# Rings without identity and non-isomorphic ring structure on $\mathbb{Z}$

I am solving Exercise 2.15 from Aluffi chapter 0.

Exercise 2.15. For $$m > 0$$ , the abelian group $$(\mathbb{Z},+)$$ and $$(m\mathbb{Z},+)$$ are manifestly isomorphic: the function $$\phi : \mathbb{Z} \rightarrow m\mathbb{Z}$$ given by $$n \mapsto nm$$ is a group isomorphism. Use this isomorphism to transfer the structure of "ring without identity" $$(m\mathbb{Z},+,*)$$ back onto $$\mathbb{Z}$$: Give an explicit formula for the "multiplication" this defines on $$\mathbb{Z}$$. Explain why structures induced by different positive integer m are non-isomorphic as "rings without 1".

Solution:

We have the following map:

$$\phi : \mathbb{Z} \rightarrow m\mathbb{Z}$$ given by $$n \mapsto nm$$. We can use this map to transfer ring structure on $$\mathbb{Z}$$ as follows:

Let $$s_1,s_2 \in \mathbb{Z}$$ and set $$s_1 s_2 = m (s_1 s_2)$$. This defines ring structure on $$\mathbb{Z}$$. This ring structure is non-isomorphic because if we have $$s_1 s_2 = m (s_1 s_2) = n (s_1 s_2)$$ then we would have n = m for $$n \neq m$$. Is my my argument valid ?

You're almost there. What you showed in the end is that the identity map is not a ring homomorphism from $$\mathbb Z$$ with $$n$$ multiplication to $$\mathbb Z$$ with $$m$$ multiplication. So indeed this is not an isomorphism. However, a priori there could be some other isomorphism. However, the only group homomorphisms $$\mathbb Z \longrightarrow \mathbb Z$$ are multiplication by some fixed integer, so the only group isomorphisms are multiplication by the units $$\pm 1$$. You've shown that the identity map is not a ring isomorphism, so it remains to consider multiplication by $$-1$$. But since you restricted to $$m > 0$$, this won't work either. Hence, these rings are not isomorphic.
Also note that by distributivity a non-unital ring structure on the group $$(\Bbb{Z},+)$$ is determined by $$1 \times 1$$. For each $$n\in \Bbb{Z}$$ there is a non-unital ring structure $$(\Bbb{Z},+,\times)$$ $$a\times b = abn$$ and all the non-unital ring structures are of this form.
For $$n\ne 0$$ it embeds into the standard ring $$\Bbb{Z}$$ through $$a\to an$$.
For any $$n$$ it embeds into the unitalization $$\Bbb{Z}[x]/(x^2-nx)$$.