Order of a principal ideal Suppose $I$ be an ideal of a ring $R$ such that $|I| = 33$. From this information, can we conclude that $I$ is principal?
I understand that with respect to addition, $I$ is a cyclic group here. This shows that $R$ is a commutative ring. From this stage I am not being able to move further.
 A: No, you cannot conclude that, at least without some mild assumptions.  Give the additive group $I = \mathbb{Z}_{33}$ the trivial multiplication $ab=0$ for all $a, b$.  Form any direct product of rings $I \times J$ and call this $R$.  Then $I$ (or really, $I \times \{0\}$) is an ideal of order $33$ in $R$, but it cannot be principal.  If $I \times \{0\}$ were principally generated by $(a,0)$, then products of the form $(x,y)(a,0)$ would eventually give all elements of $I \times \{0\}$ after multiplying by all elements $(x,y) \in R$.  But the first slot of such a product is always $xa=0$, so such products only generate the zero element (and are therefore missing 32 others).
A: If your ring has identity, then yes.
The abelian group structure of $I$ would have to be $\mathbb Z_3\times\mathbb Z_{11} $, and therefore it is cyclic of order $33$. 
So if $c$ is an element of additive order $33$, it is already enough that $\{n\cdot c\mid n\in \mathbb Z\}=I$.
This can break down when $I$ isn't a cyclic group. For example, in $F_2[x,y]/(x,y)^2$, where $(x,y)/(x,y)^2$ is not principal, the ideal has the underlying abelian group $F_2\times F_2$.
