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.

  • 2
    $\begingroup$ Your method can not work when $-s=s$. $\endgroup$
    – Zerox
    Nov 19, 2021 at 15:53
  • 6
    $\begingroup$ For a counterexample, see this MO-post. Take $R=\Bbb F_2[X,X^{-1}]$. $\endgroup$ Nov 19, 2021 at 15:57
  • 2
    $\begingroup$ $\mathbb{Z}$ is not the group of units of a ring with $1+1\ne0$. $\endgroup$
    – user26857
    Nov 19, 2021 at 16:06
  • $\begingroup$ Thanks to all of you $\endgroup$
    – tor
    Nov 19, 2021 at 16:28
  • 1
    $\begingroup$ See math.stackexchange.com/a/384492/589 $\endgroup$
    – lhf
    Nov 19, 2021 at 18:21


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