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

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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 $\pm 1$.

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  • $\begingroup$ 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
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    $\begingroup$ @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

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