If a ring $R$ has 1 then I can see how we can always form a homomorphism $\phi: \mathbb{Z} \rightarrow R$ by sending 0 to 0 and 1 to 1. However, is there an example of a ring $R$ without identity for which there is not homomorphism from $\mathbb{Z}$ to $R?$

  • $\begingroup$ Homomorphism of groups will always exists and is determined by the value of $1$ (in our case you send 1 to 1 which is fine but you can choose any other element). You should define what's an homomorphism for you, do you require $\phi(xy)=\phi(x)\phi(y)$? $\endgroup$ – Yanko Feb 11 '18 at 21:10
  • $\begingroup$ Oh a ring homomorphism so $\phi(xy) = \phi(x)\phi(y)$ and $\phi(x + y) = \phi(x) + \phi(y).$ $\endgroup$ – 伽罗瓦 Feb 11 '18 at 21:11

If you're working with rings without identity and don't need your homomorphisms to preserve 1 then there is always the zero map which is a homomorphism of rings without units.

There are rings such that the zero map is the only homomorphism: e.g. the only map $\varphi: \mathbb{Z} \rightarrow 2 \mathbb{Z}$ is the zero map, because we have $\varphi(1) = \varphi (1)^2$ and in $2 \mathbb{Z}$ the only solution to this is $0$.


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