# Are the ideals of a ring with cyclic additive group always principal?

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.

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 $$2\mathbb Z$$.

• I'm a bit confused. $8\mathbb{Z}$ is principal in $4\mathbb{Z}$ isn't it? Aug 30, 2020 at 9:54
• What ring element would generate it? Aug 30, 2020 at 9:58
• (just to cross word limit) 8? Aug 30, 2020 at 9:59
• Within $4\mathbb Z$, the ideal $(8)=32\mathbb Z\neq 8\mathbb Z$. Aug 30, 2020 at 10:00
• Oh! Nice! I see it now, thanks! Aug 30, 2020 at 10:02

Well, an ideal is an additive subgroup of the given ring. If the additive structure of the ideal is cyclic, then each element of the ideal can be written as $$rg$$, where $$r\in R$$ and $$g$$ is a generator of the additive cyclic structure. Hence, the ideal is principal.

• How do you prove the claim "If the additive structure of the ideal is cyclic, then each element of the ideal can be written as $rg$..."? This is equivalent to saying if $I$ is cyclic then it is principal, which is my question. Many thanks! Aug 30, 2020 at 9:43
• @Bellem Hmmm... it seems to me that you are mixing notation. You are using concatenation both as addition and multiplication, no? Aug 30, 2020 at 9:47
• I have been sloppy, I will answer better. Aug 30, 2020 at 9:55