Is the unit group of any finitely generated reduced $\Bbb Z$ algebra finitely generated?

If $A$ is finitely generated commutative reduced $\Bbb Z$ algebra, must the unit group $A^{\times}$ be finitely generated?

The question is motivated by the Dirichlet unit theorem which says the unit group of the algebraic integer ring of any number field is finitely generated. And for other $\Bbb Z$ algebras such as finite fields, their unit groups are even finite.

• Even if it is so, the structure of unit groups is very complicated even for group rings, e. g. $\mathcal U (\Bbb Z S_3)$ contains free group on 2 generators. – xsnl Sep 26 '17 at 10:18
• Partial solution: if algebra $A$ is an order in f. d. $\Bbb Q$-algebra, in particular an integral group ring of finite group, then $\mathcal U (A)$ is an arithmetic group which is finitely generated by Borel-Harish Chandra. Good reference is E. Kleinert. Units in classical orders: a survey, 1994. – xsnl Sep 26 '17 at 10:30
• @xsnl I think the case that $A$ is finite as $\Bbb Z$ module is always true. Firstly, as there are only finitely many minimal primes we can assume $A$ is an integer domain. Also we can assume $char A=0$ (or $A$ is a finite set), then $A$ must be an order in number field hence we are done. What you say is interesting, because it's a result about noncommutative ring which I am not quite familiar with.. – sawdada Sep 26 '17 at 13:47
• I think I need to assume $A$ is reduced, because my friends find that $\Bbb Z[x,y]/(y^2)$ is a counterexample. – sawdada Sep 27 '17 at 7:28
• – Watson Dec 4 '18 at 19:27

"I think the case that $A$ is finite as $\mathbf Z$-module is always true".
Yes, it's true, it's even presented as "a generalization of the unit theorem" in §4.7 of P. Samuel's booklet on ANT. The particular case that $A$ is an integer domain is easy, because then, as you said, $A$ would be an order of some number field in characteristic $0$, or $A$ would be finite in non zero characteristic.
But what worries me is your hint (which I can't quite grasp) at the finite number of minimal primes of $A$ to reduce to that particular case, whereas Samuel feels obliged to go on with a technical inductive proof on the nilradical $N$ of $A$. More precisely, the induction bears on the exponent $s$ such that $N^s = (0)$. The starting step is $s=0$, i.e. $A$ is a reduced ring in which ($0$) is the intersection of finitely many prime ideals $P_i$'s, and so $A^*$ injects into the direct product of the $(A/P_i)^*$'s, which are of finite type according to the particular case. Next assume $s>1$ and consider the natural map $\phi : A \to A/N^{s-1}$. By the induction hypothesis $\operatorname{Im}\phi$ is finitely generated, and Samuel shows that $\ker\phi = 1+N^{s-1}$ and that the latter group is finitely generated.
Finally, your original question, with the additional assumption that $A$ is reduced, has an affirmative answer, see P. Samuel, "A propos du théorème des unités", Bull. Sc. Math., 90 (1966), 89-96).
• When $A$ is finite as $\Bbb Z$ module, then nilradical will also be finitely generated as abelian group so it works. I don't notice this problem in general case until just now — $\Bbb Z[x,y]/(y^2)$ is a counterexample. – sawdada Sep 27 '17 at 8:07
• Once the reduced case is true, by covering with finitely many affine open subsets we can see that it's also true for any reduced scheme finite type over $\text{Spec} \Bbb Z$. By the way, I wonder whether we have similiar results for Picard group. – sawdada Sep 27 '17 at 8:09
• Hi, I can't seem to find a copy of P. Samuel's "A propos du theoreme des unites". Would it be possible for you to email me a copy? (Specifically I'm looking for a reference for the statement that the group of units of a commutative reduced finitely generated $\mathbb{Z}$-algebra is itself finitely generated) I assume by "finitely generated" you mean finitely generated as an algebra, (ie, finite type), not finitely generated as a $\mathbb{Z}$-module (ie finite) ? – oxeimon Feb 6 at 22:23