$\mathbb{Z}$ is not the group of units of some ring?

So I am trying to prove that the functor taking the group of units of a ring is not essentially surjective by showing that there is no ring with group of units isomorphic to $$\mathbb{Z}$$.

I tried to prove that no such ring exists but I am not sure my proof is correct.

Proof: Assume there exists a ring $$R$$ with $$U(R) \cong \mathbb{Z}$$. So we have some group isomorphism $$\phi: \mathbb{Z} \rightarrow U(R)$$ such that $$\phi(0) = 1_R, \phi(1) = s$$ for some $$s \in U(R)$$. It follows that $$\phi(n) = \phi(1 + ... + 1) = \phi(1) \cdot \ldots \cdot \phi(1) = s^n$$. Since we assumed that $$R$$ is a ring, there exists an additive inverse -s of s. Then we see that $$(-s) (-(s^{-1})) = 1_R$$. So $$-s$$ is also a unit in $$R$$ and we have that there exists some $$k \in \mathbb{Z}$$ such that $$-s = s^k$$ since $$\phi$$ is surjective and s generates $$U(R)$$. Furthermore, $$s^{k-1} \neq -1_R$$ because then $$s^{2k-2} = 1_R$$ would be true, which contradicts injectivity of $$\phi$$ for $$k \in \mathbb{Z} - \{1\}$$. So $$0 \neq 1 + s^{k-1} = (1 + s^{k-1})(ss^{-1}) = (s + s^k)s^{-1} = (s+ (-s))s^{-1} = 0$$ which gives a contradiction so no such ring can exist for $$k \in \mathbb{Z} - \{ 1 \}$$.

What do you think? Thank you.

edit: And i guess according to the comments such rings with $$-s = s$$ can actually exist.

• Your method can not work when $-s=s$. Nov 19, 2021 at 15:53
• For a counterexample, see this MO-post. Take $R=\Bbb F_2[X,X^{-1}]$. Nov 19, 2021 at 15:57
• $\mathbb{Z}$ is not the group of units of a ring with $1+1\ne0$. Nov 19, 2021 at 16:06
• Thanks to all of you
– tor
Nov 19, 2021 at 16:28
• – lhf
Nov 19, 2021 at 18:21