# Example of a subgroup (of a ring's additive group) that is not an ideal?

I'm looking for examples of a ring $R$ together with a subgroup $S$ of its additive group such that $S$ is not an ideal of $R$.

I imagine that there are examples of this (otherwise the definition of an ideal in a ring would be simpler than it is), but I can't readily come up any of them.

Also, I have purposely kept the specification of the ring $R$ vague so as not to constrain the answer unnecessarily. IOW, I don't require $R$ to be commutative. In fact, I don't even require it to have a multiplicative unit.

• Any additive subgroup of a field will do. For example $\mathbb{Z}$ as a subset of the rationals Commented Jan 5, 2018 at 11:28
• @leibnewtz: Thanks! I'll gladly accept your comment as the answer to my question if you care to post it as such.
– kjo
Commented Jan 5, 2018 at 12:32

A good exercise is the following : if any additive subgroup of the ring (with identity) $R$ is an ideal, then $R\simeq \mathbb{Z}/n\mathbb{Z}$ for some $n\geq 0$. So almost any ring has such a subgroup.
Therefore there are many examples : take your favourite prime number $p$, then $\mathbb{F}_p \subset \mathbb{F}_{p^2}$ is a subgroup that is not an ideal.
Similarly you have $\mathbb{Z}\subset \mathbb{Z}[\frac{1}{2}] \subset \mathbb{Q}\subset \mathbb{C}$.
Actually if you solve the above exercise in the "standard way", you will have a systematic way of producing examples. For the sake of generality, I'll provide examples that don't come from this systematic way : $\mathbb{F}_{p^2} \subset \mathbb{F}_{p^4}, X^2\mathbb{Z}[X^2] \subset \mathbb{Z}[X^2]\subset \mathbb{Z}[X]$.
The integers $\mathbb{Z} \subseteq \mathbb{Q}$ is a counterexample. For $k$ a field, one could also take any finite dimensional subspace lying in $k[x]$.
In the examples so far, the rings under consideration are actually fields. For a simple example of a ring that is not a field, notice that the group $$(\mathbf{Z},+)$$ is a subgroup but not an ideal of $$\mathbf{Z}[x]$$.