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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 ?

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2 Answers 2

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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.

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  • $\begingroup$ I understand. Thanks so much for your answer. $\endgroup$ Dec 1, 2020 at 22:51
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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 $\Bbb{Z}[x]/(x^2-nx)$.

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