I am wondering, can a non-finitely generated ring have a finite group of units? If so, what are the examples?


Consider maybe the the polynomial ring $\Bbb Z[X_1,X_2,\dots]$ with infinitely many indeterminates. This ring is not finitely generated and the only units are $\pm 1$.

  • $\begingroup$ Great example. (If I'm not mistaken, you can replace $\mathbb{Z}$ by any reduced ring $R$ whose unit group is finite too.) $\endgroup$ Nov 22 '19 at 19:03
  • 1
    $\begingroup$ @Alex I think it has to be an integral domain. There can probably be polynomials that are units if $R$ has zero divisors. $\endgroup$ Nov 22 '19 at 19:37
  • $\begingroup$ @MattSamuel: if $R$ is a commutative unital ring, then the units of $R[X]$ are those polynomials whose constant term is a unit, and whose higher terms have nilpotent coefficients. Hence, if $R$ is reduced, then $(R[X])^{\times}$ can be identified with $R^{\times}$. I am pretty sure a simple induction argument then shows that $(R[X_{1}, \ldots, X_{n}])^{\times} = R^{\times}$ for $R$ reduced, since we can view $R[X_{1}, \ldots, X_{n}]$ as $(R[X_{1}, \ldots, X_{n-1}])[X_{n}]$. I don't think infinitely many variables poses a problem either, since any two elements of $R[X_{1}, X_{2}, \ldots]$... $\endgroup$ Nov 23 '19 at 8:12
  • $\begingroup$ ...must live in some subring in finitely many variables, so if $f, g \in R[X_{1}, \ldots, X_{k}] \subset R[X_{1}, X_{2}, \ldots]$ such that $fg = 1$, then $f,g \in (R[X_{1}, \ldots, X_{k}])^{\times} = R^{\times}$. I'd welcome any corrections if you see some mistake I've made! $\endgroup$ Nov 23 '19 at 8:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.