_i = s_ix$ for $i=1,\ldots,m$. Hence, it also lies in the set.
Thus, the set in question is an ideal that contains $a$, hence it contains $(a)$. Since it is contained in the intersection, we are done.
