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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.

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2 Answers 2

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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 $2\mathbb Z$.

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

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  • $\begingroup$ 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! $\endgroup$ Aug 30, 2020 at 9:43
  • $\begingroup$ @Bellem Hmmm... it seems to me that you are mixing notation. You are using concatenation both as addition and multiplication, no? $\endgroup$ Aug 30, 2020 at 9:47
  • $\begingroup$ I have been sloppy, I will answer better. $\endgroup$
    – Bellem
    Aug 30, 2020 at 9:55

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