How can the rng $(\mathbb{N}, +, \cdot)$ be an ideal of some ring?

Question

It is known than rng is an ideal of some ring. But how can rng $(\mathbb{N}, +, \cdot)$ be an ideal to some ring? As ring has an inverse element, the first ring we get from $\mathbb{N}$ is $\mathbb{Z}$, and it seems impossible to imagine that $\mathbb{N}$ will be an ideal to $\mathbb{Z}$, for $(-3) \cdot (2) = -6$, which is not in $\mathbb{N}$.

Answer 1

It's because $(\mathbb{N}, +, \cdot)$ is not a rng. By definition if it were a rng, then $(\mathbb{N}, +)$ would be an abelian group, which it isn't: it does not have inverses.

However it is true that if $I \subset R$ is an ideal of a ring $R$, then $I$ is a rng (it's straightforward to check). And if $R$ is a rng, then $R_+ = R \oplus \mathbb{Z} \mathbf{1}$ is a ring (where $\mathbf{1}$ is the identity and the addition/multiplication extend that of $R$), and $R$ is an ideal of $R_+$.