Q:

$I+I\neq 2I$ and some pedantry about ideal/ring operations.

I've realised there is some unfortunate notation when it comes to denoting multiples of an ideal in ring theory.
For example, if we take the ideal $I=\mathbb{Z}$ in the ring $\mathbb{Z}$,
then $I+I=I=\mathbb{Z}\neq2\mathbb{Z} = 2I$.
Could anyone direct me to some kind of text which clears up the clash of notation or at least comments on it?
Remark: It would seem that $H+H=H$ holds for any subgroup of a group $(G,+)$, since $0 \in H$ and $H$ is closed under addition. On the other hand, to define $2I=\{2i \ | i \in I\}$ for the ideal $I$ of ring $R$, one needs $2\in R$, say, $R= \mathbb{Z}$ or $\mathbb{Q}$ or $\mathbb{Z}[x]$. The confusion with $I+I\neq 2I$ happens if and only if this is the case, i.e., $2\in R$. Is this reasoning correct?

A:

$2i$ can be defined as $i+i$ in any ring, even if there is no element $2$ (i.e. the ring might not have an identity).  The issue with $I+I \ne 2I$ is that not every member of the ideal might be expressible as $i+i$, as indeed is the case with $\mathbb Z$.  In fancier language, the additive group of the ideal might not be $2$-divisible.  See divisible group.

