Note for me rings need not be unital or commutative.
Let $R$ be a ring with cyclic additive group $(R, +, 0)$ and let $I$ be an ideal in $R$. Is $I$ principal?
Here's my attempt, assuming $R$ has a $1$ and $1$ generates the additive group $(R,+,0)$:
Since $(R,+,0)$ is cyclic and $(I,+,0)$ is an additive subgroup of $(R,+,0)$, it is also cyclic and generated by some $a \in R$. Best guess is $I = (a)$.
By definition, as sets $(I, +, 0 ) = (\langle a \rangle , +, 0) \subseteq (a)$ . Also if $x \in (a)$ then $x = \sum _i r_i a s_i$ for some $r_i, s_i$. Hence ( using poor notation)
$x = \sum_i r_i a (1+…+1) = \sum_i r_i (a+…+a) \\
= \sum_i (1+…+1) (a+…+a) = \sum_i ((a+…+a) +… +(a+…+a)) \in (\langle a \rangle, +, 0)$.
By double inclusion we have the desired equality. $ \blacksquare$
Firstly is this correct and also what about the case where $R$ is not unital or the case where $R$ is unital but $1$ doesn't generate the additive group?
Many thanks!
EDIT:
For future reference. It is argued here Does the unit generate the additive group in a unital ring with cyclic additive group? that the condition that $1$ generates the additive group is infact implied by $R$ being unital and is therefore not needed.
Best Answer
Edit: Assuming that we characterize principal ideals as those of the form $aR$, which is not necessarily compatible with any widely used definition, so take this with a grain of salt...
Not necessarily. Consider the ideal $8\mathbb Z$ within the ring $4\mathbb Z$.
Edit: Or, maybe this one is clearer: consider the ideal $6\mathbb Z$ within the ring $2\mathbb Z$.